ODE with variable coefficient

  • Thread starter iwasthere
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  • #1
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i want to solve the following differential equation:

y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = M*exp(-N*x)
A,B,C,M,N are constants.

-is there any solution of the above equation (except series solution)?
-is there any proper substitution that can turn the variable coefficient into constant?
-can i use method of undetermined coefficient to obtain particular integral?
 

Answers and Replies

  • #2
I'd try using laplace transforms to see if it has any analytical non-series solution. No idea if you can use undetermined coefficients, but I can't imagine why you'd want to. A transform looks much simpler, especially with the e term in there.
 
  • #3
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thanks for the reply.......

for the homogeneous equation:
y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = 0

if the substitution z =exp(-C*A*x) is used,it becomes something like the following one( if i am not wrong):

z*y''(z) + (1 + 1/C)*y'(z) - B/(C*A)^2*y(z) = 0

which has a solution containing bessel function. But how can i obtain the particular integral (without containing any integral form) for :

y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = M*exp(-N*x)

i can use method of variation of parameter for particular integral, but the wronskian becomes complicated (i have to leave the solution in integral form).
therefore i was thinking about method of undetermined coefficient to obtain the particular integral.

i need the general solution without keeping any integral form in the solution.
please help me...............
many many thanks in advance...................................
 
Last edited:
  • #4
Mute
Homework Helper
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I'd try using laplace transforms to see if it has any analytical non-series solution. No idea if you can use undetermined coefficients, but I can't imagine why you'd want to. A transform looks much simpler, especially with the e term in there.
I don't think that will work well. I don't remember Laplace transforms being very good for ODE's with variable coefficients. In particular, the [itex]\exp(-c x)y(x)[/itex] term will give a [itex]Y(s+c)[/itex] term, which means the Laplace-domain equation for the Laplace transform variable [itex]Y(s)[/itex] is a functional equation instead of an algebraic equation.

thanks for the reply.......

for the homogeneous equation:
y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = 0

if the substitution z =exp(-C*A*x) is used,it becomes something like the following one( if i am not wrong):

z*y''(z) + (1 + 1/C)*y'(z) - B/(C*A)^2*y(z) = 0

which has a solution containing bessel function. But how can i obtain the particular integral (without containing any integral form) for :

y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = M*exp(-N*x)

i can use method of variation of parameter for particular integral, but the wronskian becomes complicated (i have to leave the solution in integral form).
therefore i was thinking about method of undetermined coefficient to obtain the particular integral.

i need the general solution without keeping any integral form in the solution.
please help me...............
many many thanks in advance...................................
You may have to leave the solution in integral form. There's not necessarily a nice, closed form solution. Do you know what a Green's function is? Consider a modified form of your equation:

[tex]\mathcal L G(x;\xi) = \delta(x-\xi),[/tex]

where [itex]\delta(x-\xi)[/itex] is the dirac delta function and [itex]\mathcal L[/itex] is the differential operator,

[tex]\mathcal L = \frac{d^2}{dx^2} -A \frac{d}{dx} - B\exp(-ACx)[/tex]

If you solve that equation for [itex]G(x;\xi)[/itex], then to the solution for your equation with the [itex]M\exp(-Nx)[/itex] is just

[tex]y(x) = M\int d\xi G(x;\xi)\exp(-N\xi),[/tex]

where the integral is taken over your range of x. In a one-dimensional problem like yours, this is very similar to the variations of parameters method, I think, but this one generalizes to higher dimensions. Your solution may still be in terms of an integral, and there may be no way around that.
 
Last edited:

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