Solving 2nd order differential equation with non-constant coefficients

In summary, the conversation discusses different methods for solving differential equations, specifically the equations y''=sin(x)*y and x2y'' + x y'+(k2x2-1)y = 0. The suggested methods include using the characteristic equation, Frobenius method, and a change of variables, as well as finding the Green's function for the differential equation. The conversation also mentions the Mathieu's special functions as a possible solution for y''=sin(x)*y.
  • #1
paul143
10
0
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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  • #2
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
 
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  • #3
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
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
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
river_boy said:
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))
 
  • #7
stallionx said:
Try substitution

u=(1+aSin(x))

Thanks its really helping.
 
  • #8
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
Ceria_land said:
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.
 
  • #10
JJacquelin said:
(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)##.
 
  • #11
Mute said:
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 !
 
  • #12

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  • #13
  • #14
paul143 said:
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?
 
  • #15
Ceria_land said:
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.
 
  • #16
HallsofIvy said:
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
 

1. How do I identify a 2nd order differential equation with non-constant coefficients?

A 2nd order differential equation with non-constant coefficients is one in which the coefficients of the second derivative and the first derivative are not constants. These coefficients can be functions of the independent variable or the dependent variable.

2. What is the general form of a 2nd order differential equation with non-constant coefficients?

The general form of a 2nd order differential equation with non-constant coefficients is y''(x) + p(x)y'(x) + q(x)y(x) = f(x), where p(x) and q(x) are functions of x and f(x) is a function of x or y(x).

3. How do I solve a 2nd order differential equation with non-constant coefficients?

To solve a 2nd order differential equation with non-constant coefficients, you can use the method of undetermined coefficients or the method of variation of parameters. In the method of undetermined coefficients, you assume a particular solution based on the form of f(x), and then use substitution to find the coefficients. In the method of variation of parameters, you assume a general solution of the form y(x) = u(x)y1(x) + v(x)y2(x), where y1(x) and y2(x) are the solutions of the corresponding homogeneous equation, and then use substitution to find u(x) and v(x).

4. What are the initial conditions for a 2nd order differential equation with non-constant coefficients?

The initial conditions for a 2nd order differential equation with non-constant coefficients are two values of the dependent variable y(x) and its first derivative y'(x) at a specific x value, typically denoted as x = x0. These conditions are needed to find the particular solution of the equation.

5. Can a 2nd order differential equation with non-constant coefficients have imaginary solutions?

Yes, a 2nd order differential equation with non-constant coefficients can have imaginary solutions. This can happen when the coefficients of the equation are complex numbers. In this case, the solutions will be of the form y(x) = e^(ax) (c1 cos(bx) + c2 sin(bx)), where a and b are the real and imaginary parts of the coefficient, and c1 and c2 are arbitrary constants.

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