# Thread: Another Undetermined Coefficient question

1. ## Another Undetermined Coefficient question

y" + y' = 3x^2

Here's the work I've done:

yc:

1. r^2 + r = 0
2. r(r + 1) = 0
3. r=0, -1
4. Thus, yc= c1 +c2e^-x

How do I solve for yp? Thanks,

JN

2. Originally Posted by Jim Newt
y" + y' = 3x^2

Here's the work I've done:

yc:

1. r^2 + r = 0
2. r(r + 1) = 0
3. r=0, -1
4. Thus, yc= c1 +c2e^-x

How do I solve for yp? Thanks,

JN
To solve for $y_p$, we assume it takes on the form of a second degree polynomial : $y_p=Ax^2+Bx+C$

Now plug $y_p$ into the DE and solve for the unknown coefficients:

$y_p^/=2Ax+B$
$y_p^{//}=2A$

Therefore, we have:

$2A+2Ax+B=3x^2$

Comparing the coefficients, we have:

\begin{aligned}
2A+B &=0 \\
2A&=0
\end{aligned}

This implies that both A and B equal $\color{red}\boxed{0}$.

Therefore, $\color{red}\boxed{y_p=0}$.