y''-3y'+2y=e^t y(0)=0 y'(0)=-1(adsbygoogle = window.adsbygoogle || []).push({});

yh=solution to homogeneous equation (y''-3y'+2y=0) = Ce^t+Ae^(2t)

C and A are constants

yp=solution to particular solution (e^t)

yp=ae^t where a is a constant. It turns out that this solves the homogenuous solution so I had to multiply it by a factor of t giving

yp=a*t*e^t

yp'=a(e^t+te^t)

yp''=a(e^t+e^t+te^2)=a(2e^2+te^2)

subsitution this into the original equation gives a=-1. Thus yp=-te^t

the solution is then y(t)=yh+yp=Ce^t+Ae^(2t)-te^t

Now I solve for C and A

0=C+A C=-A

y'(t)=Ce^t+2Ae^(2t)-e^t-te^2

-1=C+2A-1 -1=-A+2A-1 A=0 C=0

The problem with this is when I solve it I get that C=A=0 which means that the homogenuous solution was a TRIVIAL solution. I've never experienced a problem like this yet. Would that just mean the solution is y=yp=-te^2?

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# Homework Help: 2nd order differential equations

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