# Diff. Eq. Problem -> y'' - 2y' - 3y = -3 t e^(-t) What is general solution?

1. May 26, 2005

### VinnyCee

Diff. Eq. Problem ---> y'' - 2y' - 3y = -3 t e^(-t) What is general solution?

This is problem number 3 in section 3.6 in Boyce, DiPrima 8th Edition "Elementary Differential Equations and Boundry Value Problems"

Find the general solution of the differential equation:

$$y'' - 2y' - 3y = -3 t e^{-t}$$

Here is what I have done so far:

The characteristic equation is $$r^2 - 2r - 3 = 0$$ and $$r_1 = 3, r_2 = -1$$

That gives $$y(t) = C_1 e^{3t} + C_2 e^{-t}$$ as the complementary solution.

Now I run into trouble when selecting the superposition equation for using the method of undetermined coefficients. I tried $$Y(t) = (A t + B) e^{-t}$$ with no luck. I get $$A = \frac{3}{4} t$$ for an answer, which is incorrect.

The book lists the answer as $$y(t) = C_1 e^{3t} + C_2 e^{-t} + \frac{3}{16} t e^{-t} + \frac{3}{8} t^2 e^{-t}$$

What am I doing wrong? Is the $$Y(t)$$ I chose correct? Please help!

Thank you.

2. May 26, 2005

### arildno

Try a particular solution $$Y(t)=(At^{2}+Bt)e^{-t}$$
and determine A, B.

3. May 26, 2005

### VinnyCee

Thanks

I tried it and it sounds more reasonable. Why must this be the particular solution instead of the one I origianlly chose?

I'm still having trouble. I have the Y, Y', Y'' all substituted and I get $$-6A - 4B = -3t$$ How do I solve this for A and B? There must be a second equation in there somewhere right?

4. May 26, 2005

### HallsofIvy

Staff Emeritus
The reason (At+B)e-t didn't work is that e-t is already a solution to the homogeneous differential equation. You should have learned that, in situations like that, you multiply by t (or higher powers of t is te-t, etc. are also solutions). Here you should try, as Arildno said, t(At+ B)e-t= (At2+ Bt)e-t.
Now you say you got -6A- 4B= -3t. How?

If y= (At2+ Bt)e-t, then y'= -(At2+ Bt)e-t+ (2At+ B)e-t and y"= (At2+ Bt)e-t- 2(2At+B)e-t+ 2Ae-t

y"= (At2+ Bt)e-t- 2(2At+B)e-t+ 2Ae-t
-2y'= 2(At2+ Bt)e-t-2(2At+ B)e-t
-3y= -3((At2+ Bt)e-t)

The first terms cancel leaving -4(2At+B)e-t+2Ae-t= -3te-t. Dividing both sides by e-t leaves
-4(2At+ B)+ 2A= -3t or -8At+ (-4B+2A)= -3t. We must have -8A= -3 and -4B+2A= 0.

5. May 26, 2005

### dextercioby

Try the method of Lagrange of varying the constants.

Daniel.