finding a solution to a second order dif eq

mathstresser

Junior Member
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Jan 28, 2006
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134
Find the solution to

\(\displaystyle \L\\ y'' +2y' +10y=0\)

\(\displaystyle \L y(0)=1,\, y'(0)=0\)

The characteristic solution is:

\(\displaystyle \L\\ r=-1 +-3i\)

The particular solution is:

\(\displaystyle \L\\ yp= C1e^{(-t)}cos(3t) + C2e^{(-t)}sin(3t)\)

\(\displaystyle \L\\ yp'= -C1e^{(-t)}cos(3t) -C1e^{(-t)}sin(3t)+C2e^{(-t)}cos(3t)-C2sin(3t)\)

\(\displaystyle \L\\ 1=C1 + 0\)

\(\displaystyle \L\\ 0= -C1 + C2,\, 0=-1 + C2\)

\(\displaystyle \L\\ C1=1,\, C2=1\)

So the solution is:

\(\displaystyle \L\\ y=e^{(-t)}cos(3t) +e^{(-t)}sin(3t)\)

Except that's not right. So what am I doing wrong?
 
oh... So is this right?

The particular solution is:

\(\displaystyle \L\\ yp= C1e^{(-t)}cos(3t) + C2e^{(-t)}sin(3t)\)

\(\displaystyle \L\\ yp'= -C1e^{(-t)}cos(3t) -3C1e^{(-t)}sin(3t)+3C2e^{(-t)}cos(3t)-C2sin(3t)\)

\(\displaystyle \L\\ 1=C1 + 0\)

\(\displaystyle \L\\ 0= -C1 + 3C2,\, 0=-1 + 3C2\)

\(\displaystyle \L\\ C1=1,\, C2=1/3\)

So the solution is:

\(\displaystyle \L\\ y=e^{(-t)}cos(3t) +(1/3)e^{(-t)}sin(3t)\)
 
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