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.