Answer to Question #312155 in Differential Equations for Haru

Question

Solve

x ^2y'′ + 2xy'− 12y = x^3log x.

Answer 1

$a_0\lambda(\lambda-1)(\lambda-2)...(\lambda-n+1)+...a_{n-2}\lambda(\lambda-1)+a_{n-1}\lambda+a_n=0$

x=e^u

$(\lambda-1)\lambda+2\lambda-12=0$

$\lambda^2+\lambda-12=0$

$\lambda_1=3$

$\lambda_2=-4$

$y''+y'-12y=ue^{3u}$

$y_0=Ce^{3u}+\frac{C_1}{e^{4u}}$

Solution for $ue^{3u}$

a+bi=3, then s=1

$y_1=u(Au+B)e^{3u}$

$y'_1=(3Au^2+(3B+2A)u+B)e^{3u}$

$y''_1=(9Au^2+(9B+12A)u+6B+2A)e^{3u}$

$14Aue^{3u}+(7B+2A)e^{3u}=ue^{3u}$

A=1/14

B=-1/49

$y_1=(\frac{u}{14}-\frac{1}{49})ue^{3u}$

$y=(\frac{u}{14}-\frac{1}{49})ue^{3u}+Ce^{3u}+\frac{C_1}{e^4u}$