# Separable Differential Equations

What are **separable** differential equations and how to solve them?

\( \)\( \)\( \)
This is a tutorial on solving **separable** differential equations of the form
\[ \dfrac{dy}{dx} = \dfrac{f(x)}{g(y)} \]
Examples with detailed solutions are presented and a set of exercises is presented after the tutorials. Depending on \( f(x) \) and
\( g(y) \), these equations may be solved analytically.

## Examples with Solutions

### Example 1:

Solve and find a general solution to the differential equation.

\[ \dfrac{dy}{dx} = 3e^{y}x^{2} \]

__Solution to Example 1:__

We first rewrite the given equations in differential form and with variables **separated**, the y's on one side and the x's on the other side as follows.

\[ e^{-y} dy = 3x^{2} dx \]

Integrate both side.

\[ \int e^{-y} dy = \int 3x^{2} dx \]

which gives

\[ -e^{-y} + C_{1} = x^{3} + C_{2} \], \( C_{1} \) and \( C_{2} \) are constant of integration.

We now solve the above equation for y

\[ -e^{-y} = x^{3} + C_{2} - C_{1} \]
\[ e^{-y} = - x^{3} - C_{2} + C_{1} \]
\[ - y = \ln(- x^{3} - C_{2} + C_{1}) \]
\[ y = -\ln(-x^{3} - C) \], where \( C = C_{2} - C_{1} \).

As practice, verify that the solution \( y = -\ln(-x^{3} - C) \) obtained satisfy the differential equation given above.

### Example 2:

Solve and find a general solution to the differential equation.

\[ \dfrac{dy}{dx} = \dfrac{\sin x}{y\cos y} \]

__Solution to Example 2:__

**Separate** variables and write in differential form.

\[ y \; \cos y \;dy = \sin x \;dx \]

Integrate both sides

\[ \int y\cos y dy = \int \sin x dx \]

The left side may be integrated by parts

\[ y\; \sin y - \int \sin y \; dy = -\cos x \]

\[ y\sin y + \cos y + C_{1} = -\cos x + C_{2} \], \( C_{1} \) and \( C_{2} \) are constants of integration.

In this case there is no simple formula for \( y \) as a function of \( x \).

\[ y = \dfrac{-\cos x - \cos y + C}{\sin y} \], where \( C = C_{2} - C_{1} \)

## Exercises:

Solve the following separable differential equations.

a) \( \dfrac{dy}{dx} = -9x^{2}y^{2} \)

b) \( \dfrac{dy}{dx} = -2xe^{y} \)

__Solutions to the above exercises__

a) \( y = \dfrac{1}{3x^{3} + C} \)

b) \( y = -\ln(x^{2} + C) \)

## More references and Links

Differential Equations

Differential Equations - Runge Kutta Method