- #1
jwxie
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Homework Statement
Find the solution of the given initial value problem.
[itex]\[y''-2y'-3y=3te^{2t}\][/itex]
Homework Equations
The Attempt at a Solution
(1) the homogenous solution is given by[itex]\[y_{h}=c_{1}e^{3t}+c_{2}e^{-t}\][/itex]
(2) the particular solution is in the form
[itex]\[y_{p}=(At+B)e^{2t}\][/itex]
(3) The first and second derivatives are then, respectively
[itex]\[y'_{p}=Ae^{2t}+2Ate^{2t}+2Be^{2t}\][/itex]
[itex]\[y''_{p}=2Ae^{2t}+2Ae^{2t}+4Ate^{2t}+4Be^{2t}\][/itex]
(3) substitutions and line up
[itex]\[y''_{p}=2Ae^{2t}+2Ae^{2t}+4Ate^{2t}+4Be^{2t}\][/itex]
[itex]\[-2y'_{p}=-2Ae^{2t}-4Ate^{2t}-4Be^{2t}\][/itex]
[itex]\[-3y_{p}=-3Ate^{2t}-3Be^{2t}\][/itex]
(4) combine like terms
[itex]\[te^{t}\left [ 4A-4A-3A \right ]=1
\][/itex]
[itex]\[e^{t}\left [ 4A+4B-2A-4B-3B \right ]=0\]
[/itex]
(5) I got A = -1/3
but this is wrong.
According to the book, Ate^{2t) has a leading coefficient of -1.
However, coincidentally, my B is 2/3, while the book gives -2/3.
The complete solution to this problem with the given IV is
[itex]y=e^{3t}+(2/3)e^{-t}-(2/3)e^{2t}-te^{2t}[/itex]
I have been working on this and other problems for hours and apparently I have been getting many answers... but I have been checking and redoing. Has anyone catch my mistake yet?
Thank you very much!