Question

how to solve this x²y''-xy'+y=2x by variation of parameter

Answer 1

This equation is euler-cauchy equation.

Substitute $y=x^{m}$ into the homogenous equation.

$$y'=mx^{m-1}$$

$$y''=m(m-1)x^{m-2}$$

$$x^2(m(m-1)x^{m-2})-x(mx^{m-1})+x^m=0$$

$$x^m(m^2-m-m+1)=0$$

$$x^m(m-1)^2=0$$

$$m=1$$

Hence the solution of the homogenous equation is

$$y=c_1x+c_2x\ln x$$

Use method of variations of parameters $c_1=c_1(x), c_2=c_2(x)$ 

$$y'=c_1+c_1'x+c_2\ln x+c_2+c_2'x\ln x$$

Suppose

$$c_1'x+c_2'x\ln x=0$$

Then

$$y'=c_1+c_2\ln x+c_2$$

$$y''=c_1'+c_2\dfrac{1}{x}+c_2'\ln x+c_2'$$

$$x^2c_1'+xc_2+x^2c_2'\ln x+x^2c_2'-xc_1-xc_2\ln x$$

$$-xc_2+c_1x+c_2x\ln x=2x$$

We have the system

$$c_1'x+c_2'x\ln x=0$$

$$x^2c_1'+x^2c_2'\ln x+x^2c_2'=2x$$

Then

$$c_1'=-c_2'\ln x$$

$$-x^2c_2'\ln x+x^2c_2'\ln x+x^2c_2'=2x$$

$$c_1'=-c_2'\ln x$$

$$c_2'x=2$$

$$c_2=\int\dfrac{2dx}{x}$$

$$c_1=-\int\dfrac{2\ln xdx}{x}$$

$$c_1=-\ln^2(x)+C_3$$

$$c_2=2\ln(x)+C_4$$

$$y=-x\ln^2(x)+C_3x+2x\ln^2(x)+C_4x\ln(x)$$

$$y=x\ln^2(x)+C_3x+C_4x\ln(x)$$