## Solving 2nd order differential equation with non-constant coefficients

Hi~

I'm having trouble with solving a certain differential equation.

x2y'' + x y'+(k2x2-1)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).

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 :)
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 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

## Solving 2nd order differential equation with non-constant coefficients

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?

 Quote by river_boy 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?
Try substitution

u=(1+aSin(x))

 Quote by stallionx Try substitution u=(1+aSin(x))
Thanks its really helping.
 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 ^^

 Quote by Ceria_land Please yhelp 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.

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 Quote by JJacquelin (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)##.

 Quote by Mute I'm afraid that sin(x) ruins the integration, and as such you're missing a term ##-\int dx y(x) \cos(x)##. .
You are right. My mistake !
Damn ODE !
 A closed form for the solutions of y''=sin(x)*y involves the Mathieu's special functions. http://mathworld.wolfram.com/MathieuFunction.html Attached Thumbnails

 Quote by JJacquelin A closed form for the solutions of y''=sin(x)*y involves the Mathieu's special functions. http://mathworld.wolfram.com/MathieuFunction.html
Thanks to JJacquelin and Mute for helping!!!! it's really helpful :)

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 Quote by paul143 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?
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?

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