undetermined coefficients

Printable View

February 6th 2010, 04:50 PM
magdaddy101

undetermined coefficients

I'm supposed to find the solution this initial value problem using the method of undetermined coefficients
$<br /> y'' - 2y' - 3y = 3te^{2t}, y(0)=1, y'(0)=0.$

My guess was to use $Y(t) = Ate^{2t} + Be^{2t}$ for the particular solution, but this led me to the wrong answer.
I have a hunch that I should use $Y(t) = t(Ate^{2t} + Be^{2t})$ but I have no idea why, since $e^{2t}$ is not part of the complimentary solution.

I appreciate any help.
February 6th 2010, 09:33 PM
Pulock2009

the P.I. should be c1.e^-t+c2.e^3t
and the A.E.=3.e^2t(t-6)
are they correct???

we get the P.I. by solving the quadratic in y' ie. D^2-2D-3=0
which gives D=-1,3

and the A.E. is calculated by separating the exponential part then........(it should be given in any text book).hope this was helpful!
February 7th 2010, 02:19 AM
HallsofIvy

Quote:

Originally Posted by magdaddy101

I'm supposed to find the solution this initial value problem using the method of undetermined coefficients
$<br /> y'' - 2y' - 3y = 3te^{2t}, y(0)=1, y'(0)=0.$

My guess was to use $Y(t) = Ate^{2t} + Be^{2t}$ for the particular solution, but this led me to the wrong answer.

What did you do that gave you a wrong answer?

If $y= Ate^{2t}+ Be^{2t}$ , then
$y"= Ae^{2t}+ 2Ate^{2t}+ 2Be^{2t}= (A+ 2B)e^{2t}+ 2Ate^{2t}$
$y"= 2(A+ 2B)e^{2t}+ 2Ae^{2t}+ 4Ate^{2t}= (6A+ 2B)e^{2t}+ 4Ate^{2t}$
Putting those into the equation gives
$[(6A+ 2B)-2(A+ 2B)- 3B]e^{2t}+ [4A- 4A- 3A]te^{2t}$ $= (4A- 5B)e^{2t}+ (-3A)te^{2t}= 3te^{2t}$

We must have 4A- 5B= 0 and -3A= 3. A= 1 so 4- 5B= 0 and B= 4/5.
That is a perfectly good solution.

Quote:

I have a hunch that I should use $Y(t) = t(Ate^{2t} + Be^{2t})<br />$ but I have no idea why, since $e^{2t}$ is not part of the complimentary solution.

I appreciate any help.