Question

Solve the ODE x^2y''-xy'+=ln(x)
I will rewrite the equation in the editor and show what I have done so far below.

Answer 1

\[x^{2}y''-xy'+y=\ln(x)\]
So I solved the homogeneous part as follows:
\[r(r-1)-r+1=0 \rightarrow r^2-2r+1=0 \rightarrow (r-1)^2=0 \rightarrow r _1=r_2=1\]
Thus \[y_H=C_1x+C_2xln(x)\]
So for the other part, I divided by x^2 to get that
\[R(x)=\ln(x)/x^2\]
Then found the Wronskian:
\[W(u_1,u_2)=W(x,xln(x))=\det(\left[\begin{matrix}x & xln(x) \\ 1 & \ln(x)+1\end{matrix}\right]=xln(x)+x-xln(x)=x \neq 0 \forall x \in (0, \infty)\]

Answer 2

So finally we get to the part I'm stuck on, my calculation of C_1
\[C_1=\int\limits_{}^{}\frac{ R(x)u_2 }{ W }dx= \int\limits_{}^{}\frac{ \frac{ \ln(x) }{ x^2 }xln(x) }{ x }dx=\int\limits_{}^{}\frac{ \ln(x)^2 }{ x^2}dx\]

Answer 3

@zepdrix could you help with ODE's?

Answer 4

oops forgot the negative above...heh heh:
\[C_1=\int\limits_{}^{}-\frac{ R(x)u_2 }{ W }dx=\int\limits_{}^{}-\frac{ \ln(x)^2 }{ x^2 }dx\]

Answer 5

@Directrix you've helped me before, any chance you could do so again?

Answer 6

@ganeshie8 it was suggested that I tag you by mimi in the chat box.

Answer 7

I'd suggest integration by parts, possibly? This is more of an integration problem than anything else at the moment. Lemme take a shot.

Answer 8

Choosing ln(x^2) as the thing that will be derived and choosing 1/x^2 as the thing that will be integrated will eventually rid you of the 1/x^2 altogether and give you another ln(x).

Answer 9

Aye, it more or less is. Integral should be set up correctly. Key being should be. Just cannot see the integration. Been working with it for about an hour now. Just cannot see it for some reason.

Answer 10

well it seems the way @Mendicant_Bias suggested works
if I didn't make a mistake

Answer 11

There is a small problem its (ln(x))^2 not ln(x^2)

Answer 12

and it it (ln(x))^2

Answer 13

it should work with (ln(x))^2 as the differentiating term and 1/x^2 as the integrating term

Answer 14

2 rounds of integration by parts is what I did

Answer 15

Let me try that again, maybe I screwed it up somewhere the first time.

Answer 16

yup just screwed up the integration. and well eliassaab, thanks but I think you did the case where the roots are imaginary not the case where they are identical. The correct answer is here http://www.wolframalpha.com/input/?i=dsolve%5Bx%5E2y%27%27-xy%27%2By%3D%3Dln%28x%29%5D

Answer 17

If I could medal both, I would but I'll medal the first person to mention the correct way of doing it.

Answer 18

I appreciate it. Did all the complex stuff just to screw up on the integral, lol. Thanks guys.

Answer 19

I did the + x y', sorry

Answer 20

No worries. I appreciate anyone who genuinely trys to help.