8.7 Separable first-order differential equations — Learn

Prerequisites: integration by substitution and standard integrals from Sections 8.1 and 8.5; factorising algebraic expressions from Section 2.8.

By the end of this section you can: recognise a separable first-order differential equation; separate the variables and integrate both sides; find a general solution; and use an initial condition to find a particular solution.

1. Separating the variables

A differential equation is an equation involving a derivative. It is separable when it can be written with all the $y$ terms on one side and all the $x$ terms on the other:

\frac{dy}{dx} = f(x)g(y).

To solve it, divide by $g(y)$ and integrate both sides with respect to $x$ :

\int \frac{1}{g(y)}\frac{dy}{dx}dx = \int f(x)dx,

which is usually written in the shorter form

\int \frac{1}{g(y)}dy = \int f(x)dx.

Then integrate each side and add one constant of integration.

Example 1. Solve $\dfrac{dy}{dx} = xy$ .

Divide by $y$ and integrate:

\int \frac{1}{y}dy = \int xdx,

\ln|y| = \frac{x^2}{2} + C.

Raise $e$ to both sides. The constant $C$ becomes a multiplier $A = e^C$ :

y = Ae^{x^2/2}.

This is the general solution. Different values of $A$ give different curves in the family.

2. A constant of integration on each side

When you integrate both sides, a constant appears on each. Subtract one from the other to leave a single constant. In Example 1 the right-hand constant absorbs the left-hand one.

Example 2. Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ .

Multiply both sides by $y$ and by $dx$ :

ydy = xdx.

Integrate:

\frac{y^2}{2} = \frac{x^2}{2} + C.

Multiply by $2$ and rename the constant $K = 2C$ :

y^2 = x^2 + K.

Taking a square root gives $y = \pm\sqrt{x^2 + K}$ . A particular solution keeps the sign fixed by the initial condition.

Example 3. Solve $\dfrac{dy}{dx} = e^{x-y}$ .

Use the index law $e^{x-y} = e^x e^{-y}$ , then separate:

e^ydy = e^xdx.

Integrate:

e^y = e^x + C, \qquad y = \ln(e^x + C).

The constant stays inside the logarithm because $e^y$ is the quantity that was integrated.

3. Particular solutions

An initial condition fixes the constant. It gives a value of $y$ at a value of $x$ , such as $y = 3$ when $x = 0$ .

Example 4. Find the particular solution of $\dfrac{dy}{dx} = 2xy$ for which $y = 3$ when $x = 0$ .

Separate and integrate:

\int \frac{1}{y}dy = \int 2xdx, \qquad \ln|y| = x^2 + C, \qquad y = Ae^{x^2}.

Substitute $x = 0$ and $y = 3$ : $3 = Ae^0 = A$ , so the particular solution is

y = 3e^{x^2}.

4. Factorising before separating

Sometimes a common factor hides the separation. Factorise it out first.

Example 5. Solve $\dfrac{dy}{dx} = xy - x$ .

Take out the common factor $x$ :

\frac{dy}{dx} = x(y - 1).

The equation is now separable. Divide by $y - 1$ :

\int \frac{1}{y - 1}dy = \int xdx, \qquad \ln|y - 1| = \frac{x^2}{2} + C.

Exponentiate both sides:

y - 1 = Ae^{x^2/2}, \qquad y = 1 + Ae^{x^2/2}.

Watch out. Four mistakes are common.

Dividing by zero. When you divide by $g(y)$ , check whether $g(y) = 0$ gives a constant solution. For $\dfrac{dy}{dx} = xy$ , the value $y = 0$ is a solution, and it is included in $y = Ae^{x^2/2}$ with $A = 0$ .

Forgetting to integrate both sides. After separating, both sides need an integral. Writing $\ln|y| = x$ instead of $\ln|y| = \dfrac{x^2}{2}$ skips the integration.

Losing the sign of the constant. The constant of integration becomes a multiplier when you exponentiate. Write $A = e^C$ so that $A$ can be positive or negative.

Not factorising first. An expression such as $xy - x$ separates only after you take out $x$ . Re-check: substitute your answer back into the original equation and confirm both sides match.

5. Try it: the family of solutions

Predict first. The figure shows the family of solutions of $\dfrac{dy}{dx} = ky$ for different starting values, with a slider for $k$ . Before you move the slider, predict what happens to the curves when $k$ changes from positive to negative. Write your prediction down.

Test. Move the slider for $k$ and watch the curves. Then move the starting value and watch which curve is highlighted.

Explain. Write two sentences. The first explains why every curve is an exponential. The second explains why a negative value of $k$ gives a curve that falls.

You should now be able to

recognise a separable first-order differential equation;
separate the variables and integrate both sides;
add one constant of integration and tidy the general solution;
use an initial condition to find a particular solution;
factorise a common factor before separating;
check a solution by substituting it into the differential equation.