Answer to Question #105854 in Differential Equations for Pappu Kumar Gupta

Question

Apply the method of variations of parameters to solve the following differential equations:
a) x^2y double prime+xyprime-y=x^2e^x

Answer 1

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

We will replace

$x=e^t,\\ t=ln|x|,\\ y'=\frac{dy}{dx}=\frac{y'_t}{x'_t}=\frac{y'_t}{e^t}=e^{-t}y'_t,\\ y''=\frac{dy'}{dx}=\frac{\frac{dy'}{dt}}{\frac{dx}{dt}}=\frac{y''_te^{-t}-e^{-t}y'_t}{e^t}=e^{-2t}(y''_t-y'_t)\\$

Input in equation

$e^{2t}e^{-2t}(y''_t-y'_t)+e^te^{-t}y'_t-y=e^{2t}e^{e^t}$

get the equation

$y''-y=e^{2t}e^{e^t}$

Consider homogenous equation

$y''-y=0\\ \lambda^2-1=0\implies\lambda_1=1, \lambda_2=-1\\ y_1=e^t, y_2=e^{-t}\\ y_0=c_1e^t+c_2e^{-t}$

Solution of equation find in form

$y=c_1(t)e^t+c_2(t)e^{-t}$

$c_1, c_2$ find from the system

$\left\{ \begin{matrix} c_1'(t)e^t+c_2'(t)e^{-t}=0 \\ c_1'(t)e^t-c_2'(t)e^{-t}=e^{2t}e^{e^t} \end{matrix} \right.\\ \left\{ \begin{matrix} c_1'(t)=\frac{1}{2}e^{t}e^{e^t}\\ c_2'(t)=-\frac{1}{2}e^{3t}e^{e^t} \end{matrix} \right.\\ c_1(t)=\int \frac{1}{2}e^{t}e^{e^t}dt=\frac{1}{2}e^{e^t}+k_1\\ c_2(t)=-\int \frac{1}{2}e^{3t}e^{e^t}dt=\\ =-\frac{1}{2}(e^{2t}e^{e^t}-2e^{t}e^{e^t}+2e^{e^t})+k_2\\ y=(\frac{1}{2}e^{e^t}+k_1)e^t+\\ +(-\frac{1}{2}(e^{2t}e^{e^t}-2e^{t}e^{e^t}+2e^{e^t})+k_2)e^{-t}$

Than

$y=(\frac{1}{2}e^x+k_1)x+(-\frac{1}{2}(x^2e^x-2xe^x+2e^x)+k_2)x^{-1}$