Math worksheet on separable equations for solving differential equations.

Here are the general solutions for each separable differential equation.

1) $\frac{dy}{dx} = \frac{3}{\tan(y)}$
Separate variables: $\tan(y) \, dy = 3 \, dx$
Integrate both sides: $\int \tan(y) \, dy = \int 3 \, dx$
Recall that $\int \tan(y) \, dy = -\ln|\cos(y)|$.
$-\ln|\cos(y)| = 3x + C_1$
$\ln|\cos(y)| = -3x - C_1$
$|\cos(y)| = e^{-3x - C_1} = e^{-C_1}e^{-3x}$
Let $C = \pm e^{-C_1}$.
Answer: $\cos(y) = Ce^{-3x}$

2) $\frac{dy}{dx} = \frac{2y}{x^2}$
Separate variables: $\frac{1}{y} \, dy = \frac{2}{x^2} \, dx$
Integrate both sides: $\ln|y| = \int 2x^{-2} \, dx$
$\ln|y| = 2(\frac{x^{-1}}{-1}) + C_1 = -\frac{2}{x} + C_1$
Exponentiate both sides: $|y| = e^{-\frac{2}{x} + C_1} = e^{C_1}e^{-\frac{2}{x}}$
Let $C = \pm e^{C_1}$.
Answer: $y = Ce^{-\frac{2}{x}}$

3) $\frac{dy}{dx} = \frac{3}{\sec^2(y)}$
Note that $\frac{1}{\sec^2(y)} = \cos^2(y)$. So, $\frac{dy}{dx} = 3\cos^2(y)$.
Separate variables: $\frac{1}{\cos^2(y)} \, dy = 3 \, dx \implies \sec^2(y) \, dy = 3 \, dx$
Integrate both sides: $\int \sec^2(y) \, dy = \int 3 \, dx$
$\tan(y) = 3x + C$
Answer: $\tan(y) = 3x + C$

4) $\frac{dy}{dx} = \frac{y}{8y}$
Simplify the right side: $\frac{dy}{dx} = \frac{1}{8}$ (assuming $y \neq 0$).
Integrate with respect to $x$: $y = \int \frac{1}{8} \, dx$
Answer: $y = \frac{1}{8}x + C$

5) $\frac{dy}{dx} = -3xy^2$
Separate variables: $\frac{1}{y^2} \, dy = -3x \, dx$
Integrate both sides: $\int y^{-2} \, dy = \int -3x \, dx$
$\frac{y^{-1}}{-1} = -\frac{3x^2}{2} + C_1$
$-\frac{1}{y} = -\frac{3x^2}{2} + C_1$
Multiply by $-1$: $\frac{1}{y} = \frac{3x^2}{2} - C_1$
Let $C = -C_1$. Invert both sides.
Answer: $y = \frac{1}{\frac{3}{2}x^2 + C}$

6) $\frac{dy}{dx} = 11e^{-x}$
This is already separated ($y$ terms are just $dy$).
Integrate both sides: $\int dy = \int 11e^{-x} \, dx$
$y = 11(-e^{-x}) + C$
Answer: $y = -11e^{-x} + C$

7) $\frac{dy}{dx} = \frac{5x}{y}$
Separate variables: $y \, dy = 5x \, dx$
Integrate both sides: $\int y \, dy = \int 5x \, dx$
$\frac{1}{2}y^2 = \frac{5}{2}x^2 + C_1$
Multiply by 2: $y^2 = 5x^2 + 2C_1$
Let $C = 2C_1$.
Answer: $y^2 = 5x^2 + C$

8) $\frac{dy}{dx} = x^2 y^{-1}$
Rewrite as $\frac{dy}{dx} = \frac{x^2}{y}$.
Separate variables: $y \, dy = x^2 \, dx$
Integrate both sides: $\int y \, dy = \int x^2 \, dx$
$\frac{1}{2}y^2 = \frac{1}{3}x^3 + C_1$
Multiply by 2: $y^2 = \frac{2}{3}x^3 + 2C_1$
Let $C = 2C_1$.
Answer: $y^2 = \frac{2}{3}x^3 + C$

9) $\frac{dy}{dx} = \frac{1}{\sin(x)}$
Rewrite as $\frac{dy}{dx} = \csc(x)$.
Integrate both sides: $y = \int \csc(x) \, dx$
The integral of $\csc(x)$ is a standard result: $-\ln|\csc(x) + \cot(x)|$ or $\ln|\csc(x) - \cot(x)|$ or $\ln|\tan(\frac{x}{2})|$. The most common textbook form is usually the first one.
Answer: $y = -\ln|\csc(x) + \cot(x)| + C$

10) $\frac{dy}{dx} = \frac{x^2}{2y}$
Separate variables: $2y \, dy = x^2 \, dx$
Integrate both sides: $\int 2y \, dy = \int x^2 \, dx$
$y^2 = \frac{1}{3}x^3 + C$
Answer: $y^2 = \frac{1}{3}x^3 + C$

11) $\frac{dy}{dx} = 9y(x - 3)$
Separate variables: $\frac{1}{y} \, dy = 9(x - 3) \, dx$
Integrate both sides: $\ln|y| = \int (9x - 27) \, dx$
$\ln|y| = \frac{9}{2}x^2 - 27x + C_1$
Exponentiate both sides: $|y| = e^{\frac{9}{2}x^2 - 27x + C_1}$
$y = \pm e^{C_1} e^{\frac{9}{2}x^2 - 27x}$
Let $C = \pm e^{C_1}$.
Answer: $y = Ce^{\frac{9}{2}x^2 - 27x}$

12) $\frac{dy}{dx} = xy^2$
Separate variables: $\frac{1}{y^2} \, dy = x \, dx$
Integrate both sides: $\int y^{-2} \, dy = \int x \, dx$
$-y^{-1} = \frac{1}{2}x^2 + C_1$
$-\frac{1}{y} = \frac{1}{2}x^2 + C_1$
Multiply by $-1$: $\frac{1}{y} = -\frac{1}{2}x^2 - C_1$
Let $C = -C_1$. Invert both sides.
Answer: $y = \frac{1}{C - \frac{1}{2}x^2}$