mathstresser
Junior Member
- Joined
- Jan 28, 2006
- Messages
- 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?
\(\displaystyle \L\\ y'' +2y' +10y=0\)
\(\displaystyle \L y(0)=1,\, y'(0)=0\)
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?