Author Topic: MT Problem 4  (Read 2638 times)

Victor Ivrii

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MT Problem 4
« on: March 06, 2013, 09:09:26 PM »
Find a particular solution of equation
\begin{equation*}
y'''-2y''+4y'-8y=e^{3x}
\end{equation*}

Branden Zipplinger

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Re: MT Problem 4
« Reply #1 on: March 06, 2013, 10:04:31 PM »
Here is the solution. I apologize for the picture instead of code, but time was of the essence

Matthew Cristoferi-Paolucci

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Re: MT Problem 4
« Reply #2 on: March 06, 2013, 10:07:10 PM »
Heres my solution

Rudolf-Harri Oberg

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Re: MT Problem 4
« Reply #3 on: March 06, 2013, 10:12:21 PM »
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 $Y_p=\frac{1}{13}e^{3x}$.
« Last Edit: March 06, 2013, 10:16:22 PM by Rudolf-Harri Oberg »

Matthew Cristoferi-Paolucci

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Re: MT Problem 2a
« Reply #4 on: March 06, 2013, 10:18:59 PM »
Im slow today...

Devangi Vaghela

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Re: MT Problem 4
« Reply #5 on: March 06, 2013, 10:20:14 PM »
This is my solution!

Jeong Yeon Yook

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Re: MT Problem 4
« Reply #6 on: March 06, 2013, 10:21:30 PM »
solutioin

Branden Zipplinger

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Re: MT Problem 4
« Reply #7 on: March 06, 2013, 10:53:05 PM »
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

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Re: MT Problem 4
« Reply #8 on: March 07, 2013, 12:11:01 AM »
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

Victor Ivrii

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Re: MT Problem 4
« Reply #9 on: March 07, 2013, 05:00:44 AM »
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