8.7 Separable first-order differential equations — Practise with help

Work each question closed book first. Separate the variables before you integrate, and keep one constant of integration. Substitute the initial condition at the end.

Warm-up: remember

Solve $e^{2x} = 5$ for $x$ , giving an exact answer.
Write $\ln a + \ln b$ as a single logarithm.
Find $\displaystyle\int \frac{1}{y}dy$ for $y > 0$ .

Answers

$2x = \ln 5$ , so $x = \dfrac{1}{2}\ln 5$ .
$\ln(ab)$ .
$\ln y + C$ .

If question 1 was hard, revise solving exponential equations in Section 6.4.

Practise with help

1. Solve $\dfrac{dy}{dx} = 3xy$ , giving the general solution.

Hint

Divide both sides by $y$ , then integrate. The right-hand side becomes $\displaystyle\int 3xdx$ .

Next step

$\ln|y| = \dfrac{3x^2}{2} + C$ . Exponentiate both sides and replace $e^C$ by a constant $A$ .

Full working

y = Ae^{3x^2/2}.

Different values of $A$ give the different curves in the family. Check by differentiating: $\dfrac{dy}{dx} = 3xAe^{3x^2/2} = 3xy$ .

2. Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ for which $y = 2$ when $x = 0$ .

Hint

Multiply both sides by $y$ and integrate. The general solution is $y^2 = x^2 + K$ .

Next step

Substitute $x = 0$ and $y = 2$ into $y^2 = x^2 + K$ : $4 = 0 + K$ , so $K = 4$ . Choose the positive root because the initial value is positive.

Full working

y^2 = x^2 + 4, \qquad y = \sqrt{x^2 + 4}.

The initial value $y(0) = 2$ is positive, so the positive square root is the correct branch.

3. Solve $\dfrac{dy}{dx} = \dfrac{e^x}{y}$ , giving the general solution.

Hint

Multiply both sides by $y$ . Both sides are then standard integrals.

Next step

$ydy = e^xdx$ , so $\dfrac{y^2}{2} = e^x + C$ . Multiply by $2$ and rename the constant.

Full working

y^2 = 2e^x + K.

The constant $K = 2C$ absorbs the factor of $2$ . The general solution can be written as $y = \pm\sqrt{2e^x + K}$ .

4. Solve $\dfrac{dy}{dx} = x(y - 1)$ for which $y = 3$ when $x = 0$ .

Hint

The right-hand side is already factorised. Divide by $y - 1$ and integrate.

Next step

$\ln|y - 1| = \dfrac{x^2}{2} + C$ , so $y = 1 + Ae^{x^2/2}$ . Substitute the initial condition.

Full working

Substitute $x = 0$ and $y = 3$ :

3 = 1 + A, \qquad A = 2.

The particular solution is

y = 1 + 2e^{x^2/2}.

The constant term $1$ is the value that makes $\dfrac{dy}{dx}$ zero when $y = 1$ .

5. Solve $\dfrac{dy}{dx} = \dfrac{y}{x}$ for $x > 0$ .

Hint

Divide by $y$ , then integrate both sides. On the right, use the standard integral of $\dfrac{1}{x}$ .

Next step

$\ln|y| = \ln x + C$ . Exponentiate to turn the sum of logarithms into a product.

Full working

y = Ax.

Every straight line through the origin with $x > 0$ is a solution. Check: $\dfrac{dy}{dx} = A = \dfrac{Ax}{x} = \dfrac{y}{x}$ .

6. A body moves so that its velocity $v$ satisfies $\dfrac{dv}{dt} = 2 - v$ , and $v = 0$ when $t = 0$ . Find $v$ in terms of $t$ .

Hint

Rearrange to $\displaystyle\int \frac{1}{2 - v}dv = \int 1dt$ . The left side gives $-\ln|2 - v|$ .

Next step

$-\ln|2 - v| = t + C$ , so $\ln|2 - v| = -t - C$ and $2 - v = Ae^{-t}$ . Substitute $t = 0$ , $v = 0$ .

Full working

At $t = 0$ : $2 - 0 = A$ , so $A = 2$ . Then

2 - v = 2e^{-t}, \qquad v = 2 - 2e^{-t}.

As $t$ grows, $e^{-t}$ tends to zero and $v$ approaches $2$ . The value $2$ is the terminal velocity, where $\dfrac{dv}{dt} = 0$ .