Please tell me how to solve this differential equation.
$$ x^2 y'' – 2 x y' + 2y = x^4 \mathrm{e}^x $$
I tried to solve it but finally I stuck tell me how to go further more if anyone gives me a hint(I want to know how to find paticular integral?).I will appreciate about it.i posted my way below please tell me any hint. Here's my work:
Best Answer
First Solve the homogeneous part: $$x^2y''-2xy'+2y=0\,.~~~~(1)$$ To solve this Euler's equation let $y=x^m$, then $m(m-1)-2m+2=0 \implies m^2-3m+2=0 m=1,2.$ So $y_{1,2}=x, x^2$ are two linearly independent solutions of (1). Next the in-homogeneous ODE $$Y''-(2/x) Y'+(2/x^2)Y=x^2e^{x}f(x) ~~~~~(2)$$ can be solved by the method of variation of parameters $C_1,C_2$ where $$C_1=-\int \frac{y_2(x) f(x)}{W(x)} dx+D_1, ~~C_2=\int \frac{y_1(x) f(x)}{W(x)}+D_2\,.$$ Here, $W(x)=y_1y'_2-y'_1 y_2=-x^{2}$ and its solution is $$Y(x)=C_1(x) y_1(x)+C_2(x) y_2(x)\,,~~~(3)$$ $$C_1=-\int x^2 e^{x} dx=e^{x}(-x^2+2x-2)\,.$$ Similarly, $$C_2=\int x e^{x} dx=e^{x}(x-1)\,.$$ Inserting $C_1,C_2,y_1,y_2$ in (3), the total solution of (2)
is $$Y(x)=D_1 x+ D_2 x^2+ e^x(x^2-2x)\,.$$