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

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Discussion Overview

The discussion revolves around finding the solution to the differential equation 4*(x^2)*y'' + y = 0 with initial conditions y(-1) = 2 and y'(-1) = 0. Participants explore various methods for solving this equation, including series solutions and specific forms of y.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to start solving the differential equation and questions whether it is a series solution problem.
  • Another suggests trying a solution of the form x^n.
  • A participant elaborates on the implications of using the form y = a(n)x^n, discussing how to derive y' and y'' from this assumption.
  • It is noted that the constant a(n) will cancel out during substitution but is still important for applying boundary conditions.
  • One participant attempts to derive a recurrence relation but is corrected by another who suggests a simpler approach by directly writing y = x^n.
  • A calculation is presented showing that n = 1/2 provides a working solution, leading to the conclusion that y = A*x^(1/2) is a solution.
  • Another participant introduces the idea of resonance, suggesting that y = B*x^(1/2)*ln(x) will also be a solution, leading to a general solution of y = x^(1/2)*[A + Bln(x)].

Areas of Agreement / Disagreement

Participants explore different approaches to solving the differential equation, with some agreeing on the form of the solution while others propose alternative methods. No consensus is reached on a single solution method, and the discussion remains open to various interpretations.

Contextual Notes

The discussion involves assumptions about the form of the solution and the implications of boundary conditions, which may not be fully resolved. The approach to resonance and the specific values of constants remain dependent on further analysis.

BobMarly
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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?
 
Last edited:
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Try a solution of the form x^n
 


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?
 
Last edited:


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.
 


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?
 


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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