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Solving 2nd order differential equation with nonconstant coefficients 
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#1
Sep209, 03:38 AM

P: 10

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 :) 


#2
Sep209, 08:44 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568

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 


#3
Sep209, 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.



#4
Sep209, 09:41 AM

P: 10

Solving 2nd order differential equation with nonconstant 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? 


#5
Sep811, 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? 


#6
Sep811, 11:50 AM

P: 100

u=(1+aSin(x)) 


#7
Sep1211, 11:22 AM

P: 2




#8
Nov2712, 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 ^^ 


#9
Nov2812, 09:24 AM

P: 759

You can find the solutions in the particular case C=0 in terms of exponential of Incomplete elliptic integral of the second kind. 


#10
Nov2912, 08:17 PM

HW Helper
P: 1,391

$$\frac{d}{dx}(y')^2 = \left[\frac{d}{dx}(y^2)\right] \sin(x),$$ which can't be integrated exactly  you get ##(y')^2 = y^2\sin(x) + C  \int dx~y(x) \cos(x)##. 


#11
Nov3012, 03:13 AM

P: 759

Damn ODE ! 


#12
Nov3012, 04:13 AM

P: 759

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


#13
Nov3012, 05:53 AM

P: 6




#14
Nov3012, 07:23 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568




#15
Nov3012, 07:25 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568




#16
Nov3012, 07:30 AM

P: 6




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