MT Problem 4

[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...

[Devangi Vaghela]

This is my solution!

[Jeong Yeon Yook]

solutioin

[Branden Zipplinger]

there needs to be a correction in my solution: I said r = 2i when its other part of the conjugate pair, -2i, was not mentioned. and I should have subbed the r's with a 1,2, and 3 respectively for good notation

[Victor Lam]

I'm a lil late , but here's a quick way to find Y(particular). A slower "coefficients" method can also verify that A = 1/13

[Victor Ivrii]

Guys, are you showing that you have no idea how to scan or to post (Matthew posted it first in the wrong forum)? I decided to distribute several karma points just to encourage those with very few if any