Differential Equations, Undetermined Coefficients

[Darkmisc]

Homework Statement

y'' + 3y' + 2y = 24exp(-2x)

Homework Equations

y'' + 3y' + 2y = 24exp(-2x)

The Attempt at a Solution

I've solved the characteristic equation and am after the particular solution.

I've let y_p = Aexp(-2x), y_p' = -2Axexp(-2x) and y_p'' = 4Axexp(-2x)

When substituting these terms into the original D.E., I end up with A = 0.

What have I done wrong?

 

[gomunkul51]

Your homogeneous solution affects the way you look for your spesific answer.
Homogeneous solution: y(x) = A*exp(-x) + B*exp(-2x).

you search the specific solution of the form: Y(x) = C*x*exp(-2x)
(multiplication by x because the specific part is a part of your homogeneous answer)

use the Chain Rule to differentiate C*x*exp(-2x) !

Tell us what you get :)

 

[Darkmisc]

Thanks.

I get y_p = -24xexp(-2x).

I suspect I may be ignorant about some basic theory. Why does the particular solution in this case need to be Cxexp(-2x), rather than Aexp(-2x)?

 

[gomunkul51]

The rule is that if you don't look at the homogeneous answer and build you template for finding the specific answer using the undetermined coefficients (I assume you know how to format it) and the look at you homogeneous answer and notice that part of your template is linearly dependent on part of your homogeneous solution (i.e. is a constant multiplicand of the other) then you multiply by x[/] and then you check again to see if it is linearly dependent so some other part of your homogeneous solution, if yes you multiply by x again repeat until no part of the template is linearly dependent on any part of the homogeneous answer. then search insert the template and its derivatives to find the coefficients.

This is the method. the simple answer why you multiply by x is that if you don't you will just get 0 :) for a more detailed answer ask you teacher I'm sure he will point you to the right source (your textbook may be)

Good Luck.

 

[HallsofIvy]

Thanks.

I get y_p = -24xexp(-2x).

I suspect I may be ignorant about some basic theory. Why does the particular solution in this case need to be Cxexp(-2x), rather than Aexp(-2x)?

Presumably, when you solve the associated homogeneous equation, y'' + 3y' + 2y= 0, which has characteristic equation $r^2+ 3r+ 2= (r+ 2)(r+ 1)= 0$, you found that two independent solutions were $e^{-2x}$ and $e^{-3x}$.

Trying a particular solution of the form $Ae^{-2x}$ can't work. That's already a solution to the homogeneous equation- any multiple will make the equation equal to 0.

 

[Darkmisc]

Thanks, that cleared it up for me.