# ODE with variable coefficient

1. Nov 22, 2011

### iwasthere

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?

2. Nov 22, 2011

### Angry Citizen

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. Nov 23, 2011

### iwasthere

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.

Last edited: Nov 23, 2011
4. Nov 23, 2011

### Mute

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 $\exp(-c x)y(x)$ term will give a $Y(s+c)$ term, which means the Laplace-domain equation for the Laplace transform variable $Y(s)$ is a functional equation instead of an algebraic equation.

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:

$$\mathcal L G(x;\xi) = \delta(x-\xi),$$

where $\delta(x-\xi)$ is the dirac delta function and $\mathcal L$ is the differential operator,

$$\mathcal L = \frac{d^2}{dx^2} -A \frac{d}{dx} - B\exp(-ACx)$$

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

$$y(x) = M\int d\xi G(x;\xi)\exp(-N\xi),$$

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: Nov 23, 2011