Answer to Question #312156 in Differential Equations for haru

Question

Solve

x^2y′′+ xy′− y =1/x + 1

Answer 1

$a_0x^ny^n+a_1x^{n-1}y^n+...+a_{n-1}xy'+a^ny=f(x)$

$x=e^u$

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

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

$\lambda^2-1=0$

$y''-y=\frac{1}{e^u}+1$

$\lambda_1=1$

$\lambda_2=-1$

$y=Ce^u+\frac{C_1}{e^u}$

$y_i=u^se^{au}(R_m(u)cos(bu)+T_m(u)sin(bu))$

s=0 if a+bi is not a root or s=k if it is.

Solution for 1:

a+bi=0, then s=0

$y_0=A$

$y''_0=0$

A=-1

$y_0=-1$

Solution for $\frac{1}{e^u}$

a+bi=-1

s=1

$y_1=\frac{Au}{e^u}$

$y''_1=\frac{Au-2A}{e^u}$

$\frac{-2A}{e^u}=\frac{1}{e^u}$

A=-0.5

$y_1=-\frac{u}{2e^u}$

$y=Ce^u-\frac{u}{2e^u}+\frac{C_1}{e^u}-1$

u=ln(x)

$y=\frac{-ln(x)}{2x}+Cx+\frac{C_1}{x}-1$