Toronto Math Forum
MAT244-2013S => MAT244 Math--Tests => MidTerm => Topic started by: Victor Ivrii on March 06, 2013, 09:09:26 PM
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Find a particular solution of equation
\begin{equation*}
y'''-2y''+4y'-8y=e^{3x}
\end{equation*}
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Here is the solution. I apologize for the picture instead of code, but time was of the essence
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Heres my solution
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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 $Y_p=\frac{1}{13}e^{3x}$.
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Im slow today...
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This is my solution!
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solutioin
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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
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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
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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