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

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- May 27th 2008, 10:42 AMJim NewtAnother 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 - May 27th 2008, 11:11 AMChris L T521
To solve for $\displaystyle y_p$, we assume it takes on the form of a second degree polynomial : $\displaystyle y_p=Ax^2+Bx+C$

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

$\displaystyle y_p^/=2Ax+B$

$\displaystyle y_p^{//}=2A$

Therefore, we have:

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

Comparing the coefficients, we have:

$\displaystyle \begin{aligned}

2A+B &=0 \\

2A&=0

\end{aligned}

$

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

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