Homework Help: Differential Equations - complementary function

1. May 13, 2008

Gwilim

1. The problem statement, all variables and given/known data

y'' - 4y = -4e^x

a) find the complementary function
b) find the particular integral
c) write down the solution, given y(0) = 2 and y'(0) = 0

2. The attempt at a solution

Yesterday I learned how to do simpler differential equations of the form y'' + ay' + by = 0.
I Tried some others this morning and solved them quite happily without using any reference material.

Here is what I have so far for this problem, I'm quite sure I'm going about it the wrong way:

y(x)=e^rx
y''(x)=r^2*e^rx
r^2*e^rx - 4e^rx = -4e^x

This seemed to be the the most straightforward way to determine r. Next I factorised the left hand side and divided through by e^x as follows:

(r^2 - 4)e^rx = -4e^x
(r^2 - 4)e^r = -4

I am unsure where to go from there. Would I take the natural logarithm? It's been a long time since I've used logs so if that is the correct way to find r I'd be grateful if somebody took me through it. I'm still not quite clear on what exactly is meant by the particular integral, if someone could show me this too it would be very helpful. Thanks in advance.

2. May 13, 2008

Defennder

First do you know what "complementary function" and "particular integral" means? Because I don't.

3. May 13, 2008

Gwilim

Wikipedia has this to say:

4. May 13, 2008

Defennder

Oh I see. Well for such problems you have to find the homogenous solution first. Solve for the case where the RHS=0.

For the particular integral, there are a number of ways to find it. The method which works almost every time is variation of parameters. Have you learnt that yet?

Last edited: May 13, 2008
5. May 13, 2008

dx

Gwilim, any linear combination of derivatives of $$e^{rx}$$ will be of the form $$Ae^{rx}$$. But the exponent of $$e$$ on the right side is just $$x$$, not $$rx$$. Can you see now what function you should try for the particular integral?

6. May 13, 2008

Gwilim

Okay, so r^2 - 4 = 0 => r = 2 or -2, so the complementary function is e^2x + e^(-2x)

I don't know the variation of parameters method, can you provide a link to a page explaining this or give a brief explanation yourself?

The exponent of e in right side of all the equations in the (incorrect) workings I gave was x, rather than rx. Have I misunderstood your reply?

7. May 13, 2008

dx

You included the right hand side in your calculation, so I assume youre trying to find the particular integral. In that case, yes you misunderstood my reply. I was saying that the function you tried for the particular integral doesnt have the correct form.

If you were trying to find the complementary function, then you shouldnt have the right hand side in the equation. The complementary function is a solution to the equation y'' - 4y = 0.

8. May 13, 2008

Defennder

Try this:
http://www.sosmath.com/diffeq/second/variation/variation.html

You don't have to use the method of variation of parameters if the RHS is an expression of the form which can be 'guessed' using the method of undetermined coefficients.
http://www.sosmath.com/diffeq/second/guessing/guessing.html

9. May 13, 2008

lurflurf

10. May 14, 2008

tiny-tim

Hi Gwilim!

The particular solution usually has no connection whatever with the complementary solution.

I was never much good at particular solutions, but I vaguely recall that one just makes intelligent guesses … in this case (only), it seems obvious to me that the p.s. must be a polynomial times the RHS … which of course is the way lurflurf has shown you.

Generally, in future problems, get the c.s first … then forget it and start again … make an intelligent guess for the p.s., and keep plugging away.