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November 27th, 2017, 07:53 AM  #1 
Newbie Joined: Nov 2017 From: NY Posts: 1 Thanks: 0  Method Of Undet. Coefficients question
Hi everyone! I'm really struggling with this problem Solve X' = AX + F(t) with A = ( 1 2 , 2 4) and F(t) = ( t , 2e^(5t) ) A and F(t) are both matrices, I'm not sure if there was a better way for me to write them. I think I need to use method of undetermined coefficients, and I already solved the complementary solution but I'm having a lot of trouble with the particular solution. Can anyone help?? Thanks 
November 27th, 2017, 09:45 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,939 Thanks: 794 
You want to find a set of functions, $\displaystyle \begin{pmatrix}x(t) \\ y(t)\end{pmatrix}$, that satisfy the entire equation by "undertermined coefficients". In order to use "undetermined coefficients", even in the case of a single equation in a single unknown, you need to be able to make a "guess" at the form of the function. Since the "nonhomogenous" part consists of the functions $\displaystyle t$ and $\displaystyle 2e^{5t}$ I would look for functions of the form $\displaystyle x= At+ B+ Ce^{5t}$ and $\displaystyle y= Et+ F+ Ge^{5t}$. Then $\displaystyle \frac{dx}{dt}= A+ 5Ce^{5t}= x+ 2y= (A+ 2E+ 1)t+ (B+ 2F)+ (C+ 2G)e^{5t}$ and $\displaystyle \frac{dy}{dt}= E+ 5Ge^{5t}= t+ 5GTe^{5t}= (2A+ 4E)t+ (2B+4F)+ (2C+ 4G+ 1)e^{5t}$. That gives the equations $\displaystyle A+ 2E+ 1= 0$, $\displaystyle B+ 2F= A$, $\displaystyle C+ 2G= 5$, $\displaystyle 2A+ 4E= 1$, $\displaystyle 2B+ 4F= =0$, and $\displaystyle 2C+ 4G+ 1= 5G$, six equations to solve for the 6 constants, A, B, C, D, E, and F. 

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coefficients, method, question, undet 
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