Separable Differential Equations

Section 2.4 Separable Differential Equations

A differential equation is an equation for an unknown function that involves the derivative of the unknown function. Differential equations play a central role in modelling a huge number of different phenomena. Here is a table giving a bunch of named differential equations and what they are used for. It is far from complete.

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Newton’s Law of Motion	describes motion of particles
Maxwell’s equations	describes electromagnetic radiation
Navier-Stokes equations	describes fluid motion
Heat equation	describes heat flow
Wave equation	describes wave motion
Schrödinger equation	describes atoms, molecules and crystals
Stress-strain equations	describes elastic materials
Black-Scholes models	used for pricing financial options
Predator-prey equations	describes ecosystem populations
Einstein’s equations	connects gravity and geometry
Ludwig-Jones-Holling’s equation	models spruce budworm/Balsam fir ecosystem
Zeeman’s model	models heart beats and nerve impulses
Sherman-Rinzel-Keizer model	for electrical activity in Pancreatic $\beta$-cells
Hodgkin-Huxley equations	models nerve action potentials

We are just going to scratch the surface of the study of differential equations. Most universities offer half a dozen different undergraduate courses on various aspects of differential equations. We will just look at one special, but important, type of equation.

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Subsection 2.4.1 Separate and integrate

Definition 2.4.1.

A separable differential equation is an equation for a function $y(x)$ of the form

\begin{equation*} \diff{y}{x}(x) = f(x)\ g\big(y(x)\big) \end{equation*}

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We’ll start by developing a recipe for solving separable differential equations. Then we’ll look at many examples. Usually one suppresses the argument of $y(x)$ and writes the equation

Look at the right hand side of the equation. The $x$-dependence is separated from the $y$-dependence. That’s the reason for the name “separable”.

\begin{equation*} \diff{y}{x} = f(x)\ g(y) \end{equation*}

and solves such an equation by cross multiplying/dividing to get all of the $y$’s, including the $\dee{y}$ on one side of the equation and all of the $x$’s, including the $\dee{x}\text{,}$ on the other side of the equation.

\begin{equation*} \frac{\dee{y}}{g(y)}=f(x)\,\dee{x} \end{equation*}

(We are of course assuming that $g(y)$ is nonzero.) Then you integrate both sides

\begin{gather} \int\frac{\dee{y}}{g(y)}=\int f(x)\,\dee{x}\tag{✶} \end{gather}

This looks illegal, and indeed is illegal — $\diff{y}{x}$ is not a fraction. But we’ll now see that the answer is still correct. This procedure is simply a mnemonic device to help you remember that answer (✶).

Our goal is to find all functions $y(x)$ that obey $\diff{y}{x}(x) = f(x)\ g\big(y(x)\big)\text{.}$

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Assuming that $g$ is nonzero,

\begin{align*} y'(x) = f(x)\ g(y(x)) \amp \iff \frac{y'(x)}{g(y(x))}=f(x)\iff \int\frac{y'(x)}{g(y(x))}\,\dee{x}= \int f(x)\,\dee{x}\\ \amp \iff \int\frac{\dee{y}}{g(y)}\bigg|_{y=y(x)}= \int f(x)\,\dee{x}\\ \amp\hskip0.5in \text{with the substitution }y=y(x), \dee y = y'(x)\,\dee{x} \end{align*}

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That’s our answer (✶) again.

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Let $G(y)$ be an antiderivative of $\frac{1}{g(y)}$ (i.e. $G'(y)=\frac{1}{g(y)}$) and $F(x)$ be an antiderivative of $f(x)$ (i.e. $F'(x)=f(x)$). If we reinstate the argument of $y\text{,}$ (✶) is

\begin{equation} G\big(y(x)\big) = F(x) + C\tag{2.4.1} \end{equation}

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Observe that the solution equation (2.4.1) contains an arbitrary constant, $C\text{.}$ The value of this arbitrary constant can not be determined by the differential equation. You need additional data to determine it. Often this data consists of the value of the unknown function for one value of $x\text{.}$ That is, often the problem you have to solve is of the form

\begin{equation*} \diff{y}{x}(x) = f(x)\ g\big(y(x)\big)\qquad y(x_0)=y_0 \end{equation*}

where $f(x)$ and $g(y)$ are given functions and $x_0$ and $y_0$ are given numbers. This type of problem is called an “initial value problem”. It is solved by first using the method above to find the general solution to the differential equation, including the arbitrary constant $C\text{,}$ and then using the “initial condition” $y(x_0)=y_0$ to determine the value of $C\text{.}$ We’ll see examples of this shortly.

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Example 2.4.2. A separable warm-up.

The differential equation

\begin{equation*} \diff{y}{x} = xe^{-y} \end{equation*}

is separable, and we now find all of its solutions by using our mnemonic device. We start by cross-multiplying so as to move all $y$’s to the left hand side and all $x$’s to the right hand side.

\begin{equation*} e^y\,\dee{y} = x\,\dee{x} \end{equation*}

Then we integrate both sides

\begin{equation*} \int e^y\dee{y} = \int x\dee{x} \iff e^y = \frac{x^2}{2}+C \end{equation*}

The $C$ on the right hand side contains both the arbitrary constant for the indefinite integral $\int e^y\dee{y}$ and the arbitrary constant for the indefinite integral $\int x\dee{x}\text{.}$ Finally, we solve for $y\text{,}$ which is really a function of $x\text{.}$

\begin{equation*} y(x) = \log\Big(\frac{x^2}{2}+C\Big) \end{equation*}

Recall that we are using $\log$ to refer to the natural (base $e$) logarithm.

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Note that $C$ is an arbitrary constant. It can take any value. It cannot be determined by the differential equation itself. In applications $C$ is usually determined by a requirement that $y$ take some prescribed value (determined by the application) when $x$ is some prescribed value. For example, suppose that we wish to find a function $y(x)$ that obeys both

\begin{equation*} \diff{y}{x} = xe^{-y}\qquad\text{and}\qquad y(0)=1 \end{equation*}

We know that, to have $\diff{y}{x} = xe^{-y}$ satisfied, we must have $y(x) = \log\big(\frac{x^2}{2}+C\big)\text{,}$ for some constant $C\text{.}$ To also have $y(0)=1\text{,}$ we must have

\begin{gather*} 1=y(0)=\log\Big(\frac{x^2}{2}+C\Big)\bigg|_{x=0}=\log C \iff \log C =1 \iff C=e \end{gather*}

So our final solution is $y(x) = \log\big(\frac{x^2}{2}+e\big)\text{.}$

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Example 2.4.3. A little more warm-up.

Let $a$ and $b$ be any two constants. We’ll now solve the family of differential equations

\begin{equation*} \diff{y}{x} =a(y-b) \end{equation*}

using our mnemonic device.

\begin{align*} \frac{\dee{y}}{y-b} =a\,\dee{x} &\implies \int \frac{\dee{y}}{y-b}= \int a\,\dee{x}\\ & \implies \log|y-b|= ax+c \implies |y-b|= e^{ax+c} =e^c e^{ax}\\ &\implies y-b = C e^{ax} \end{align*}

where $C$ is either $+e^c$ or $-e^c\text{.}$ Note that as $c$ runs over all real numbers, $+e^c$ runs over all strictly positive real numbers and $-e^c$ runs over all strictly negative real numbers. So, so far, $C$ can be any real number except $0\text{.}$ But we were a bit sloppy here. We implicitly assumed that $y-b$ was nonzero, so that we could divide it across. None-the-less, the constant function $y=b\text{,}$ which corresponds to $C=0\text{,}$ is a perfectly good solution — when $y$ is the constant function $y=b\text{,}$ both $\diff{y}{x}$ and $a(y-b)$ are zero. So the general solution to $\diff{y}{x} =a(y-b)$ is $y(x)=C e^{ax}+b\text{,}$ where the constant $C$ can be any real number. Note that when $y(x)=C e^{ax}+b$ we have $y(0)=C+b\text{.}$ So $C=y(0)-b$ and the general solution is

\begin{equation*} y(x) = \{y(0)-b\}\,e^{ax} + b \end{equation*}

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This is worth stating as a theorem.

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Theorem 2.4.4.

Let $a$ and $b$ be constants. The differentiable function $y(x$) obeys the differential equation

\begin{equation*} \diff{y}{x} =a(y-b) \end{equation*}

if and only if

\begin{equation*} y(x) = \{y(0)-b\}\,e^{ax} + b \end{equation*}

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Example 2.4.5. Solve $\diff{y}{x}=y^2$.

Solve $\diff{y}{x}=y^2$

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Solution: When $y\ne 0\text{,}$

\begin{equation*} \diff{y}{x}=y^2 \implies \frac{\dee{y}}{y^2}=\dee{x} \implies \frac{y^{-1}}{-1}=x+C \implies y=-\frac{1}{x+C} \end{equation*}

When $y=0\text{,}$ this computation breaks down because $\frac{\dee{y}}{y^2}$ contains a division by 0. We can check if the function $y(x)=0$ satisfies the differential equation by just subbing it in:

\begin{equation*} y(x)=0\implies y'(x)=0,\ y(x)^2=0\implies y'(x)=y(x)^2 \end{equation*}

So $y(x)=0$ is a solution and the full solution is

\begin{equation*} y(x)=0 \text{ or } y(x)=-\frac{1}{x+C}, \text{ for any constant $C$} \end{equation*}

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Example 2.4.6. A falling raindrop.

When a raindrop falls it increases in size so that its mass $m(t)\text{,}$ is a function of time $t\text{.}$ The rate of growth of mass, i.e. $\frac{dm}{dt}\text{,}$ is $km(t)$ for some positive constant $k\text{.}$ According to Newton’s law of motion, $\diff{}{t} (mv)=gm\text{,}$ where $v$ is the velocity of the raindrop (with $v$ being positive for downward motion) and $g$ is the acceleration due to gravity. Find the terminal velocity, $\lim\limits_{t\rightarrow\infty}v(t)\text{,}$ of a raindrop.

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Solution: In this problem we have two unknown functions, $m(t)$ and $v(t)\text{,}$ and two differential equations, $\frac{dm}{dt}=km$ and $\diff{}{t} (mv)=gm\text{.}$ The first differential equation, $\frac{dm}{dt}=km\text{,}$ involves only $m(t)\text{,}$ not $v(t)\text{,}$ so we use it to determine $m(t)\text{.}$ By Theorem 2.4.4, with $b=0\text{,}$ $a=k\text{,}$ $y$ replaced by $m$ and $x$ replaced by $t\text{,}$

\begin{equation*} \diff{m}{t}=km \implies m(t) = m(0) e^{kt} \end{equation*}

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Now that we know $m(t)$ (except for the value of the constant $m(0)$), we can substitute it into the second differential equation, which we can then use to determine the remaining unknown function $v(t)\text{.}$ Observe that the second equation, $\diff{}{t} (mv)=gm(t)=gm(0)e^{kt}$ tells that the derivative of the function $y(t)=m(t)v(t)$ is $gm(0)e^{kt}\text{.}$ So $y(t)$ is just an antiderivative of $gm(0)e^{kt}\text{.}$

\begin{equation*} \diff{y}{t}=gm(t)=gm(0)e^{kt} \qquad \implies y(t)=\int gm(0)e^{kt}\,\dee{t} = gm(0)\frac{e^{kt}}{k}+C \end{equation*}

Now that we know $y(t)=m(t)v(t)=m(0)e^{kt}v(t)\text{,}$ we can get $v(t)$ just by dividing out the $m(0)e^{kt}\text{.}$

\begin{align*} y(t) =gm(0)\frac{e^{kt}}{k}+C & \implies m(0)e^{kt}v(t)=gm(0)\frac{e^{kt}}{k}+C\\ & \implies v(t)=\frac{g}{k}+\frac{C}{m(0)e^{kt}} \end{align*}

Our solution, $v(t)\text{,}$ contains two arbitrary constants, namely $C$ and $m(0)\text{.}$ They will be determined by, for example, the mass and velocity at time $t=0\text{.}$ But since we are only interested in the terminal velocity $\lim\limits_{t\rightarrow\infty}v(t)\text{,}$ we don’t need to know $C$ and $m(0)\text{.}$ Since $k \gt 0\text{,}$ $\lim\limits_{t\rightarrow\infty}\frac{C}{e^{kt}}=0$ and the terminal velocity $\lim\limits_{t\rightarrow\infty}v(t)=\frac{g}{k}\text{.}$

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Example 2.4.7. Intravenous glucose.

A glucose solution is administered intravenously into the bloodstream at a constant rate $r\text{.}$ As the glucose is added, it is converted into other substances at a rate that is proportional to the concentration at that time. The concentration, $C(t)\text{,}$ of the glucose in the bloodstream at time $t$ obeys the differential equation

\begin{equation*} \diff{C}{t}=r-kC \end{equation*}

where $k$ is a positive constant of proportionality.

Express $C(t)$ in terms of $k$ and $C(0)\text{.}$

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Find $\lim\limits_{t\rightarrow\infty} C(t)\text{.}$

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Solution: (a) Since $r-kC=-k\big(C-\frac{r}{k}\big)$ the given equation is

\begin{equation*} \diff{C}{t}=-k\big(C-\frac{r}{k}\big) \end{equation*}

which is of the form solved in Theorem 2.4.4 with $a=-k$ and $b=\frac{r}{k}\text{.}$ So the solution is

\begin{equation*} C(t)=\frac{r}{k}+\Big(C(0)-\frac{r}{k}\Big)e^{-kt} \end{equation*}

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For any $k \gt 0\text{,}$ $\lim\limits_{t\rightarrow\infty} e^{-kt}=0\text{.}$ Consequently, for any $C(0)$ and any $k \gt 0\text{,}$ $\lim\limits_{t\rightarrow\infty} C(t)=\frac{r}{k}, \text{.}$ We could have predicted this limit without solving for $C(t)\text{.}$ If we assume that $C(t)$ approaches some equilibrium value $C_e$ as $t$ approaches infinity, then taking the limits of both sides of $\frac{dC}{dt}=r-kC$ as $t\rightarrow\infty$ gives

\begin{equation*} 0=r-kC_e\implies C_e=\frac{r}{k} \end{equation*}

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Subsection 2.4.2 Optional — Carbon Dating

Scientists can determine the age of objects containing organic material by a method called carbon dating or radiocarbon dating

Willard Libby, of Chicago University was awarded the Nobel Prize in Chemistry in 1960, for developing radiocarbon dating.

. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, ${}^{14}C\text{,}$ with a half-life of about 5730 years. Vegetation absorbs carbon dioxide from the atmosphere through photosynthesis and animals acquire ${}^{14}C$ by eating plants. When a plant or animal dies, it stops replacing its carbon and the amount of ${}^{14}C$ begins to decrease through radioactive decay. Therefore the level of radioactivity also decreases. More precisely, let $Q(t)$ denote the amount of ${}^{14}C$ in the plant or animal $t$ years after it dies. The number of radioactive decays per unit time, at time $t\text{,}$ is proportional to the amount of ${}^{14}C$ present at time t, which is $Q(t)\text{.}$ Thus

Equation 2.4.8. Radioactive decay.

\begin{equation*} \diff{Q}{t}(t)=-k Q(t) \end{equation*}

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Here $k$ is a constant of proportionality that is determined by the half-life. We shall explain what half-life is, and also determine the value of $k\text{,}$ in Example 2.4.9, below.

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Before we do so, let’s think about the sign in equation (2.4.8).

Recall that $Q(t)$ denotes a quantity, namely the amount of ${}^{14}C$ present at time $t\text{.}$ There cannot be a negative amount of ${}^{14}C\text{.}$ Nor can this quantity be zero. (We would not use carbon dating when there is no ${}^{14}C$ present.) Consequently, $Q(t) \gt 0\text{.}$

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As the time $t$ increases, $Q(t)$ decreases, because ${}^{14}C$ is being continuously converted into ${}^{14}N$ by radioactive decay
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The precise transition is ${}^{14}C\rightarrow {}^{14}N+ e^- + \bar{\nu}_e$ where $e^-$ is an electron and $\bar{\nu}_e $ is an electron neutrino.
. Thus $\diff{Q}{t}(t) \lt 0\text{.}$

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The signs $Q(t) \gt 0$ and $\diff{Q}{t}(t) \lt 0$ are consistent with (2.4.8) provided the constant of proportionality $k \gt 0\text{.}$

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In (2.4.8), we chose to call the constant of proportionality “$-k$”. We did so in order to make $k \gt 0\text{.}$ We could just as well have chosen to call the constant of proportionality “$K$”. That is, we could have replaced equation (2.4.8) by $\diff{Q}{t}(t)=K Q(t)\text{.}$ The constant of proportionality $K$ would have to be negative, (and $K$ and $k$ would be related by $K=-k$).

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Example 2.4.9. Half life and the constant $k$.

In this example, we determine the value of the constant of proportionality $k$ in (2.4.8) that corresponds to the half-life of ${}^{14}C\text{,}$ which is 5730 years.

Imagine that some plant or animal contains a quantity $Q_0$ of ${}^{14}C$ at its time of death. Let’s choose the zero point of time $t=0$ to be the instant that the plant or animal died.

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Denote by $Q(t)$ the amount of ${}^{14}C$ in the plant or animal $t$ years after it died. Then $Q(t)$ must obey both (2.4.8) and $Q(0)=Q_0\text{.}$

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Theorem 2.4.4, with $b=0$ and $a=-k\text{,}$ then tells us that $Q(t) = Q_0 e^{-kt}$ for all $t\ge 0\text{.}$

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By definition, the half-life of ${}^{14}C$ is the length of time that it takes for half of the ${}^{14}C$ to decay. That is, the half-life $t_{1/2}$ is determined by
\begin{align*} Q(t_{1/2})=\half Q(0)&=\half Q_0 &\text{but we know that $Q(t)=Q_0e^{-kt}$}\\ Q_0 e^{-kt_{1/2}}&=\half Q_0&\text{now cancel $Q_0$}\\ e^{-kt_{1/2}}&=\half \end{align*}
Taking the logarithm of both sides gives
\begin{gather*} -k t_{1/2}=\log \frac{1}{2} = -\log 2 \implies k=\frac{\log 2}{t_{1/2}} \end{gather*}
Recall that, in this text, we use $\log x$ to indicate the natural logarithm. That is,
\begin{equation*} \log x = \log_e x=\log x \end{equation*}

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We are told that, for ${}^{14}C\text{,}$ the half-life $t_{1/2}=5730\text{,}$ so

\begin{align*} k=\frac{\log 2}{5730} &= 0.000121 &\text{to 6 decimal places} \end{align*}

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From the work in the above example we have accumulated enough new facts to make a corollary to Theorem 2.4.4.

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Corollary 2.4.10.

The function $Q(t)$ satisfies the equation

\begin{gather*} \diff{Q}{t} = -k Q(t) \end{gather*}

if and only if

\begin{gather*} Q(t) = Q(0)\, e^{-kt} \end{gather*}

The half-life is defined to be the time $t_{1/2}$ which obeys

\begin{gather*} Q\big(t_{1/2}\big) = \frac{1}{2}\,Q(0) \end{gather*}

The half-life is related to the constant $k$ by

\begin{gather*} t_{1/2} =\frac{\log 2}{k} \end{gather*}

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Now here is a typical problem that is solved using Corollary 2.4.10.

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Example 2.4.11. The age of a piece of parchment.

A particular piece of parchment contains about $64\%$ as much ${}^{14}C$ as plants do today. Estimate the age of the parchment.

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Solution: Let $Q(t)$ denote the amount of ${}^{14}C$ in the parchment $t$ years after it was first created. By equation (2.4.8) and Example 2.4.9

\begin{gather*} \diff{Q}{t}(t)=-k Q(t)\qquad\text{with } k=\frac{\log 2}{5730} = 0.000121 \end{gather*}

By Corollary 2.4.10

\begin{gather*} Q(t) = Q(0)\, e^{-kt} \end{gather*}

The time at which $Q(t)$ reaches $0.64\,Q(0)$ is determined by

\begin{align*} Q(t)&=0.64\, Q(0) &\text{but $Q(t) = Q(0)\, e^{-kt}$}\\ Q(0)\,e^{-kt}&=0.64\, Q(0) &\text{cancel $Q(0)$}\\ e^{-kt}&=0.64&\text{take logarithms}\\ -kt&=\log 0.64\\ t&=\frac{\log 0.64}{-k} =\frac{\log 0.64}{-0.000121} = 3700 \qquad&\text{to 2 significant digits} \end{align*}

That is, the parchment

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The British Museum has an Egyptian mathematical text from the seventeenth century B.C.

is about 37 centuries old.

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We have stated that the half-life of ${}^{14}C$ is 5730 years. How can this be determined? We can explain this using the following example.

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Example 2.4.12. Half life of implausium.

A scientist in a B-grade science fiction film is studying a sample of the rare and fictitious element, implausium. With great effort he has produced a sample of pure implausium. The next day — 17 hours later — he comes back to his lab and discovers that his sample is now only $37\%$ pure. What is the half-life of the element?

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Solution: We can again set up our problem using Corollary 2.4.10. Let $Q(t)$ denote the quantity of implausium at time $t\text{,}$ measured in hours. Then we know

\begin{align*} Q(t)&= Q(0) \cdot e^{-kt} \end{align*}

We also know that

\begin{align*} Q(17) &= 0.37 Q(0). \end{align*}

That enables us to determine $k$ via

\begin{align*} Q(17) = 0.37 Q(0) &= Q(0) e^{-17k} & \text{ divide both sides by $Q(0)$}\\ 0.37 &= e^{-17k}\\ \end{align*}

and so

\begin{align*} k &= -\frac{\log 0.37}{17} = 0.05849 \end{align*}

We can then convert this to the half life using Corollary 2.4.10:

\begin{align*} t_{1/2} &= \frac{\log 2}{k} \approx 11.85 \text{ hours} \end{align*}

While this example is entirely fictitious, one really can use this approach to measure the half-life of materials.

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Subsection 2.4.3 Optional — Newton’s Law of Cooling

Newton’s law of cooling says:

The rate of change of temperature of an object is proportional to the difference in temperature between the object and its surroundings. The temperature of the surroundings is sometimes called the ambient temperature.

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If we denote by $T(t)$ the temperature of the object at time $t$ and by $A$ the temperature of its surroundings, Newton’s law of cooling says that there is some constant of proportionality, $K\text{,}$ such that

Equation 2.4.13. Newton’s law of cooling.

\begin{equation*} \diff{T}{t}(t) = K\big[T(t)-A\big] \end{equation*}

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This mathematical model of temperature change works well when studying a small object in a large, fixed temperature, environment. For example, a hot cup of coffee in a large room

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It does not work so well when the object is of a similar size to its surroundings since the temperature of the surroundings will rise as the object cools. It also fails when there are phase transitions involved — for example, an ice-cube melting in a warm room does not obey Newton’s law of cooling.

. Let’s start by thinking a little about the sign of the constant of proportionality. At any time $t\text{,}$ there are three possibilities.

If $T(t) \gt A\text{,}$ that is, if the body is warmer than its surroundings, we would expect heat to flow from the body into its surroundings and so we would expect the body to cool off so that $\diff{T}{t}(t) \lt 0\text{.}$ For this expectation to be consistent with (2.4.13), we need $K \lt 0\text{.}$

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If $T(t) \lt A\text{,}$ that is the body is cooler than its surroundings, we would expect heat to flow from the surroundings into the body and so we would expect the body to warm up so that $\diff{T}{t}(t) \gt 0\text{.}$ For this expectation to be consistent with (2.4.13), we again need $K \lt 0\text{.}$

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Finally if $T(t)=A\text{,}$ that is the body and its environment have the same temperature, we would not expect any heat to flow between the two and so we would expect that $\diff{T}{t}(t)=0\text{.}$ This does not impose any condition on $K\text{.}$

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In conclusion, we would expect $K \lt 0\text{.}$ Of course, we could have chosen to call the constant of proportionality $-k\text{,}$ rather than $K\text{.}$ Then the differential equation would be $\diff{T}{t} = -k\big(T-A\big)$ and we would expect $k \gt 0\text{.}$

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Example 2.4.14. Warming iced tea.

The temperature of a glass of iced tea is initially $5^\circ\text{.}$ After 5 minutes, the tea has heated to $10^\circ$ in a room where the air temperature is $30^\circ\text{.}$

Determine the temperature as a function of time.

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What is the temperature after 10 minutes?

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Determine when the tea will reach a temperature of $20^\circ\text{.}$

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Solution: (a)

Denote by $T(t)$ the temperature of the tea $t$ minutes after it was removed from the fridge, and let $A=30$ be the ambient temperature.

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By Newton’s law of cooling,
\begin{equation*} \diff{T}{t}=K(T-A) = K(T-30) \end{equation*}
for some, as yet unknown, constant of proportionality $K\text{.}$

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By Theorem 2.4.4 with $a=K$ and $b=30\text{,}$
\begin{equation*} T(t) = [T(0)-30]\,e^{Kt} + 30 =30-25 e^{Kt} \end{equation*}
since the initial temperature $T(0)=5\text{.}$

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This solution is not complete because it still contains an unknown constant, namely $K\text{.}$ We have not yet used the given data that $T(5)=10\text{.}$ We can use it to determine $K\text{.}$ At $t=5\text{,}$
\begin{align*} T(5)=30-25 e^{5K}=10 &\implies e^{5K}=\frac{20}{25} \implies 5K=\log\frac{20}{25}\\ &\implies K=\frac{1}{5}\log\frac{4}{5}=-0.044629 \end{align*}
to six decimal places.

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(b) To find the temperature at 10 minutes we can just use the solution we have determined above.

\begin{align*} T(10)&=30-25 e^{10K}\\ &=30-25 e^{10\times\frac{1}{5}\log\frac{4}{5}}\\ &=30-25 e^{2\log\frac{4}{5}} = 30-25 e^{\log\frac{16}{25}}\\ &=30-16=\text{$14^\circ$} \end{align*}

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\begin{align*} 30-25 e^{Kt}=20 &\implies e^{Kt}=\frac{10}{25} \implies Kt=\log\frac{10}{25}\\ &\implies t=\frac{1}{K}\log\frac{2}{5}=\hbox{20.5 min} \end{align*}

to one decimal place.

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Example 2.4.15. Temperature back in time.

A dead body is discovered at 3:45pm in a room where the temperature is 20$^\circ$C. At that time the temperature of the body is 27$^\circ$C. Two hours later, at 5:45pm, the temperature of the body is 25.3$^\circ$C. What was the time of death? Note that the normal (adult human) body temperature is $37^\circ$C.

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Solution: We will assume that the body’s temperature obeys Newton’s law of cooling.

Denote by $T(t)$ the temperature of the body at time $t\text{,}$ with $t=0$ corresponding to 3:45pm. We wish to find the time of death — call it $t_d\text{.}$

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There is a lot of data in the statement of the problem. We are told
1. the ambient temperature: $A=20$
  
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2. the temperature of the body when discovered: $T(0)=27$
  
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3. the temperature of the body 2 hours later: $T(2)=25.3$
  
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4. assuming the person was a healthy adult right up until he died, the temperature at the time of death: $T(t_d)=37\text{.}$
  
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Theorem 2.4.4 with $a=K$ and $b=A=20$
\begin{equation*} T(t) = [T(0)-A]\,e^{Kt} + A =20+7 e^{Kt} \end{equation*}
Two unknowns remain, $K$ and $t_d\text{.}$

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We can find the first, $K\text{,}$ by using the condition (3), which says $T(2)=25.3\text{.}$
\begin{align*} 25.3=T(2) = 20+7 e^{2K} &\implies 7 e^{2K}=5.3 \implies 2K = \log\big(\tfrac{5.3}{7}\big)\\ &\implies K = \tfrac{1}{2} \log\big(\tfrac{5.3}{7}\big) = -0.139 \end{align*}

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Finally, $t_d$ is determined by the condition (4).
\begin{align*} 37 = T(t_d) = 20+7 e^{-0.139 t_d} &\implies e^{-0.139 t_d} = \tfrac{17}{7}\\ \amp\implies -0.139 t_d =\log\big(\tfrac{17}{7}\big)\\ &\implies t_d = -\tfrac{1}{0.139}\log\big(\tfrac{17}{7}\big) = - 6.38 \end{align*}
to two decimal places. Now $6.38$ hours is $6$ hours and $0.38\times 60 = 23$ minutes. So the time of death was $6$ hours and $23$ minutes before 3:45pm, which is 9:22am.

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A slightly tricky example — we need to determine the ambient temperature from three measurements at different times.

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Example 2.4.16. Finding the ambient temperature.

A glass of room-temperature water is carried out onto a balcony from an apartment where the temperature is $22^\circ$C. After one minute the water has temperature $26^\circ$C and after two minutes it has temperature $28^\circ$C. What is the outdoor temperature?

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Solution: We will assume that the temperature of the water obeys Newton’s law of cooling.

Let $A$ be the outdoor temperature and $T(t)$ be the temperature of the water $t$ minutes after it is taken outside.

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By Newton’s law of cooling,
\begin{gather*} T(t)=A+\big(T(0)-A\big)e^{Kt} \end{gather*}
Theorem 2.4.4 with $a=K$ and $b=A\text{.}$ Notice there are 3 unknowns here — $A\text{,}$ $T(0)$ and $K$ — so we need three pieces of information to find them all.

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We are told $T(0)=22\text{,}$ so
\begin{align*} T(t) &=A+\big(22-A\big)e^{Kt}. \end{align*}

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We are also told $T(1)=26\text{,}$ which gives
\begin{align*} 26 &=A+\big(22-A\big)e^{K} & \text{rearrange things}\\ e^K&=\frac{26-A}{22-A} \end{align*}

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Finally, $T(2)=28\text{,}$ so
\begin{align*} 28&=A+\big(22-A\big)e^{2K} & \text{rearrange}\\ e^{2K} &= \frac{28-A}{22-A} & \text{but $e^K=\frac{26-A}{22-A}$, so}\\ \left(\frac{26-A}{22-A}\right)^2 &=\frac{28-A}{22-A} & \text{multiply through by $(22-A)^2$}\\ (26-A)^2 &= (28-A)(22-A) \end{align*}
We can expand out both sides and collect up terms to get
\begin{align*} \underbrace{26^2}_{=676}-52A+A^2 &= \underbrace{28\times22}_{=616}-50A+A^2\\ 60 &= 2A\\ 30 &= A \end{align*}
So the temperature outside is $30^\circ\text{.}$

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Subsection 2.4.4 Optional — Population Growth

Suppose that we wish to predict the size $P(t)$ of a population as a function of the time $t\text{.}$ In the most naive model of population growth, each couple produces $\beta$ offspring (for some constant $\beta$) and then dies. Thus over the course of one generation $\beta\tfrac{P(t)}{2}$ children are produced and $P(t)$ parents die so that the size of the population grows from $P(t)$ to

\begin{gather*} P(t+t_g)= \underbrace{P(t) +\beta\frac{P(t)}{2}}_{\text{parents+offspring}} -\underbrace{P(t)}_{\text{parents die}} =\frac{\beta}{ 2 } P(t) \end{gather*}

where $t_g$ denotes the lifespan of one generation. The rate of change of the size of the population per unit time is

\begin{gather*} \frac{P(t+t_g)-P(t)}{t_g} =\frac{1}{t_g}\Big[\frac{\beta}{2}P(t) -P(t)\Big] = b P(t) \end{gather*}

where $b=\tfrac{\beta-2}{2t_g}$ is the net birthrate per member of the population per unit time. If we approximate

\begin{gather*} \frac{P(t+t_g)-P(t)}{t_g}\approx\diff{P}{t}(t) \end{gather*}

we get the differential equation

Equation 2.4.17. Population growth.

\begin{equation*} \diff{P}{t} = bP(t) \end{equation*}

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By Corollary 2.4.10, with $-k$ replaced by $b\text{,}$

Equation 2.4.18. Malthusian growth model.

\begin{equation*} P(t) = P(0)\cdot e^{bt} \end{equation*}

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This is called the Malthusian

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This is named after Rev. Thomas Robert Malthus. He described this model in a 1798 paper called “An essay on the principle of population”.

growth model. It is, of course, very simplistic. One of its main characteristics is that, since $P(t+T) = P(0)\cdot e^{b(t+T)} = P(t)\cdot e^{bT}\text{,}$ every time you add $T$ to the time, the population size is multiplied by $e^{bT}\text{.}$ In particular, the population size doubles every $\frac{\log 2}{b}$ units of time. The Malthusian growth model can be a reasonably good model only when the population size is very small compared to its environment

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That is, the population has plenty of food and space to grow.

. A more sophisticated model of population growth, that takes into account the “carrying capacity of the environment” is considered below.

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Example 2.4.19. A rough estimate of the earth’s population.

In 1927 the population of the world was about 2 billion. In 1974 it was about 4 billion. Estimate when it reached 6 billion. What will the population of the world be in 2100, assuming the Malthusian growth model?

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Solution: We follow our usual pattern for dealing with such problems.

Let $P(t)$ be the world’s population, in billions, $t$ years after 1927. Note that 1974 corresponds to $t=1974-1927 = 47\text{.}$

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We are assuming that $P(t)$ obeys equation (2.4.17). So, by (2.4.18)
\begin{gather*} P(t)=P(0)\cdot e^{bt} \end{gather*}
Notice that there are 2 unknowns here — $b$ and $P(0)$ — so we need two pieces of information to find them.

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We are told $P(0)=2\text{,}$ so
\begin{gather*} P(t)=2\cdot e^{bt} \end{gather*}

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We are also told $P(47)=4\text{,}$ which gives
\begin{align*} 4 &=2\cdot e^{47b} & \text{clean up}\\ e^{47b}&=2 & \text{take the log and clean up}\\ b&=\frac{\log 2}{47} = 0.0147 & \text{to 3 decimal places} \end{align*}

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We now know $P(t)$ completely, so we can easily determine the predicted population
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The 2015 Revision of World Population, a publication of the United Nations, predicts that the world’s population in 2100 will be about 11 billion. But “about” covers a pretty large range. They give an $80\%$ confidence interval running from 10 billion to 12.5 billion. Our prediction of a population of 25.4 billion at time $t=173$ looks pretty poor.
in 2100, i.e. at $t=2100-1927 = 173\text{.}$
\begin{gather*} P(173) = 2 e^{173 b} = 2 e^{173\times 0.0147} = 25.4\text{ billion} \end{gather*}

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Finally, our crude model predicts that the population is 6 billion at the time $t$ that obeys
\begin{align*} P(t) &= 2 e^{b t} = 6 & \text{clean up}\\ e^{b t}&=3 & \text{take the log and clean up}\\ t&=\frac{\log 3}{b} = 47\frac{\log 3}{\log 2} = 74.5 \end{align*}
which corresponds
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The world population really reached 6 billion in about 1999. So our prediction that the population would reach 6 billion at about time $t=74.5$ is not so bad.
to the middle of 2001.

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Logistic growth adds one more wrinkle to the simple population model. It assumes that the population only has access to limited resources. As the size of the population grows the amount of food available to each member decreases. This in turn causes the net birth rate $b$ to decrease. In the logistic growth model $b=b_0\left(1-\tfrac{P}{K}\right)\text{,}$ where $K$ is called the carrying capacity of the environment, so that

\begin{gather*} P'(t) =b_0\left(1-\frac{P(t)}{K}\right)P(t) \end{gather*}

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This is a separable differential equation and we can solve it explicitly. We shall do so shortly. See Example 2.4.20, below. But, before doing that, we’ll see what we can learn about the behaviour of solutions to differential equations like this without finding formulae for the solutions. It turns out that we can learn a lot just by watching the sign of $P'(t)\text{.}$ For concreteness, we’ll look at solutions of the differential equation

\begin{gather*} \diff{P}{t}(t)=\big(\,6000-3P(t)\,\big)\,P(t) \end{gather*}

We’ll sketch the graphs of four functions $P(t)$ that obey this equation.

For the first function, $P(0)=0\text{.}$

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For the second function, $P(0)=1000\text{.}$

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For the third function, $P(0)=2000\text{.}$

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For the fourth function, $P(0)=3000\text{.}$

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The sketches will be based on the observation that $(6000-3P)\,P=3(2000-P)\,P$

is zero for $P=0,\ 2000\text{,}$

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is strictly positive for $0 \lt P \lt 2000$ and

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is strictly negative for $P \gt 2000\text{.}$

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Consequently

\begin{align*} \diff{P}{t}(t)\ \begin{cases} =0 & \text{if }P(t)=0\\ \gt 0 & \text{if }0 \lt P(t) \lt 2000\\ =0 & \text{if }P(t)=2000\\ \lt 0 & \text{if }P(t) \gt 2000 \end{cases} \end{align*}

Thus if $P(t)$ is some function that obeys $\diff{P}{t}(t)=\big(6000-3P(t)\big)P(t)\text{,}$ then as the graph of $P(t)$ passes through the point $\big(t,P(t)\big)$

\begin{align*} \text{the graph has } \begin{cases} \text{slope zero,}& \text{i.e. is horizontal, \ \ if }P(t)=0 \\ \text{positive slope,}& \text{i.e. is increasing, \ \ if } 0 \lt P(t) \lt 2000 \\ \text{slope zero,}& \text{i.e. is horizontal, \ \ if }P(t)=2000 \\ \text{negative slope,}& \text{i.e. is decreasing, \ \ if }0 \lt P(t) \lt 2000 \end{cases} \end{align*}

as illustrated in the figure

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As a result,

if $P(0)=0\text{,}$ the graph starts out horizontally. In other words, as $t$ starts to increase, $P(t)$ remains at zero, so the slope of the graph remains at zero. The population size remains zero for all time. As a check, observe that the function $P(t)=0$ obeys $\diff{P}{t}(t)=\big(6000-3P(t)\big)P(t)$ for all $t\text{.}$

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Similarly, if $P(0)=2000\text{,}$ the graph again starts out horizontally. So $P(t)$ remains at $2000$ and the slope remains at zero. The population size remains 2000 for all time. Again, the function $P(t)=2000$ obeys $\diff{P}{t}(t)=\big(6000-3P(t)\big)P(t)$ for all $t\text{.}$

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If $P(0)=1000\text{,}$ the graph starts out with positive slope. So $P(t)$ increases with $t\text{.}$ As $P(t)$ increases towards 2000, the slope $(6000-3P(t)\big)P(t)\text{,}$ while remaining positive, gets closer and closer to zero. As the graph approaches height 2000, it becomes more and more horizontal. The graph cannot actually cross from below 2000 to above 2000, because to do so it would have to have strictly positive slope for some value of $P$ above 2000, which is not allowed.

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If $P(0)=3000\text{,}$ the graph starts out with negative slope. So $P(t)$ decreases with $t\text{.}$ As $P(t)$ decreases towards 2000, the slope $(6000-3P(t)\big)P(t)\text{,}$ while remaining negative, gets closer and closer to zero. As the graph approaches height 2000, it becomes more and more horizontal. The graph cannot actually cross from above 2000 to below 2000, because to do so it would have to have negative slope for some value of $P$ below 2000, which is not allowed.

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These curves are sketched in the figure below. We conclude that for any initial population size $P(0)\text{,}$ except $P(0)=0\text{,}$ the population size approaches $2000$ as $t\rightarrow\infty\text{.}$

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Now we’ll do an example in which we explicitly solve the logistic growth equation.

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Example 2.4.20. Population predictions using logistic growth.

In 1986, the population of the world was 5 billion and was increasing at a rate of $2\%$ per year. Using the logistic growth model with an assumed maximum population of 100 billion, predict the population of the world in the years 2000, 2100 and 2500.

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Solution: Let $y(t)$ be the population of the world, in billions of people, at time $1986+t\text{.}$ The logistic growth model assumes

\begin{equation*} y'=ay(K-y) \end{equation*}

where $K$ is the carrying capacity and $a=\frac{b_0}{K}\text{.}$

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First we’ll determine the values of the constants $a$ and $K$ from the given data.

We know that, if at time zero the population is below $K\text{,}$ then as time increases the population increases, approaching the limit $K$ as $t$ tends to infinity. So in this problem $K$ is the maximum population. That is, $K=100\text{.}$

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We are also told that, at time zero, the percentage rate of change of population, $100\frac{y'}{y}\text{,}$ is 2, so that, at time zero, $\frac{y'}{y}=0.02\text{.}$ But, from the differential equation, $\frac{y'}{y}=a(K-y)\text{.}$ Hence at time zero, $0.02=a(100-5)\text{,}$ so that $a= \frac{2}{9500}\text{.}$

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We now know $a$ and $K$ and can solve the (separable) differential equation

\begin{align*} \frac{\dee{y}}{dt}=ay(K-y)&\implies \frac{\dee{y}}{y(K-y)}=a\,\dee{t}\\ \amp\implies \int\frac{1}{K}\Big[\frac{1}{y}-\frac{1}{y-K}\Big]\dee{y}=\int a\,\dee{t}\\ &\implies \frac{1}{K}[\log |y|-\log|y-K|]=at+C\\ &\implies \log\frac{|y|}{|y-K|}=aKt+CK\implies \Big|\frac{y}{y-K}\Big|=De^{aKt} \end{align*}

with $D=e^{CK}\text{.}$ We know that $y$ remains between $0$ and $K\text{,}$ so that $\Big|\frac{y}{y-K}\Big|=\frac{y}{K-y}$ and our solution obeys

\begin{equation*} \frac{y}{K-y}=De^{aKt} \end{equation*}

At this stage, we know the values of the constants $a$ and $K\text{,}$ but not the value of the constant $D\text{.}$ We are given that at $t=0\text{,}$ $y=5\text{.}$ Subbing in this, and the values of $K$ and $a\text{,}$

\begin{equation*} \frac{5}{100-5}=De^{0} \implies D=\frac{5}{95} \end{equation*}

So the solution obeys the algebraic equation

\begin{equation*} \frac{y}{100-y}=\frac{5}{95}e^{2t/95} \end{equation*}

which we can solve to get $y$ as a function of $t\text{.}$

\begin{align*} \amp y=(100-y)\frac{5}{95}e^{2t/95} \implies 95y=(500-5y)e^{2t/95}\\ &\hskip0.5in\implies \big(95+5e^{2t/95}\big)y=500 e^{2t/95}\\ &\hskip0.5in\implies y=\frac{500e^{2t/95}}{95+5e^{2t/95}} =\frac{100e^{2t/95}}{19+e^{2t/95}} =\frac{100}{1+19e^{-2t/95}} \end{align*}

Finally,

In the year 2000, $t=14$ and $y=\frac{100}{1+19e^{-28/95}}\approx6.6$ billion.

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In the year 2100, $t=114$ and $y=\frac{100}{1+19e^{-228/95}}\approx 36.7$ billion.

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In the year 2200, $t=514$ and $y=\frac{100}{1+19e^{-1028/95}}\approx 100$ billion.

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Subsection 2.4.5 Optional — Mixing Problems

Example 2.4.21. Dissolving salt.

At time $t=0\text{,}$ where $t$ is measured in minutes, a tank with a 5-litre capacity contains 3 litres of water in which 1 kg of salt is dissolved. Fresh water enters the tank at a rate of 2 litres per minute and the fully mixed solution leaks out of the tank at the varying rate of $2t$ litres per minute.

Determine the volume of solution $V(t)$ in the tank at time $t\text{.}$

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Determine the amount of salt $Q(t)$ in solution when the amount of water in the tank is at maximum.

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Solution
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No pun intended (sorry).
: (a) The rate of change of the volume in the tank, at time $t\text{,}$ is $2-2t\text{,}$ because water is entering at a rate $2$ and solution is leaking out at a rate $2t\text{.}$ Thus

\begin{equation*} \diff{V}{t}=2-2t \implies \dee{V}=(2-2t)\,\dee{t} \implies V=\int (2-2t)\,\dee{t} =2t-t^2+C \end{equation*}

at least until $V(t)$ reaches either the capacity of the tank or zero. When $t=0\text{,}$ $V=3$ so $C=3$ and $V(t)=3+2t-t^2\text{.}$ Observe that $V(t)$ is at a maximum when $\diff{V}{t}=2-2t=0\text{,}$ or $t=1\text{.}$

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(b) In the very short time interval from time $t$ to time $t+\dee{t}\text{,}$ $2t\,\dee{t}$ litres of brine leaves the tank. That is, the fraction $\frac{2t\,\dee{t}}{V(t)}$ of the total salt in the tank, namely $Q(t)\frac{2t\,\dee{t}}{V(t)}$ kilograms, leaves. Thus salt is leaving the tank at the rate

\begin{equation*} \frac{Q(t)\frac{2t\,\dee{t}}{V(t)}}{\dee{t}} =\frac{2tQ(t)}{V(t)} =\frac{2tQ(t)}{3+2t-t^2} \text{ kilograms per minute} \end{equation*}

\begin{align*} \diff{Q}{t}=-\frac{2tQ(t)}{3+2t-t^2} &\implies \frac{dQ}{Q}=-\frac{2t}{3+2t-t^2}\dee{t}=-\frac{2t}{(3-t)(1+t)}\dee{t}\\ & \qquad\qquad\qquad =\Big[\frac{3/2}{t-3}+\frac{1/2}{t+1}\Big]\dee{t}\\ &\implies \log Q=\frac{3}{2}\log|t-3|+\frac{1}{2}\log|t+1|+C \end{align*}

We are interested in the time interval $0\le t\le 1\text{.}$ In this time interval $|t-3|=3-t$ and $|t+1|=t+1$ so

\begin{equation*} \log Q=\frac{3}{2}\log(3-t)+\frac{1}{2}\log(t+1)+C \end{equation*}

At $t=0\text{,}$ $Q$ is 1 so

\begin{align*} \log 1 & =\frac{3}{2}\log(3-0)+\frac{1}{2}\log(0+1)+C\\ &\implies C=\log 1-\frac{3}{2}\log 3-\frac{1}{2}\log 1=-\frac{3}{2}\log 3 \end{align*}

At $t=1$

\begin{align*} \log Q&=\frac{3}{2}\log(3-1)+\frac{1}{2}\log(1+1)-\frac{3}{2}\log 3\\ & =2\log 2-\frac{3}{2}\log 3 =\log 4 -\log 3^{\frac{3}{2}} \end{align*}

so $Q=\frac{4}{3^{\frac{3}{2}}}\text{.}$

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Example 2.4.22. Mixing brines.

A tank contains 1500 liters of brine with a concentration of $0.3$ kg of salt per liter. Another brine solution, this with a concentration of $0.1$ kg of salt per liter is poured into the tank at a rate of $20$ li/min. At the same time, $20$ li/min of the solution in the tank, which is stirred continuously, is drained from the tank.

How many kilograms of salt will remain in the tank after half an hour?

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How long will it take to reduce the concentration to $0.2$ kg/li?

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Solution: Denote by $Q(t)$ the amount of salt in the tank at time $t\text{.}$ In a very short time interval $\dee{t}\text{,}$ the incoming solution adds $20\, \dee{t}$ liters of a solution carrying $0.1$ kg/li. So the incoming solution adds $0.1\times 20\, \dee{t}=2\, \dee{t}$ kg of salt. In the same time interval $20\, \dee{t}$ liters is drained from the tank. The concentration of the drained brine is $\frac{Q(t)}{1500}\text{.}$ So $\frac{Q(t)}{1500} 20\, \dee{t}$ kg were removed. All together, the change in the salt content of the tank during the short time interval is

\begin{equation*} dQ=2\, \dee{t}-\frac{Q(t)}{1500} 20\, \dee{t} =\Big(2-\frac{Q(t)}{75}\Big) \dee{t} \end{equation*}

The rate of change of salt content per unit time is

\begin{equation*} \frac{dQ}{dt}=2-\frac{Q(t)}{75} =-\frac{1}{75}\big(Q(t)-150\big) \end{equation*}

The solution of this equation is

\begin{equation*} Q(t)=\big\{Q(0)-150\big\}e^{-t/75} + 150 \end{equation*}

by Theorem 2.4.4, with $a=-\frac{1}{75}$ and $b=150\text{.}$ At time $0\text{,}$ $Q(0)=1500\times 0.3=450\text{.}$ So

\begin{equation*} Q(t)=150+300e^{-t/75} \end{equation*}

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(a) At $t=30$

\begin{equation*} Q(30)=150+300e^{-30/75}=\text{351.1 kg} \end{equation*}

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(b) $Q(t)=0.2\times 1500=300$ kg is achieved when

\begin{align*} &150+300e^{-t/75}=300\implies 300e^{-t/75}=150 \implies e^{-t/75}=0.5\cr &\implies -\frac{t}{75}=\log (0.5) \implies t=-75\log (0.5)=\text{51.99 min} \end{align*}

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Subsection 2.4.6 Optional — Interest on Investments

Suppose that you deposit $\$P$ in a bank account at time $t=0\text{.}$ The account pays $r\%$ interest per year compounded $n$ times per year.

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The first interest payment is made at time $t=\frac{1}{n}\text{.}$ Because the balance in the account during the time interval $0 \lt t \lt \frac{1}{n}$ is $\$P$ and interest is being paid for $\big(\frac{1}{n}\big)^{\rm th}$ of a year, that first interest payment is $\frac{1}{n}\times\frac{r}{100}\times P\text{.}$ After the first interest payment, the balance in the account is $P+\frac{1}{n}\times\frac{r}{100}\times P = \big(1+\frac{r}{100n}\big)P\text{.}$

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The second interest payment is made at time $t=\frac{2}{n}\text{.}$ Because the balance in the account during the time interval $\frac{1}{n} \lt t \lt \frac{2}{n}$ is $\big(1+\frac{r}{100n}\big)P$ and interest is being paid for $\big(\frac{1}{n}\big)^{\rm th}$ of a year, the second interest payment is $\frac{1}{n}\times\frac{r}{100}\times \big(1+\frac{r}{100n}\big)P\text{.}$ After the second interest payment, the balance in the account is $\big(1+\frac{r}{100n}\big)P+\frac{1}{n}\times\frac{r}{100}\times \big(1+\frac{r}{100n}\big)P = \big(1+\frac{r}{100n}\big)^2P\text{.}$

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And so on.

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In general, at time $t=\frac{m}{n}$ (just after the $m^{\rm th}$ interest payment), the balance in the account is

Equation 2.4.23. Discrete compounding interest.

\begin{equation*} B(t) = \Big(1+\frac{r}{100n}\Big)^m P = \Big(1+\frac{r}{100n}\Big)^{nt}P \end{equation*}

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Three common values of $n$ are $1$ (interest is paid once a year), $12$ (i.e. interest is paid once a month) and 365 (i.e. interest is paid daily). The limit $n\rightarrow\infty$ is called continuous compounding

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There are banks that advertise continuous compounding. You can find some by googling “interest is compounded continuously and paid”

. Under continuous compounding, the balance at time $t$ is

\begin{align*} B(t) &= \lim_{n\rightarrow\infty} \Big(1+\frac{r}{100n}\Big)^{nt}P \end{align*}

You may have already seen the limit

Equation 2.4.24. A useful limit.

\begin{equation*} \lim_{x\rightarrow 0}(1+x)^{a/x}=e^a \end{equation*}

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If so, you can evaluate $B(t)$ by applying 2.4.24 with $x=\frac{r}{100n}$ and $a=\frac{rt}{100}$ (so that $\frac{a}{x}=nt$). As $n\rightarrow \infty\text{,}$ $x\rightarrow 0$ so that

Equation 2.4.25. Continuous compound interest.

\begin{equation*} B(t) = \lim_{n\rightarrow\infty} \Big(1+\frac{r}{100n}\Big)^{nt}P =\lim_{x\rightarrow 0}(1+x)^{a/x}P=e^aP = e^{rt/100}P \end{equation*}

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If you haven’t seen (2.4.24) before, that’s OK. In the following example, we rederive (2.4.25) using a differential equation instead of (2.4.24).

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Example 2.4.26. Computing my future bank balance.

Suppose, again, that you deposit $\$P$ in a bank account at time $t=0\text{,}$ and that the account pays $r\%$ interest per year compounded $n$ times per year, and denote by $B(t)$ the balance at time $t\text{.}$ Suppose that you have just received an interest payment at time $t\text{.}$ Then the next interest payment will be made at time $t+\frac{1}{n}$ and will be $\frac{1}{n}\times\frac{r}{100}\times B(t)=\frac{r}{100n}B(t)\text{.}$ So, calling $\frac{1}{n}=h\text{,}$

\begin{equation*} B(t+h)=B(t) + \frac{r}{100}B(t)h\qquad\text{or}\qquad \frac{B(t+h)-B(t)}{h} = \frac{r}{100}B(t) \end{equation*}

To get continuous compounding we take the limit $n\rightarrow\infty$ or, equivalently, $h\rightarrow 0\text{.}$ This gives

\begin{equation*} \lim_{h\rightarrow 0}\frac{B(t+h)-B(t)}{h} = \frac{r}{100}B(t) \qquad\text{or}\qquad \diff{B}{t}(t) = \frac{r}{100}B(t) \end{equation*}

By Theorem 2.4.4, with $a=\frac{r}{100}$ and $b=0\text{,}$ (or Corollary 2.4.10 with $k=-\frac{r}{100}$),

\begin{gather*} B(t)=e^{rt/100}B(0)=e^{rt/100}P \end{gather*}

once again.

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Example 2.4.27. Double your money.

A bank advertises that it compounds interest continuously and that it will double your money in ten years. What is the annual interest rate?

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A bank advertises that it compounds monthly and that it will double your money in ten years. What is the annual interest rate?

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Solution: (a) Let the interest rate be $r\%$ per year. If you start with $\$P\text{,}$ then after $t$ years, you have $Pe^{rt/100}\text{,}$ under continuous compounding. This was equation (2.4.25). After 10 years you have $Pe^{r/10}\text{.}$ This is supposed to be $2P\text{,}$ so

\begin{align*} Pe^{r/10}=2P\quad &\Longrightarrow\quad e^{r/10}=2\\ & \Longrightarrow\quad \frac{r}{10}=\log 2\quad\Longrightarrow\quad r=10\log2 =6.93\% \end{align*}

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(b) Let the interest rate be $r\%$ per year. If you start with $\$P\text{,}$ then after $t$ years, you have $P\big(1+\frac{r}{100\times 12}\big)^{12 t}\text{,}$ under monthly compounding. This was (2.4.23). After 10 years you have $P\big(1+\frac{r}{100\times 12}\big)^{120}\text{.}$ This is supposed to be $2P\text{,}$ so

\begin{align*} &P\big(1+\frac{r}{100\times 12}\big)^{120}=2P &&\Longrightarrow\quad \big(1+\frac{r}{1200}\big)^{120}=2\\ &&&\Longrightarrow\quad 1+\frac{r}{1200}=2^{1/120}\\ &&&\Longrightarrow\quad \frac{r}{1200}=2^{1/120}-1\\ &&&\Longrightarrow\quad r=1200\big(2^{1/120}-1\big) =6.95\% \end{align*}

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Example 2.4.28. Pension planning.

A 25 year old graduate of UBC is given $50,000 which is invested at 5% per year compounded continuously. The graduate also intends to deposit money continuously at the rate of $2000 per year.

Find a differential equation that $A(t)$ obeys, assuming that the interest rate remains 5%.

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Determine the amount of money in the account when the graduate is 65.

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At age 65, the graduate will start withdrawing money continuously at the rate of $W$ dollars per year. If the money must last until the person is 85, what is the largest possible value of $W\text{?}$

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Solution: (a) Let’s consider what happens to $A$ over a very short time interval from time $t$ to time $t + \De t\text{.}$ At time $t$ the account balance is $A(t)\text{.}$ During the (really short) specified time interval the balance remains very close to $A(t)$ and so earns interest of $\frac{5}{100}\times\De t\times A(t)\text{.}$ During the same time interval, the graduate also deposits an additional $\$ 2000\De t\text{.}$ So

\begin{align*} A(t+\De t)& \approx A(t) + 0.05 A(t)\De t+ 2000\De t\\ & \implies \frac{A(t+\De t)- A(t)}{\De t} \approx 0.05 A(t) + 2000 \end{align*}

In the limit $\De t\rightarrow 0\text{,}$ the approximation becomes exact and we get

\begin{equation*} \diff{A}{t}= 0.05 A+2000 \end{equation*}

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(b) The amount of money at time $t$ obeys

\begin{equation*} \diff{A}{t}= 0.05 A(t)+2,\!000=0.05\big(A(t)+40,\!000\big) \end{equation*}

So by Theorem 2.4.4 (with $a=0.05$ and $b=-40,\!000$),

\begin{equation*} A(t)=\big(A(0)+40,\!000\big)e^{0.05 t} -40,\!000 \end{equation*}

At time 0 (when the graduate is 25), $A(0)=50,\!000\text{,}$ so the amount of money at time $t$ is

\begin{equation*} A(t)=90,\!000\, e^{0.05 t}-40,000 \end{equation*}

In particular, when the graduate is 65 years old, $t=40$ and

\begin{equation*} A(40)=90,\!000\, e^{0.05 \times 40}-40,000=\text{\$625,015.05 } \end{equation*}

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(c) When the graduate stops depositing money and instead starts withdrawing money at a rate $W\text{,}$ the equation for $A$ becomes

\begin{equation*} \diff{A}{t}= 0.05 A-W= 0.05 (A-20 W) \end{equation*}

assuming that the interest rate remains 5%. This time, Theorem 2.4.4 (with $a=0.05$ and $b=20W$) gives

\begin{equation*} A(t)=\big(A(0)-20W\big)e^{0.05 t} + 20W \end{equation*}

If we now reset our clock so that $t=0$ when the graduate is 65, $A(0)=625,015.05\text{.}$ So the amount of money at time $t$ is

\begin{equation*} A(t)=20W+ e^{0.05 t}(625,015.05-20W) \end{equation*}

We want the account to be depleted when the graduate is 85. So, we want $A(20)=0\text{.}$ This is the case if

\begin{align*} 20W+ e^{0.05\times 20}(625,015.05-20W) & =0\\ \implies& 20W+ e(625,015.05-20W)=0\\ \implies& 20(e-1)W= 625,015.05e\\ \implies& W=\frac{625,015.05e}{20(e-1)}=\$49,437.96 \end{align*}

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Exercises 2.4.7 Exercises

Recall that we are using $\log x$ to denote the logarithm of $x$ with base $e\text{.}$ In other courses it is often denoted $\ln x\text{.}$

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Exercises — Stage 1 .

1.

Below are pairs of functions $y=f(x)$ and differential equations. For each pair, decide whether the function is a solution of the differential equation.

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	function	differential equation
(a)	$y=5(e^x-3x^2-6x-6)$	$\displaystyle \diff{y}{x}=y+15x^2$
(b)	$y=\dfrac{-2}{x^2+1}$	$y'(x)=xy^2$
(c)	$y=x^{3/2}+x$	$\displaystyle\left(\diff{y}{x}\right)^2 + \diff{y}{x}=y$

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2.

Following Definition 2.4.1, a separable differential equation has the form

\begin{equation*} \diff{y}{x}(x) = f(x)\ g\big(y(x)\big). \end{equation*}

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Show that each of the following equations can be written in this form, identifying $f(x)$ and $g(y)\text{.}$

$\displaystyle 3y\diff{y}{x}=x\sin y$

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$\displaystyle \diff{y}{x} = e^{x+y}$

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$\displaystyle \diff{y}{x}+1=x$

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$\displaystyle \left(\diff{y}{x}\right)^2-2x\diff{y}{x}+x^2=0$

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3.

Suppose we have the following functions:

$y$ is a differentiable function of $x$

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$f$ is a function of $x\text{,}$ with $\int f(x)\,\dee{x}=F(x)$

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$g$ is a nonzero function of $y\text{,}$ with $\int \frac{1}{g(y)} \,\dee{y}=G(y)=G(y(x))\text{.}$

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In the work below, we set up a solution to the separable differential equation

\begin{equation*} \diff{y}{x}=f(x)\,g(y)=f(x)\,g(y(x)) \end{equation*}

without using the mnemonic of Equation (✶).

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By deleting some portion of our work, we can create the solution as it would look using the mnemonic. What portion can be deleted?

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Remark: the purpose of this exercise is to illuminate what, exactly, the mnemonic is a shortcut for. Despite its peculiar look, it agrees with what we already know about integration.

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\begin{align*} \diff{y}{x}&=f(x)g(y(x))\\ \end{align*}
Since $g(y(x))$ is a nonzero function, we can divide both sides by it.
\begin{align*} \frac{1}{g(y(x))}\cdot\diff{y}{x}&=f(x)\\ \end{align*}
If these functions of $x$ are the same, then they have the same antiderivative with respect to $x\text{.}$
\begin{align*} \int \frac{1}{g(y(x))}\cdot\diff{y}{x}\,\dee{x}&=\int f(x)\,\dee{x}\\ \end{align*}
The left integral is in the correct form for a change of variables to $y\text{.}$ To make this easier to see, we’ll use a $u$-substitution, since it’s a little more familiar than a $y$-substitution. If $u=y\text{,}$ then $\diff{u}{x}=\diff{y}{x}\text{,}$ so $\dee{u}=\diff{y}{x}\dee{x}\text{.}$
\begin{align*} \int\frac{1}{g(u)}\,\dee{u}&=\int f(x)\,\dee{x}\\ \end{align*}
Since $u$ was just the same as $y\text{,}$ again for cosmetic reasons, we can swap it back. (Formally, you could have skipped the step above--we just included it to be extra clear that we’re not using any integration techniques we haven’t seen before.)
\begin{align*} \int\frac{1}{g(y)}\,\dee{y}&=\int f(x)\,\dee{x}\\ \end{align*}
We’re given the antiderivatives in question.
\begin{align*} G(y)+C_1&=F(x)+C_2\\ G(y)&=F(x)+(C_2-C_1)\\ \end{align*}
where $C_1$ and $C_2$ are arbitrary constants. Then also $C_2-C_1$ is an arbitrary constant, so we might as well call it $C\text{.}$
\begin{align*} G(y)&=F(x)+C \end{align*}

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4.

Suppose $y=f(x)$ is a solution to the differential equation $\diff{y}{x}=xy\text{.}$

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True or false: $f(x)+C$ is also a solution, for any constant $C\text{.}$

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5.

Suppose a function $y=f(x)$ satisfies $|y| = Cx\text{,}$ for some constant $C \gt 0\text{.}$

What is the largest possible domain of $f(x)\text{,}$ given the information at hand?

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Give an example of function $y=f(x)$ with the following properties, or show that none exists:
- $|y| = Cx\text{,}$
  
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- $\diff{y}{x}$ exists for all $x \gt 0\text{,}$ and
  
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- $y \gt 0$ for some values of $x\text{,}$ and $y \lt 0$ for others.
  
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6.

Express the following sentence

¹²

The sentence is paraphrased from the Pharmakokinetics website of Université de Lausanne, Elimination Kinetics. The half-life of morphine is given on the same website. Accessed 12 August 2017.

as a differential equation. You don’t have to solve the equation.

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About 0.3 percent of the total quantity of morphine in the bloodstream is eliminated every minute.
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7.

Suppose a particular change is occurring in a language, from an old form to a new form.

¹³

An example is the change in German from “wollt” to “wollst” for the second-person conjugation of the verb “wollen.” This example is provided by the site Laws in Quantitative Linguistics, Change in Language, accessed 18 August 2017.

Let $p(t)$ be the proportion (measured as a number between 0, meaning none, and 1, meaning all) of the time that speakers use the new form. Piotrowski’s law

¹⁴

Piotrowski’s law is paraphrased from the page Piotrowski-Gesetz on Glottopedia, accessed 18 August 2017. According to this source, the law was based on work by the married couple R. G. Piotrowski and A. A. Piotrowskaja, later generalized by G. Altmann.

predicts the following.

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Use of the new form over time spreads at a rate that is proportional to the product of the proportion of the new form and the proportion of the old form.
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Express this as a differential equation. You do not need to solve the differential equation.

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8.

Consider the differential equation $y'=\frac{y}{2}-1\text{.}$

When $y=0\text{,}$ what is $y'\text{?}$

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When $y=2\text{,}$ what is $y'\text{?}$

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When $y=3\text{,}$ what is $y'\text{?}$

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On the axes below, interpret the marks we have made, and use them to sketch a possible solution to the differential equation.

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9.

Consider the differential equation $y'=y-\frac{x}{2}\text{.}$

If $y(1)=0\text{,}$ what is $y'(1)\text{?}$

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If $y(1)=2\text{,}$ what is $y'(1)\text{?}$

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If $y(1)=-2\text{,}$ what is $y'(1)\text{?}$

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Draw a sketch similar to that of Question 8(d) showing the derivatives of $y$ at the points with integer values for $x$ in $[0,6]$ and $y$ in $[-3,3]\text{.}$

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Sketch a possible graph of $y\text{.}$

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Exercises — Stage 2 .

10. (✳).

Find the solution to the separable initial value problem:

\begin{align*} \diff{y}{x} &= \frac{2x}{e^y}, & y(0) = \log 2 \end{align*}

Express your solution explicitly as $y = y(x)\text{.}$

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11. (✳).

Find the solution $y(x)$ of $\displaystyle\diff{y}{x}=\frac{xy}{x^2+1}, \quad y(0)=3\text{.}$

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12. (✳).

Solve the differential equation $y'(t)=e^{\frac{y}{3}}\cos t\text{.}$ You should express the solution $y(t)$ in terms of $t$ explicitly.

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13. (✳).

Solve the differential equation

\begin{equation*} \diff{y}{x} = x e^{x^2-\log(y^2)} \end{equation*}

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14. (✳).

Let $y=y(x)\text{.}$ Find the general solution of the differential equation $y'=xe^y\text{.}$

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15. (✳).

Find the solution to the differential equation $\displaystyle\frac{y y'}{e^x -2x} = \frac{1}{y}$ that satisfies $y(0) = 3\text{.}$ Solve completely for $y$ as a function of $x\text{.}$

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16. (✳).

Find the function $y=f(x)$ that satisfies

\begin{gather*} \diff{y}{x} = -xy^3 \qquad\text{and}\qquad f(0)=-\frac{1}{4} \end{gather*}

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17. (✳).

Find the function $y=y(x)$ that satisfies $y(1)=4$ and

\begin{equation*} \diff{y}{x} = \frac{15x^2 + 4x + 3}{y} \end{equation*}

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18. (✳).

Find the solution $y(x)$ of $y'=x^3y$ with $y(0)=1\text{.}$

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19. (✳).

Find the solution of the initial value problem

\begin{equation*} x\diff{y}{x} + y = y^2\qquad y(1) = -1 \end{equation*}

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20. (✳).

A function $f(x)$ is always positive, has $f(0)=e$ and satisfies $f'(x) = x\,f(x)$ for all $x\text{.}$ Find this function.

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21. (✳).

Solve the following initial value problem:

\begin{gather*} \diff{y}{x} =\frac{1}{(x^2+x)y}\qquad y(1)=2 \end{gather*}

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22. (✳).

Find the solution of the differential equation $\displaystyle \frac{1+\sqrt{y^2-4}}{\tan x} y' = \frac{\sec x}y$ that satisfies $y(0)=2\text{.}$ You don’t have to solve for $y$ in terms of $x\text{.}$

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23. (✳).

The fish population in a lake is attacked by a disease at time $t=0\text{,}$ with the result that the size $P(t)$ of the population at time $t\ge 0$ satisfies

\begin{gather*} \diff{P}{t}=-k\sqrt{P} \end{gather*}

where $k$ is a positive constant. If there were initially 90,000 fish in the lake and 40,000 were left after 6 weeks, when will the fish population be reduced to 10,000?

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24. (✳).

An object of mass $m$ is projected straight upward at time $t=0$ with initial speed $v_0\text{.}$ While it is going up, the only forces acting on it are gravity (assumed constant) and a drag force proportional to the square of the object’s speed $v(t)\text{.}$ It follows that the differential equation of motion is

\begin{gather*} m\diff{v}{t}=-(mg+kv^2) \end{gather*}

where $g$ and $k$ are positive constants. At what time does the object reach its highest point?

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25. (✳).

A motor boat is traveling with a velocity of 40 ft/sec when its motor shuts off at time $t=0\text{.}$ Thereafter, its deceleration due to water resistance is given by

\begin{gather*} \diff{v}{t}=-k\,v^2 \end{gather*}

where $k$ is a positive constant. After 10 seconds, the boat’s velocity is 20 ft/sec.

What is the value of $k\text{?}$

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When will the boat’s velocity be 5 ft/sec?

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26. (✳).

Consider the initial value problem $\diff{x}{t}= k(3-x)(2-x)\text{,}$ $x(0)=1\text{,}$ where $k$ is a positive constant. (This kind of problem occurs in the analysis of certain chemical reactions.)

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Solve the initial value problem. That is, find $x$ as a function of $t\text{.}$

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What value will $x(t)$ approach as $t$ approaches $+\infty\text{.}$

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27. (✳).

The quantity $P=P(t)\text{,}$ which is a function of time $t\text{,}$ satisfies the differential equation

\begin{gather*} \diff{P}{t}=4P-P^2 \end{gather*}

and the initial condition $P(0)=2\text{.}$

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Solve this equation for $P(t)\text{.}$

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What is $P$ when $t=0.5\text{?}$ What is the limiting value of $P$ as $t$ becomes large?

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28. (✳).

An object moving in a fluid has an initial velocity $v$ of 400 m/min. The velocity is decreasing at a rate proportional to the square of the velocity. After 1 minute the velocity is 200 m/min.

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Give a differential equation for the velocity $v=v(t)$ where $t$ is time.

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Solve this differential equation.

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When will the object be moving at 50 m/min?

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Exercises — Stage 3 .

29. (✳).

An investor places some money in a mutual fund where the interest is compounded continuously and where the interest rate fluctuates between $4\%$ and $8\%\text{.}$ Assume that the amount of money $B=B(t)$ in the account in dollars after $t$ years satisfies the differential equation

\begin{gather*} \diff{B}{t}=\big(0.06+0.02\sin t\big)B \end{gather*}

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Solve this differential equation for $B$ as a function of $t\text{.}$

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If the initial investment is $\$1000\text{,}$ what will the balance be at the end of two years?

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30. (✳).

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem $B'(t) = aB - m$ for $t \ge 0\text{,}$ with $B(0) = B_0$ . The constant $a$ reflects the annual interest rate, $m$ is the annual rate of withdrawal, and $B_0$ is the initial balance in the account.

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Solve the initial value problem with $a = 0.02$ and $B(0) = B_0 = \$30,000\text{.}$ Note that your answer depends on the constant $m\text{.}$

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If $a = 0.02$ and $B(0) = B_0 = \$30,000\text{,}$ what is the annual withdrawal rate $m$ that ensures a constant balance in the account?

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31. (✳).

A certain continuous function $y=y(x)$ satisfies the integral equation

\begin{equation*} y(x)=3+\int_0^x\big(y(t)^2-3y(t)+2\big)\sin t \dee{t} \tag{$*$} \end{equation*}

for all $x$ in some open interval containing $0\text{.}$ Find $y(x)$ and the largest interval for which $(*)$ holds.

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32. (✳).

A cylindrical water tank, of radius 3 meters and height 6 meters, is full of water when its bottom is punctured. Water drains out through a hole of radius 1 centimeter. If

$h(t)$ is the height of the water in the tank at time $t$ (in meters) and

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$v(t)$ is the velocity of the escaping water at time $t$ (in meters per second) then

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Torricelli’s law states that $v(t)=\sqrt{2gh(t)}$ where $g=9.8\ {\rm m/sec^2}\text{.}$ Determine how long it takes for the tank to empty.

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33. (✳).

A spherical tank of radius 6 feet is full of mercury when a circular hole of radius 1 inch is opened in the bottom. How long will it take for all of the mercury to drain from the tank?

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Use the value $g=32\ {\rm feet}/{\rm sec}^2\text{.}$ Also use Torricelli’s law, which states when the height of mercury in the tank is $h\text{,}$ the speed of the mercury escaping from the tank is $v=\sqrt{2gh}\text{.}$

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34. (✳).

Consider the equation

\begin{gather*} f(x)=3+\int_0^x\big(f(t)-1\big)\big(f(t)-2\big)\dee{t} \end{gather*}

What is $f(0)\text{?}$

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Find the differential equation satisfied by $f(x)\text{.}$

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Solve the initial value problem determined in (a) and (b).

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35. (✳).

A tank 2 m tall is to be made with circular cross--sections with radius $r=y^p\text{.}$ Here $y$ measures the vertical distance from the bottom of the tank and $p$ is a positive constant to be determined. You may assume that when the tank drains, it obeys Torricelli’s law, that is

\begin{gather*} A(y)\diff{y}{t}=-c\sqrt{y} \end{gather*}

for some constant $c$ where $A(y)$ is the cross--sectional area of the tank at height $y\text{.}$ It is desired that the tank be constructed so that the top half ($y=2$ to $y=1$) takes exactly the same amount of time to drain as the bottom half ($y=1$ to $y=0$). Determine the value of $p$ so that the tank has this property. Note: it is not possible or necessary to find $c$ for this question.

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36.

Suppose $f(t)$ is a continuous, differentiable function and the root mean square of $f(t)$ on $[a,x]$ is equal to the average of $f(t)$ on $[a,x]$ for all $x\text{.}$ That is,

\begin{equation*} \frac{1}{x-a}\int_a^xf(t)\,\dee(t)=\sqrt{\frac{1}{x-a}\int_a^x f^2(t)\,\dee{t}}\tag{$*$} \end{equation*}

You may assume $x \gt a\text{.}$

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Guess a function $f(t)$ for which the average of $f(t)$ is the same as the root mean square of $f(t)$ on any interval.

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Differentiate both sides of the given equation.

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Simplify your answer from (b) by using Equation ($*$) to replace all terms containing $\int_a^x f^2(t)\,\dee{t}$ with terms containing $\int_a^x f(t)\,\dee{t}\text{.}$

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Let $Y(x) = \int_a^x f(t)\,\dee{t}\text{,}$ so the equation from (c) becomes a differential equation. Find all functions that satisfy it.

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What is $f(t)\text{?}$

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37.

Find the function $y(x)$ such that

\begin{equation*} \ddiff{2}{y}{x}=\frac{2}{y^3}\cdot\diff{y}{x} \end{equation*}

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and if $x=-\frac{1}{16}\log 3\text{,}$ then $y=1$ and $\diff{y}{x}=3\text{.}$

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You do not need to solve for $y$ explicitly.

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Prev Top Next

	function	differential equation
(a)	\(y=5(e^x-3x^2-6x-6)\)	\(\displaystyle \diff{y}{x}=y+15x^2\)
(b)	\(y=\dfrac{-2}{x^2+1}\)	\(y'(x)=xy^2\)
(c)	\(y=x^{3/2}+x\)	\(\displaystyle\left(\diff{y}{x}\right)^2 + \diff{y}{x}=y\)