I'd like to solve $y''+\frac{2}{x}y'=\frac{1}{x^2}$
I identify this as Euler-Cauchy and rewrite to $x^2y''+2xy'=1$
So I have a homogenous solution of $y_h=C_1+\frac{C_2}{x}$
Second order Variation of parameters:
$C_1(x)'=\frac{1\frac{1}{x}}{\frac{1}{x^2}}=x \rightarrow C_1 = \frac{x^2}{2}$
$C_1(x)'=-\frac{1\cdot1}{\frac{1}{x^2}}=-x^2 \rightarrow C_2 = -\frac{x^3}{3}$
Is this procedure correct?
On
By using $x=e^z$ and $D_1=d/dz$, the equation becomes. $$(D_1(D_1-1)+2D_1)y=1$$ So the particular integral is $$\frac{1}{D_1(D_1+1)}1=\frac1{D_1}1=z=\log x$$ So the solution is $$y=C_1+\frac{C_2}{x}+\log x$$
On
$$y''+\frac{2}{x}y'=\frac{1}{x^2}$$ Substitute $z=y'$ $$x^2z'+2xz=1$$ Using variation of parameter
Solving the homogeneous equation $$x^2z'+2xz=0$$ $$(x^2z)'=0 \implies x^2z=K_1$$ $$z=\frac {K_1}{x^2}$$ $$x^2z'+2xz=1$$ $$x^2(-2K_1x^{-3})+K'_1+2x\frac {K_1}{x^2}=1$$ $$K'_1=1$$ $$K_1=x+C$$ $$\implies z=\frac {x+C}{x^2}$$ $$y'=\frac {x+C}{x^2}$$ Simply integrate $$y=\int \frac {x+C}{x^2}dx$$ $$y=\ln|x|+\frac {C_1}x+C_2$$
No. You can see that $C_1+\frac {C_2} x$ is not a solution to the DE for the values of $C_1,C_2$ you have computed.. What you have to do is compute first and second derivatives of $C_1+\frac {C_2} x$, plug them in to the DE. There will be some cancellations and you will get a first order DE in $C_1'$ and $C_2'$. Solving these and integrating you can find the solutions.