Find Solution 4*(x^2)*y"+y=0, y(-1)=2, y'(-1)=0

BobMarly

Not even sure where to start on this one???
Is this a series solution problem?

Marioeden

Try a solution of the form x^n

BobMarly

x^n?, does that mean y=a(n)x^n, leading to y'=n*a(n)*x^(n-1), y"=n*(n-1)*a(n)*x(n-2), then substitute?

Marioeden

Well, you can throw in an a(n) if you want but once you do your substitution you'll see that it cancels out. That said, the a(n) is important as it is a constant that will be determined once you use your boundary conditions.

Basically, set y = a(n)x^n and find which values of n will make 4*(x^2)*y'' + y equal zero.

BobMarly

I end up with 4*x^2*(sum(n=0))(n+2)*(n+1)*a(n-2)*(x^n)+(sum(n=0))a(n)*(x^n)
I'm working toward the recurance relation, correct?
What about the 4*x^2 for the first summation?

Marioeden

You're not working towards a recurrence relation, let's see if we can get away with something easier!

Literally just write y = x^n
Then 4*(x^2)*y'' + y = 4n(n-1)x^n + x^n = [4n^2 - 4n + 1]x^n = ([2n - 1]^2)x^n

So we know that n=1/2 will give us a working solution i.e. y = A*x^(1/2)

Unfortunately, we need two degrees of freedom and we just have one. Luckily, we know how to deal with resonance of this form as we know that y = B*x^(1/2)*ln(x) will be a solution.

Thus, our general solution is y = x^(1/2)*[A + Bln(x)]