Hi~ I'm having trouble with solving a certain differential equation. x^{2}y'' + x y'+(k^{2}x^{2}-1)y = 0 I'm tasked to find a solution that satisfies the boundary conditions: y(0)=0 and y(1)=0 I have tried solving this using the characteristic equation, but i arrived at a solution that is unable to satisfy the boundary conditions except for when y(x)=0 (which is trivial). Any pointers on how i should go about this? Actually, i am trying to find the Green's function for this differential equation, which is why I need the solution to the said equation first. Thanks so much for any help :)
There is at least one obvious solution: y= 0 for all x. Since that is a regular singular equation at 0, you probably will want to use "Frobenius' method", using power series. That is much too complicated to explain here- try http://en.wikipedia.org/wiki/Frobenius_method
See the "Bessel" differential equation. Change variables to convert yours to that. So what you need to do is select k so that your solution has a zero at 1.
Thanks for the tip! Are there any possible way to solve this? Like possibly an out of this world change in variables (ex let u = lnx)? or series substitution?
Hi every body, I have another kind of equation which seems rather difficult to solve (1 + a Sin(x)) y'' + a Cos(x) y' + b Sin(x) y = 0 Should I first expand sin and cos in their series and then try Laplace or serious solution methods? Or there is a method that can solve it directly? Please help?
Please help me to solve this DE: y''=ysinx (I think i should multiply both sides with 2y' but I don't know how to do next) thanks in advance ^^
(y')² = y²sin(x)+C You can find the solutions in the particular case C=0 in terms of exponential of Incomplete elliptic integral of the second kind.
I'm afraid that sin(x) ruins the integration, and as such you're missing a term ##-\int dx y(x) \cos(x)##. After multiplying by 2y' you get $$\frac{d}{dx}(y')^2 = \left[\frac{d}{dx}(y^2)\right] \sin(x),$$ which can't be integrated exactly - you get ##(y')^2 = y^2\sin(x) + C - \int dx~y(x) \cos(x)##.
A closed form for the solutions of y''=sin(x)*y involves the Mathieu's special functions. http://mathworld.wolfram.com/MathieuFunction.html
You have already been given two methods, Frobenius and a change of variables that changes this to a Bessel equation, and a complete solution: y(x)= 0. What more do you want?
In future, however, please do not "hi-jack" someone else's thread to ask a new question. Use the "new thread" button to start your own thread.