Solution of Linear Differential Equation

Here you will learn how to find solution of linear differential equation of first order first degree with examples.

Let’s begin –

Solution of Linear Differential Equation

(1) Linear Differential Equation of the form $dy\over dx$ + Py = Q

A differential equation is linear if the dependent variable (y) and its derivative appear only in first degree.

The general form of a linear differential equation is 

$dy\over dx$  + Py = Q

where P and Q are functions of x (or constants)

This type of differential equations are solved when they are multiplied a factor, which is called integrating factor, because by multiplication of this factor the left hand side of the differential equation (i) becomes exact differential of some function.

Algorithm :

1). Write the differential equation in the form $dy\over dx$  + Py = Q and obtain P and Q

2). find the integrating factor (I. f.) given by I.f = $e^{\int Pdx}$

3). Multiply both sides of equation in step 1 by I.f.

4). Integrate both sides of the equation obtained in step 3 with respect to x to obtain 

y(I.f) = $\int$ Q(I.f) dx + C, which gives the required solution.

Example : Solve the differential equation : $dy\over dx$ – $y\over x$ = $2x^2$

Solution : We are given that,

$dy\over dx$ – $y\over x$ = $2x^2$

Clearly it is a differential equation of the form

$dy\over dx$ + Py = Q , where P = $-1\over x$ and Q = $2x^2$

Now, I.f = $e^{\int Pdx}$ = $e^{\int (-1/x)dx}$ = $e^{-log x}$ = $1\over x$

By algorithm, the solution is

y$1\over x$ = $\int$ 2x dx + C

$\implies$ $y\over x$ = $x^2$ + C

$\implies$ y = $x^3$ + Cx, which is the required solution.

(2) Linear Differential Equation of the form $dx\over dy$ + Rx = S

Sometimes a linear differential equation can be put in the form $dx\over dy$ + Rx = S where R and S are functions of y or constants

Note that here y is independent variable and x  is a dependent variable.

Algorithm :

1). Write the differential equation in the form $dx\over dy$ + Rx = S and obtain R and S

2). find the integrating factor (I. f.) given by I.f = $e^{\int Rdy}$

3). Multiply both sides of equation in step 1 by I.f.

4). Integrate both sides of the equation obtained in step 3 with respect to x to obtain 

x(I.f) = $\int$ S(I.f) dy + C, which gives the required solution.

Example : Solve the differential equation : ydx + $x – y^3$ dy = 0

Solution : We are given that,

ydx + $x – y^3$ dy = 0

$\implies$ $dy\over dx$ + $x\over y$ = $y^2$

Clearly it is a differential equation of the form

$dx\over dy$ + Ry = S , where R = $1\over y$ and S = $y^2$

Now, I.f = $e^{\int Rdy}$ = $e^{\int (1/y)dy}$ = $e^{log y}$ = y

By algorithm, the solution is

xy = $\int$ $y^3$ dy + C

$\implies$ xy = $y^4\over 4$ + C, which is the required solution.