Crap, nevermind, I left b^2 out of the quadratic formula, thanks anyways.(adsbygoogle = window.adsbygoogle || []).push({});

Here's the question:

Find y as a function of t from the diff eq:

y''+6y'+25y=0 with the initial conditions y(0)=8 and y'(0)=8

I used the form r^2+6r+25=0 to solve for r and through the quadratic equation got r = -3+/-5i

so my equation now looks like

c1*e^(-3t)*cos(5t)+c2*e^(-3t)*sin(5t)=8

The second part of the equation cancels and I'm left with c1=8

Now to find y'(0) I have:

-3*c1*e^(-3t)*cos(5t)+c1*e^(-3t)*(-5*sin(5t))+-3*c2*e^(-3t)*sin(5t)

+c2*e^(-3t)*(5*cos(5t))

The second and third terms in that cancel out and I'm left with

-3*c1*e^(-3t)*cos(5t)+c2*e^(-3t)*5*cos(5t), and plugging in 0 for t and 8 for c1 I get:

-24+5*c2=8

5*c2=32

c2=32/5

So the final equation looks like

8*e^(-3t)*cos(5t)+32/5*e^(-3t)*sin(5t)

This isn't right and I can't figure out what I did wrong. Any help would be awesome.

Thanks a lot.

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# Homework Help: Second order diff eq with i

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