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# Separable

This section contains worked examples of the type of differential equation which can be solved by integration

## Separable Differential Equations

This section contains worked examples of the type of differential equation which can be solved by direct Integration.

### Definition

Separable Differential Equations are differential equations which respect one of the following forms :
• $\inline&space;\displaystyle&space;\frac{dy}{dx}&space;=&space;F(x,y)$ where $\inline&space;F$ is a two variable function, also continuous.

• $\inline&space;\displaystyle&space;f(y)dy=g(x)dx$, where $\inline&space;f$ and $\inline&space;g$ are two real continuous functions.

### Rational Functions

A rational function on $\inline&space;\mathhf{R}$ is a function $\inline&space;f:\mathhf{R}\to\mathhf{R}$ which can be expressed as $\inline&space;\displaystyle&space;f(x)=\frac{P(x)}{Q(x)}$ where $\inline&space;P,Q$ are two polynomials.
Example:
##### Example - Simple Differential Equation
Problem
Solve:
$\frac{d^2y}{dx^2}&space;=&space;2&space;+&space;\frac{1}{x}$

Workings
As the equation is of first order, integrate the function twice, i.e.
$\frac{dy}{dx}&space;=&space;2x&space;+&space;ln\,x&space;+&space;C$
and
$y&space;=&space;x^2&space;+&space;x\,ln\,x&space;+&space;Cx&space;+&space;K$
Solution
$y&space;=&space;x^2&space;+&space;x\,ln\,x&space;+&space;Cx&space;+&space;K$

### Trigonometric Functions

A rational function on $\inline&space;\mathhf{R}$ is a function $\inline&space;f:\mathhf{R}\to\mathhf{R}$ which can be expressed as a combination of trigonometric functions ($\inline&space;sinx,cosx,tanx,cotanx$).
Example:
##### Example - Simple Cosine
Problem
$\frac{dy}{dx}=\frac{1}{cos^2&space;\tfrac{1}{2}x}$

Workings
This is the same as
$y=\int&space;\frac{1}{\cos^2&space;\tfrac{1}{2}x}dx$

which we integrate in the normal way to yield
$y=2\tan&space;\tfrac{1}{2}&space;x&space;+&space;C$
Solution
$y=2\tan&space;\tfrac{1}{2}&space;x&space;+&space;C$

### Physics Examples

Example:
##### Example - Potential example
Problem
If a and b are the radii of concentric spherical conductors at potentials of $\inline&space;V_1$ respectively, then V is the potential at a distance r from the centre. Find the value of V if:
$\frac{d}{dr}\left(r^2&space;\frac{dV}{dr}&space;\right)=0}$
and $\inline&space;V=V_1$ at r=a and $\inline&space;V=0$ at r=b

Workings
$\frac{d}{dr}\left(r^2\frac{dV}{dr}&space;\right)=0$

$\therefore\;\;\;\;\;\;r^2\frac{dV}{dr}&space;\right)=A$

$\frac{dV}{A}=\frac{dr}{r^2}$

$\therefore\;\;\;\;\;\;\frac{V}{A}=-\frac{1}{r}\;+\;B$

Substituting in the given values for V and r
$\frac{V_1}{A}=-\frac{1}{a}+B$
and
$\frac{0}{A}=-\frac{1}{b}+B$
$\therefore\;\;\;\;\;\;B=\frac{1}{b}$
$\therefore\;\;\;\;\;\;\frac{V_1}{A}=-\frac{1}{a}+\frac{1}{b}=-\frac{b-a}{ab}$
Thus
$A=-\frac{abV_1}{b-a}$
Solution
$V=\left(\frac{1}{r}-\frac{1}{b}&space;\right)\left(\frac{abV_1}{b-a}&space;\right)$