Toronto Math Forum
MAT2442013S => MAT244 MathTests => MidTerm => Topic started by: Victor Ivrii on March 06, 2013, 09:09:26 PM

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

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

Heres my solution

We use Milman's method:
$$L\left[A\frac{x^m}{m!}e^{rx}\right]=Ae^{rx}\left(Q(r)\frac{x^m}{m!}+Q'(r)\frac{x^{m1}}{(m1)!}+Q''(r)\frac{x^{m2}}{2!(m2)!}+... \right)$$
In this case $Q=r^32r^2+4r8$. 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 $Y_p=\frac{1}{13}e^{3x}$.

Im slow today...

This is my solution!

solutioin

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

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

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