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Solving 2nd order differential equation with non-constant coefficients

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paul143
#1
Sep2-09, 03:38 AM
P: 10
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 :)
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HallsofIvy
#2
Sep2-09, 08:44 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,304
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
#3
Sep2-09, 09:34 AM
P: 607
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
#4
Sep2-09, 09:41 AM
P: 10
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?
river_boy
#5
Sep8-11, 11:06 AM
P: 2
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
#6
Sep8-11, 11:50 AM
P: 100
Quote Quote by river_boy View Post
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
#7
Sep12-11, 11:22 AM
P: 2
Quote Quote by stallionx View Post
Try substitution

u=(1+aSin(x))
Thanks its really helping.
Ceria_land
#8
Nov27-12, 10:26 PM
P: 6
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
#9
Nov28-12, 09:24 AM
P: 756
Quote Quote by Ceria_land View Post
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
#10
Nov29-12, 08:17 PM
HW Helper
P: 1,391
Quote Quote by JJacquelin View Post
(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
#11
Nov30-12, 03:13 AM
P: 756
Quote Quote by Mute View Post
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
#12
Nov30-12, 04:13 AM
P: 756
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
Mathieu.JPG  
Ceria_land
#13
Nov30-12, 05:53 AM
P: 6
Quote Quote by JJacquelin View Post
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
#14
Nov30-12, 07:23 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,304
Quote Quote by paul143 View Post
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
#15
Nov30-12, 07:25 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,304
Quote Quote by Ceria_land View Post
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
#16
Nov30-12, 07:30 AM
P: 6
Quote Quote by HallsofIvy View Post
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


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