8.7 Separable first-order differential equations — Practise

Work through these on your own. Separate the variables, integrate both sides, and add one constant. For a particular solution, substitute the initial condition.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Solve $\dfrac{dy}{dx} = xy$ , giving the general solution.
Solve $\dfrac{dy}{dx} = 3xy$ , giving the general solution.
Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ , giving the general solution.
Solve $\dfrac{dy}{dx} = e^{x-y}$ , giving the general solution.
Find the particular solution of $\dfrac{dy}{dx} = 2xy$ for which $y = 3$ when $x = 0$ .
Find the particular solution of $\dfrac{dy}{dx} = \dfrac{x}{y}$ for which $y = 2$ when $x = 0$ .
Solve $\dfrac{dy}{dx} = xy - x$ , giving the general solution.
Solve $\dfrac{dy}{dx} = \dfrac{y}{x}$ for $x > 0$ , giving the general solution.
A body moves so that $\dfrac{dv}{dt} = 2 - v$ and $v = 0$ when $t = 0$ . Find $v$ in terms of $t$ .
A population $P$ grows so that $\dfrac{dP}{dt} = 0.05P$ and $P = 200$ when $t = 0$ . Find $P$ in terms of $t$ .
A mass $m$ decays so that $\dfrac{dm}{dt} = -0.05m$ and $m = 200$ when $t = 0$ . Find $m$ in terms of $t$ .
Find the particular solution of $\dfrac{dy}{dx} = xy - x$ for which $y = 3$ when $x = 0$ .

Answers and concise solutions

1. The general solution

Answer: Separate: $\displaystyle\int \dfrac{1}{y}\,dy = \int x\,dx$ , so $\ln|y| = \dfrac{x^2}{2} + C$ and $y = Ae^{x^2/2}$ .

2. The general solution

Answer: Separate: $\ln|y| = \dfrac{3x^2}{2} + C$ , so $y = Ae^{3x^2/2}$ .

3. The general solution

Answer: Separate: $y\,dy = x\,dx$ , so $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$ and $y^2 = x^2 + A$ . Hence $y = \pm\sqrt{x^2 + A}$ .

4. The general solution

Answer: Write $e^{x-y} = e^xe^{-y}$ , then $e^y\,dy = e^x\,dx$ . Integrating gives $e^y = e^x + A$ , so $y = \ln(e^x + A)$ .

5. The particular solution

Answer: The general solution is $y = Ae^{x^2}$ . Substituting $x = 0$ , $y = 3$ gives $A = 3$ , so $y = 3e^{x^2}$ .

6. The particular solution

Answer: Separating and integrating gives $y^2 = x^2 + A$ . Substituting $x = 0$ , $y = 2$ gives $A = 4$ . The initial value is positive, so $y = \sqrt{x^2 + 4}$ .

7. The general solution

Answer: Factorise: $\dfrac{dy}{dx} = x(y - 1)$ . Then $\ln|y - 1| = \dfrac{x^2}{2} + C$ , so $y = 1 + Ae^{x^2/2}$ .

8. The general solution

Answer: Separate: $\displaystyle\int \dfrac{1}{y}\,dy = \int \dfrac{1}{x}\,dx$ , so $\ln|y| = \ln x + C$ and $y = Ax$ .

9. The velocity

Answer: Separate to $-\ln|2 - v| = t + C$ , so $2 - v = Ae^{-t}$ . The condition $v(0) = 0$ gives $A = 2$ , so $v = 2 - 2e^{-t}$ . The velocity approaches the terminal value $2$ .

10. The population

Answer: Separate and integrate: $\ln P = 0.05t + C$ , so $P = Ae^{0.05t}$ . The condition $P(0) = 200$ gives $A = 200$ , so $P = 200e^{0.05t}$ .

11. The mass

Answer: Separate and integrate: $\ln m = -0.05t + C$ , so $m = Ae^{-0.05t}$ . The condition $m(0) = 200$ gives $A = 200$ , so $m = 200e^{-0.05t}$ .

12. The particular solution

Answer: Factorise to $\dfrac{dy}{dx} = x(y - 1)$ . The general solution is $y = 1 + Ae^{x^2/2}$ . Substituting $x = 0$ , $y = 3$ gives $A = 2$ , so $y = 1 + 2e^{x^2/2}$ .