MAT244-2013S > MidTerm

MT Problem 4

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Victor Ivrii:
Find a particular solution of equation
\begin{equation*}
y'''-2y''+4y'-8y=e^{3x}
\end{equation*}

Branden Zipplinger:
Here is the solution. I apologize for the picture instead of code, but time was of the essence

Matthew Cristoferi-Paolucci:
Heres my solution

Rudolf-Harri Oberg:
We use Milman's method:

$$L\left[A\frac{x^m}{m!}e^{rx}\right]=Ae^{rx}\left(Q(r)\frac{x^m}{m!}+Q'(r)\frac{x^{m-1}}{(m-1)!}+Q''(r)\frac{x^{m-2}}{2!(m-2)!}+... \right)$$
In this case $Q=r^3-2r^2+4r-8$. As we see an exponent in the power of three, let $r=3$. We need to evaluate $Q(3)=13$. As there are no polynomial terms, let $m=0$.

Then $L(Ae^{3x})=13Ae^{3x}$ which implies $A=\frac{1}{13}$.

Solution is $Y_p=\frac{1}{13}e^{3x}$.

Matthew Cristoferi-Paolucci:
Im slow today...