- #1

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

Not even sure where to start on this one???

Is this a series solution problem?

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- Thread starter BobMarly
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- #1

- 19

- 0

Not even sure where to start on this one???

Is this a series solution problem?

Last edited:

- #2

- 64

- 1

Try a solution of the form x^n

- #3

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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?

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- #4

- 64

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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.

- #5

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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?

- #6

- 64

- 1

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)]

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