Answer to Question #248514 in Differential Equations for nishant

Question

Solve the Cauchys Linear Differential Equations :

x^2.d^2y/dx^2 + x.dy/dx - y = x^3.e^4

Answer 1

The homogeneous differential equation

$$x^2y''+xy'-y=0$$

Substitute $y=x^\lambda$

$$x^2\lambda(\lambda-1)x^{\lambda-2}+x\lambda x^{\lambda-1}-x^\lambda=0$$

$$\lambda^2-1=0$$

$$\lambda_1=1, \lambda_2=-1$$

The general solution to the homogeneous differential equation is

$$y_h=c_1x+\dfrac{c_2}{x}$$

Find the particular solution to the non homogeneous differential equation

$$y_p=Ae^4x^3$$

$$y_p'=3Ae^4x^2$$

$$y_p''=6Ae^4x$$

Substitute

$$x^2(6Ae^4x)+x(3Ae^4x^2)-Ae^4x^3=e^4x^3$$

$$A=\dfrac{1}{8}$$

The particularsolution to the non homogeneous differential equation is

$$y_p=\dfrac{1}{8}e^4x^3$$

The general solution to the homogeneous differential equation is

$$y=y_h+y_p$$

$$y(x)=c_1x+\dfrac{c_2}{x}+\dfrac{1}{8}e^4x^3$$