Can, y=sin(x^2) be a solution to,

y''+py'+qy=0

With initial conditions, y(0)=A and y'(0)=B

Where p and q are continous in an open interval containing 0?

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- September 30th 2006, 08:37 PMThePerfectHackerNon-existenant solution to differencial equation
Can, y=sin(x^2) be a solution to,

y''+py'+qy=0

With initial conditions, y(0)=A and y'(0)=B

Where p and q are continous in an open interval containing 0? - October 1st 2006, 08:40 AMThePerfectHacker
I solved it, thank you anyways.

If you are interested how it was solved:

The equation,

y''+py'+qy=0

Has basis of dimension 2 if p,q are continous in an neigborhood of x=0.

Thus, there exists two linearly independent solutions that span the set. One is given as sin(x^2). Let the other linearly independent solution be y.

Then we compute the Wronskian,

[sin(x^2) y]

[2xcos(x^2) y']

Thus,

y'sin(x^2)-2xycos(x^2)

But at x=0 the Wronskian is zero.

Thus, there does not exist a linearly independant solution with sin(x^2), a contradiction.