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