# Can this equation be solved with the method of undetermined coefficiants?

1. Aug 12, 2009

### horefaen

y''-2y'+y=6/(x^3)*e^x

the sollution to the homogenous equation is e^x and x*e^x, so how do I solve for the particual sollutioin, how do I choose the undetermined coefficients? I have tried:

(c + c1/x+c2/x^2+c3/x^4)*e^x and (cx^2+c1x+c2+c3/x), but none of them seems to give a good answer, should one of these work?

Maybe this should be in the homework section, but school hasn't started yet, so this isn't really homework. :P

2. Aug 12, 2009

### Mute

You might try including a logarithm in your guess. That could be the missing piece.

The problem with the method of undetermined coefficients is that really it's the "method of educated guessing". For certain inhomogeneous terms it's easy to guess the correct form of the particular solution and plug it in, but that won't always work.

There are more general methods, namely the "variation of parameters" method or "green's function" method.

I forget how the variation of parameters method works (it's been a while since I used it) - it's something like plugging u(x)y1(x) + v(x)y2(x) into the DE, and then demanding a certain term is zero: something like u'(x)y1(x) + v'(x)y2(x) = 0. This then takes this term out of the DE and reduces the number of undetermined functions by 1, so you can solve the DE. Try looking up how this actually works on wikipedia or something.

The green's function method works by, in this case, considering the equation

$$G''(x;\xi) - 2G'(x;\xi) + G(x;\xi) = \delta (x-\xi),$$

where the derivative is with respect to x and $\xi$ is a parameter introduced that will be integrated over: Solving this equation, the function y is then given as an integral over the Green's function G:

$$y(x) = \int_a^b d\xi G(x;\xi) \frac{6e^\xi}{\xi^3}$$.

The limits a and b should be taken as the endpoints of the range your problem is defined on. The green's function method is the one I've used most. I learned the variation of parameters back in second year of undergrad, but in practice the green's function method tends to be more widely used, as it can be generalized to higher dimensions, for example. If you haven't dealt with the delta function, $\delta(x)$, though, you may opt for the other method.

Anywho, it may turn out to be the case that the integrals you get in doing these methods may be expressed in closed form, in which case in principle you could have gotten the solution by undetermined coefficients, but the functions may be ones that you never would have been able to guess. Sometimes the integrals have to be left as integrals, in which case you certainly wouldn't have been able to guess them.

Last edited: Aug 12, 2009
3. Aug 12, 2009

### rubrix

Nope.

you can apply method of undetermined coefficients only when right hand side is of the form C*t^m*e^rt*cosbt*sinbt

your equation has 6/(x^3)*e^x ....e is in denominator so you can't use undetermined coefficients.

^ that's not a picture perfect definition but i hope you get the idea. If you don't, check your text, it should be in your text.

Variation of parameters should work though.

4. Aug 12, 2009

### horefaen

e^x is not in the denominator but in the numerator

5. Aug 12, 2009

### g_edgar

The thing that rules it out is x in the denominator.
The thing that makes this method work is that the RHS must be the solution of some other homogeneous linear DE with constant coefficients. So: C*t^m*e^(r*t)*cos(b*t)*sin(b*t)

6. Aug 13, 2009

### rubrix

ah okay but x is in the denominator so you still can't use method of undetermined coefficients.

7. Aug 14, 2009

### g_edgar

In this particular case, a particular solution is $$\frac{6e^x}{x}$$

8. Aug 14, 2009

### coomast

No, it isn't according to my calculations. I got:

$$\frac{3}{x}\cdot e^x$$

I checked by putting the complete solution

$$y=\left(K_1 \cdot x + K_2 +\frac{3}{x}\right) \cdot e^x$$

back into the original DE.

coomast