Solving 2nd order differential equation with non-constant coefficients

[paul143]

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).

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 :)

 

[HallsofIvy]

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

 

[g_edgar]

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.

 

[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?

 

[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?

 

[stallionx]

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))

 

[river_boy]

Try substitution

u=(1+aSin(x))

Thanks its really helping.

 

[Ceria_land]

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 ^^

 

[JJacquelin]

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.

 

[Mute]

(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 $(y')^2 = y^2\sin(x) + C - \int dx~y(x) \cos(x)$.

 

[JJacquelin]

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 !

 

[JJacquelin]

A closed form for the solutions of y''=sin(x)*y involves the Mathieu's special functions.
http://mathworld.wolfram.com/MathieuFunction.html

 

[Ceria_land]

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 :)

 

[HallsofIvy]

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?

 

[HallsofIvy]

Thanks to JJacquelin and Mute for helping! it's really helpful :)

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.

 

[Ceria_land]

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.

Thanks for your warning. I won't do it again :D