Answer to Question #299374 in Differential Equations for Pankaj

Question

Solve:

x².d²y/dx² -x.dy/dx +y=ln x

Answer 1

Let us use the transformation $x=e^t.$ Then 

$$y'_x=y'_te^{-t},\ y''_{x^2}=(y''_{t^2}-y'_t)e^{-2t}.$$

We get the equation for the function $y(t)$

$$e^{2t}(y''-y')e^{-2t} -e^ty'e^{-t} +y=t,$$

 which is equivalent to 

$$y''-2y'+y=t$$

Corresponding homogeneous differential equation

$$y''-2y'+y=0$$

Characteristic (auxiliary) equation

$$r^2-2r+1=0$$

$$r_1=r_2=1$$

The general solution of the homogeneous differential equation is

$$y_h=c_1e^t+c_2 te^t$$

Find the particular solution of the non homogeneous differential equation

$$y_p=At+B$$

$$y_p '=A$$

$$y''_p=0$$

Substitute

$$0-2A+At+B=t$$

$$A=1, B=2$$

The general solution of the non homogeneous differential equation is

$$y=c_1e^t+c_2 te^t+t+2$$

$$y(x)=c_1x+c_2x\ln x+\ln x+2$$