Separable differential equations

Section 1.13 Separable differential equations

Definition 1.13.1. Exponential growth and decay.

Suppose the function \(f(x)\) satisfies the differential equation

\begin{equation} f'(x)=k\, f(x)\text{,}\tag{1.13.1} \end{equation}

where \(k\) is a fixed constant.

If \(k> 0\) then \(f(x)\) is said to undergo exponential growth.

If \(k< 0\) then \(f(x)\) is said to undergo exponential decay.

Example 1.13.2. Exponential growth/decay.

Fix a nonzero constant \(k\text{.}\) Find a general formula for all functions \(f\) with domain \(\R=(-\infty,\infty)\) satisfying

\begin{equation*} f'(x)=k\, f(x)\text{.} \end{equation*}

Solution.

The following technique for solving for \(f\) is called separation of variables, as Procedure 1.13.5 makes clear.

First let’s assume that \(f(x)\ne 0\) for all \(x\in \R\text{.}\) In this case we have

Definition 1.13.3. First-order differential equation.

A first-order differential equation for the function \(f\) is an equation that can be written in the form

\begin{equation} f'(x)=F(x,f(x)) \tag{1.13.2} \end{equation}

where \(F(x,f(x))\) denotes an arbitrary expression involving \(x\) and \(f(x)\text{.}\)

A solution to a differential equation is any function \(f\) that satisfies equation (1.13.2); as such it is useful to think of the function \(f\) in this context as an unknown that we are trying to solve for. The general solution to a differential equation is a formula, possibly containing undetermined constants, describing all solutions to the differential equation.

Definition 1.13.4. Separable first-order differential equation.

A separable differential equation for the function \(f\) is a differential equation that can be written in the form

\begin{equation} f'(x)=g(x)h(f(x))\text{,}\tag{1.13.3} \end{equation}

or equivalently, letting \(y=f(x)\text{,}\)

\begin{equation} \frac{dy}{dx}=g(x)h(y)\text{.}\tag{1.13.4} \end{equation}

Procedure 1.13.5. Separation of variables (algebraic form).

To solve a separable differential equation of the form

\begin{equation} \frac{dy}{dx}=g(x)h(y)\tag{1.13.5} \end{equation}

for the function \(y=f(x)\text{,}\) proceed as follows.

Separation.

Write the equation as

\begin{equation*} \frac{1}{h(y)}\, dy=g(x)\, dx \end{equation*}

and take take the indefinite integral of both sides.

\begin{equation*} \int \frac{1}{h(y)}\, dy=\int g(x)\, dx\text{.} \end{equation*}
Integration.
Attempt to find an antiderivative \(F(y)\) of \(1/h(y)\) and an antiderivative \(G(x)\) for \(g(x)\text{.}\)
Algebra.

The resulting general equation

\begin{equation} F(y)=G(x)+C\tag{1.13.6} \end{equation}

is said to be an implicit solution of the differential equation (1.13.5). An explicit solution is obtained from (1.13.6) by solving for \(y\) in terms of \(x\) and \(C\text{,}\) if possible.

Example 1.13.6. Separation of variables.

Solve the following differential equations using separation of variables. If an initial condition is given, provide the corresponding particular solution. Otherwise, give the general solution.

\(\displaystyle \displaystyle f'(x)=xf(x)+x\)
\(\displaystyle \displaystyle\frac{dy}{dx}=\frac{x^3}{y^2}\)
\(\cot x \cdot f'(x)+f(x)=2\text{,}\) \(f(0)=0\text{.}\)

Solution.

Let \(y=f(x)\text{.}\) Following Procedure 1.13.5 we compute

\begin{align*} \frac{dy}{dx} \amp = x(y+1)\\ \frac{dy}{y+1} \amp = x\, dx\\ \int \frac{1}{y+1}\, dy \amp = \int x\, dx\\ \ln\vert y+1\vert \amp =\frac{1}{2}x^2+C\\ \vert y+1\vert \amp = e^{x^2/2+C}\\ y+1\amp =\pm e^{x^2/2}e^C \\ y \amp = (\pm e^C)e^{x^2/2}-1\text{.} \end{align*}

We conclude that

\begin{equation*} f(x)=De^{x^2/2}-1 \end{equation*}

is the general solution, where \(D=(\pm e^C)\text{.}\) How arbitrary is \(D\text{?}\) Since \(C\) was arbitrary, the expression \(\pm e^C\) can assume any value except 0. Setting \(D=0\) in our formula yields the function \(f(x)=-1\text{,}\) which by inspection also satisfies the differential equation. We conclude that the general solution is \(f(x)=De^{x^2/2}-1\) where \(D\) is any real number.
Following Procedure 1.13.5 we compute

\begin{align*} \int y^2\, dy\amp = \int x^3\, dx \\ \frac{1}{3}y^3 \amp = \frac{1}{4}x^4+C\\ y^3 \amp = \frac{3}{4}x^4+3C\\ y \amp = \sqrt[3]{\frac{3}{4}x^4+D} \end{align*}

where \(D\) is any real number.
Let \(y=f(x)\text{.}\) Following Procedure 1.13.5 we compute

\begin{align*} \cot x\cdot \frac{dy}{dx}\amp =2-y \\ \frac{dy}{2-y} \amp= \tan x\, dx \\ \int \frac{1}{2-y}\, dy \amp = \int \tan x\, dx\\ -\ln\vert 2-y\vert \amp = -\ln\vert \cos x\vert+C \\ \ln\vert 2-y\vert \amp = \ln\vert \cos x\vert-C \\ \vert 2-f(x)\vert \amp = e^{\ln\vert \cos x\vert}\, e^{-C}=\vert \cos x\vert e^{-C} \\ f(x) \amp = 2-D\cos x \text{.} \end{align*}

Lastly, using the initial condition, we see that

\begin{align*} f(0)=0 \amp \implies 2-D=0\\ \amp \implies D=2 \end{align*}

Thus \(f(x)=2-2\cos x\text{.}\)

Example 1.13.7. Newton’s law of cooling.

Suppose a hot object cools in a room kept at constant temperature of \(T_0\) (in celcius). Newton’s law of cooling states that the rate at which the object cools (with respect to time) is proportional to the difference between its current temperature and the room temperature \(T_0\text{.}\)

Write a differential equation that describes Newton’s law of cooling in this setting.
Find the general solution to this differential equation.
Find a the particular solution to the situation where \(T_0=15 ^\circ\)C, the object’s initial temperature is \(100 ^\circ\)C, and after \(5\) minutes the object’s temperature is \(80\)\(^\circ\)C.

Solution.

Let temperature \(T\) be given by the function \(T=f(t)\text{.}\) The rate of change of \(T\) with respect to \(t\) is then \(f'(t)\text{.}\) Newton’s law says this rate of change is proportional to the difference \(f(t)-T_0\) of the current temperature of the object and the room temperature. Mathematically, we write this proportionality relation as

\begin{equation} f'(t)=k(f(t)-T_0)\text{,}\tag{1.13.7} \end{equation}

where \(k\) is some fixed constant, called the constant of proportionality.
Equation (1.13.7) can be solved using Procedure 1.13.5 in much the same manner as Example 1.13.6, yielding a general solution of the form

\begin{equation*} f(t)=De^{kt}+T_0\text{,} \end{equation*}

where \(D\) is any real number.
Assuming \(T_0=15\text{,}\) we have \(f(t)=De^{kt}+15\text{.}\) We use the two additional pieces of information to determine the undetermined constants \(D\) and \(k\text{.}\) We have

\begin{align*} f(0)=100 \amp \implies De^0+15=100\implies D=85\\ f(5)=80 \amp \implies 85e^{k5}+15=80\\ \amp \implies e^{k5}=\frac{65}{85}=\frac{13}{17}\\ \amp\implies 5k=\ln(13/17) \\ k \amp =\frac{1}{5}\ln (13/17)\text{.} \end{align*}

We conclude that \(f(t)=85e^{kt}+15\text{,}\) where \(k=\frac{1}{5}\ln (13/17)\text{.}\) Below you find a graph of \(f\) along with its horizontal asymptote corresponding to the steady room temperature \(T_0=15\text{.}\)

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