# Math Help - Stuck on a diff EQ: Undetermined Coefficient

1. ## Stuck on a diff EQ: Undetermined Coefficient

Find this general solution:

y" + 3y' + 2y = 36xe^x

Here's my work:

1. r^2 + 3r + 2 = 0; r=-1, -2
2. yc = c1e^-x + c2^e^-2x

3. yp=Axe^x
y'p=Axe^x + Ae^x
y"p=Axe^x + 2Ae^x

4. (Axe^x + 2Ae^x) + 3(Axe^x + Ae^x) +2(Axe^x) = 36xe^x

5. 6Axe^x + 5Ae^x = 36xe^x

And here's where I get fuzzy. I need to find this value of A and then put in the form y = yc + yp. The answer in the back of the book is:

y= c1e^-x + c2^e^-2x + (6x-5)e^x

Any suggestions? Thanks, JN

2. Originally Posted by Jim Newt
Find this general solution:

y" + 3y' + 2y = 36xe^x

Here's my work:

1. r^2 + 3r + 2 = 0; r=-1, -2
2. yc = c1e^-x + c2^e^-2x

3. yp=Axe^x
y'p=Axe^x + Ae^x
y"p=Axe^x + 2Ae^x

4. (Axe^x + 2Ae^x) + 3(Axe^x + Ae^x) +2(Axe^x) = 36xe^x

5. 6Axe^x + 5Ae^x = 36xe^x

And here's where I get fuzzy. I need to find this value of A and then put in the form y = yc + yp. The answer in the back of the book is:

y= c1e^-x + c2^e^-2x + (6x-5)e^x

Any suggestions? Thanks, JN
Since you have a polynomial multiplied by an exponential function your particular solution should look like

$y_p=(Ax+B)e^x$

Take a few derivatives and plug into the original equation to to solve for A and B.

Good luck

3. Originally Posted by TheEmptySet
Since you have a polynomial multiplied by an exponential function your particular solution should look like

$y_p=(Ax+B)e^x$

Take a few derivatives and plug into the original equation to to solve for A and B.

Good luck
Thanks for the input EmptySet. So from this point:

1. yp = (Ax + B)e^x
y'p= Ae^x + (Ax + B )e^x
y"p=2Ae^x + Axe^x + Be^x

2. yp= 2Ae^x + Axe^x + Be^x +3Ae^x + 3xAe^x + 3Be^x + 2Axe^x + 2Be^x 2Be^x

So then 5Ae^x + 6Axe^x + 6Be^x = 36xe^x

how do I make the jump to yp = (6x-5)e^x?

4. Originally Posted by Jim Newt
Thanks for the input EmptySet. So from this point:

1. yp = (Ax + B)e^x
y'p= Ae^x + (Ax + B )e^x
y"p=2Ae^x + Axe^x + Be^x

2. yp= 2Ae^x + Axe^x + Be^x +3Ae^x + 3xAe^x + 3Be^x + 2Axe^x + 2Be^x 2Be^x

So then 5Ae^x + 6Axe^x + 6Be^x = 36xe^x

how do I make the jump to yp = (6x-5)e^x?
$5Ae^x + 6Axe^x + 6Be^x = 36xe^x
$

Rewritting as (adding a $0e^{x}$) may help you see it

So then
$(5A + 6B)e^x + 6Axe^x = 36xe^x + 0e^{x}
$

We get a "system" of equations for A and B

$5A+6B=0$
$6A=36$

So we can see A=6 and B=-5.

I hope this helps.

5. Originally Posted by Jim Newt
Thanks for the input EmptySet. So from this point:

1. yp = (Ax + B)e^x
y'p= Ae^x + (Ax + B )e^x
y"p=2Ae^x + Axe^x + Be^x

2. yp= 2Ae^x + Axe^x + Be^x +3Ae^x + 3xAe^x + 3Be^x + 2Axe^x + 2Be^x 2Be^x

So then 5Ae^x + 6Axe^x + 6Be^x = 36xe^x

how do I make the jump to yp = (6x-5)e^x?
Compare the coefficients:

\begin{aligned}
5A+6B&=0 \\
6A&=36
\end{aligned}

This implies that A=6 and B=-5

Therefore, $y_p=(6x-5)e^x$.

Hope this clarified things