Solving 2nd order differential equation with non-constant coefficients


by paul143
Tags: coefficients, differential, equation, nonconstant, order, solving
paul143
paul143 is offline
#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
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#2
Sep2-09, 08:44 AM
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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
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#3
Sep2-09, 09:34 AM
P: 608
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
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#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
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#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
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#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
river_boy is offline
#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
Ceria_land is offline
#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
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#9
Nov28-12, 09:24 AM
P: 745
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
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#10
Nov29-12, 08:17 PM
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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
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#11
Nov30-12, 03:13 AM
P: 745
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
JJacquelin is offline
#12
Nov30-12, 04:13 AM
P: 745
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
Ceria_land is offline
#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
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#14
Nov30-12, 07:23 AM
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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
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#15
Nov30-12, 07:25 AM
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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
Ceria_land is offline
#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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