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Solution to 2nd order diff eq

  • Thread starter glid02
  • Start date
  • #1
54
0
Never mind, I figured it out.

Here's the question:
Find the general solution to the homogeneous differential equation
https://webwork.math.uga.edu/webwork2_files/tmp/equations/59/540a7a16e5c4e841a098d9d2a72f0a1.png

The solution has the form https://webwork.math.uga.edu/webwork2_files/tmp/equations/06/69c97d88bd2a92e464b45652c75c181.png

enter your answers so that https://webwork.math.uga.edu/webwork2_files/tmp/equations/b6/bf0051fdc1775f5fe9263992f485f41.png

I'm supposed to find f1(t) and f2(t).

I know the form ar^2+br+c=0 but in this case it's only r^2=0 so r=0.
Also y=c1f1(t)+c2f2(t)=c1e^(r1t)+c2e^(r2t)

f1(t)=e^(r1t) and f1(0)=1 so f1(0)=e^(r1*0)=1
I know that's right

I'm stuck on f2(t)
f2(t)=e^(r2t) and f2(2)=0 so f2(2)=e^(r2*2)=2
I tried solving for r2
r2*2=ln(2)
r2=ln(2)/2=.3466
and then plugged it into e^(rt), so it was e^(.3466*t)
This isn't right.

I don't know what else to try. Both r1 and r2 should be equal to 0 to satisfy r^2=0, but then f2(t)=e^(rt)=e^(0*t)=1 and that wouldn't satisfy f(2)=2.

Can anyone tell me what else I can try?

Thanks a lot.
 
Last edited by a moderator:

Answers and Replies

  • #2
150
0
There is a problem with your attempt. You have repeated roots. The exponential form doesn't work. If that does not get you anywhere, you can just use general anti-derivatives (integrate both sides) to get the general form of the solution.

This ought to give you the idea of the problem. I hope nobody tries to one-up my answer and gives it all away like you couldn't get it.
 

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