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

[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 boundary 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.

 

[arildno]

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

 

[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?

 

[HallsofIvy]

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.

 

[dextercioby]

Try the method of Lagrange of varying the constants.

Daniel.