Answer to Question #105857 in Differential Equations for Pappu Kumar Gupta

Question

apply method of variation of parameters to solve D.E.
a)x^2y''+xy'-y=x^2e^x

Answer 1

This equation is euler-cauchy equation.

Substitute $y=x^{\lambda}$ into the homogenous equation

$x^2(x^\lambda)''+x(x^\lambda)'-x^\lambda=0$

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

$\lambda^2-1=0; \lambda=\pm1$

Hence the solution of the homogenous equation

$y=c_1x+c_2x^{-1}$

Using method of variations of parameters $c_1=c_1(x);c_2=c_2(x)$ will give us the system

$\begin{matrix} c_1'x+c_2'x^{-1}=0 \\ c_1'-c_2'x^{-2}=e^x \end{matrix}\implies \begin{matrix} c_1'=c_2'x^{-2}+e^x \\ xe^x+c_2'x*x^{-2}+c_2'x^{-1}=0 \end{matrix}$

$c_2'x^{-1}=-xe^x\implies c_2'=-\frac{1}{2}x^2e^x$

$c_2=-\frac{1}{2}e^x(x^2-2x+2)$

$c_1'=-\frac{1}{2}x^2e^x*x^{-2}+e^x=\frac{1}{2}e^x$

$c_1=\frac{1}{2}e^x$

Therefore

$y=c_1x+c_2x^{-1}+\frac{1}{2}xe^x-\frac{1}{2}x^{-1}e^x(x^2-2x+2)$