CodeCogs

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

View versions (2)

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 :

$\displaystyle \frac{dy}{dx} = F(x,y)$ where $F$ is a two variable function, also continuous.

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

Rational Functions

A rational function on $\mathhf{R}$ is a function $f:\mathhf{R}\to\mathhf{R}$ which can be expressed as $\displaystyle f(x)=\frac{P(x)}{Q(x)}$ where $P,Q$ are two polynomials.

Example 1

Problem

Solve: $\frac{d^2y}{dx^2} = 2 + \frac{1}{x}$

Workings

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

Solution

$y = x^2 + x\,ln\,x + Cx + K$

Trigonometric Functions

A rational function on $\mathhf{R}$ is a function $f:\mathhf{R}\to\mathhf{R}$ which can be expressed as a combination of trigonometric functions ( $sinx,cosx,tanx,cotanx$ ).

Example 1

Problem

$\frac{dy}{dx}=\frac{1}{cos^2 \tfrac{1}{2}x}$

Workings

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

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

Solution

$y=2\tan \tfrac{1}{2} x + C$

Physics Examples

Example 1

Problem

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

Workings

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

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

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

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

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 $A=-\frac{abV_1}{b-a}$