Question

Solve
y''+(1/x)y'−(4/x^2)y = g(x),

y(0) bounded, y(1) = 0 for g(x) = x.
Bonus: solve the equation for g(x) = 1

Answer 1

Solve for what?

Answer 2

for g(x) = x

Answer 3

We're trying to find y, in this case

Answer 4

That's not a very easy diffeq :/

Answer 5

Lets try this:
\[y_p(x)=\alpha x^3\]
\[\implies y'_p(x)=3 \alpha x^2; y''_p(x)=6 \alpha x \implies 6\alpha x+3 \alpha x-4 \alpha x=\]
\[x \implies 5 \alpha = 1 \implies \alpha = \frac{1}{5}\]
So one solution is:
\[\frac{1}{5}x^3\]
From inspection but there may be others...

Answer 6

No it's not! If it was I wouldn't be posting it here :P
I have the solution but I don't understand it completely...

Answer 7

What about the general solution? How would you know if there are other particular solutions?

Answer 8

Well my particular solution is correct. For the general solution I'm not honestly sure...Because you have polynomial coefficients. I mean I can get an answer:
http://www.wolframalpha.com/input/?i=solve+y%27%27%2B%281%2Fx%29y%27-%284%2Fx^2%29y%3Dx
But I'm not sure how you'd get there

Answer 9

Maybe the solution will help?

Answer 10

Ahhhh, assume a solution of the form:
\[x^{\lambda}\]
Then solve for what lambda is to get +-2 which gives you an x^2 and a x^(-2) solution!

Answer 11

Hmm... I understand that, but how could I figure something like that out for a different kind of problem?

Answer 12

I'm also not entirely sure how they got that c2 equals 0 for the particular solution... would you be able to explain that step?

Answer 13

You wouldn't. I got the particular solution by inspection and seeing that if I chose that as x^3 then the powers on the left all worked out to give me an "x" at the end. So then the constant needed to be determined which is my alpha. But in general these problems can't be solved.

No problem: If y(0) is bounded then it can't be x^{-2} right? Because x^{-2} blows up at x=0 :P So the constant in front of it has to be zero to avoid it blowing up (i.e., the term isn't even there).

Answer 14

Ohh, totally didn't think about that... thanks!
You made the problem a little less intimidating.