Answer to Question #145007 in Differential Equations for Tama

Question

Solve the equation by method of variation of parameters: x^(2)(d^(2)y)/(dx^(2))+x(dy)/(dx)-y=x^(2)e^(x)

Answer 1

x²y'' +xy' - y=x²e^x

homogeneous equation:
x2y'' +xy' - y=0
It can be noted that one of the solutions of the homogeneous equation is

y₁=x

Find the second independent solution
according to the Liouville-Ostrogradsky formula:

$W_{y_1y_2}=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} =C_1e^{-\int{\frac{a_1(x)}{a_2(x)}}}$

$y_2'y_1 - y_2y_1'=C_1e^{-\int{\frac{-x}{x^2}dx}}= C_1e^{lnx}=C_1x$

$\frac{y_2'y_1-y_2y_1'}{y_1^2}=\frac{C_1x}{y_1^2}=\frac{C_1x}{x^2}=\frac{C_1}{x}$

$(\frac{y_2}{y_1})'=\frac{C_1}{x}$

$\frac{y_2}{y_1}=C_1ln(x)+C_2$

y₂= xC₁ln(x)+xC₂

_{Common solution:}

y₀(x)= xC₁ln(x) +xC₂

system of equations:

$xC_1'ln(x) + xC_2'=0$

$C_1'(xln(x))' + C_2'(x)'=e^x$

C₂'= -C₁ln(x)

C₁'(ln(x)+1) + C₂'= e^x

C₁'(ln(x) +1) - C₁'ln(x) = e^x

C₁' =e^x

C₂'= - e^xln(x)=-1

C₁=e^x + A₁

C₂=-x+A₂

y(x)= C₁(x)xln(x) + C₂x= (e^x+A₁)xln(x) + (-x+A₂)x=x+A₁xln(x) - x² +xA₂