mathstresser
Junior Member
- Joined
- Jan 28, 2006
- Messages
- 134
Find the solution to y''+ 12y'+ 36y=0
y(0)=1, y'(0)=-1
the characteristic solution is
yc= r^2 + 12r +36=0
(r+6)(r+6)=0
r= -6, -6
So, the general solution is
y=Ae^(rt)+Bte^(rt)
and the general solution for this equation is
y=Ae^(-6y) + Bte^(-6t)
Now how do I find the solution?
I find the particular solution and get
yp= Ae^(-6t)+ Bte^(-6t)
yp'= -6Ae^(-6t)- 6Be^(-6t) - 6Bte^(-6t)
yp''= 36Ae(-6t)- 12Be^(-6t)+ 36Bte^(-6t)
So I sub thos back into the original equation to find A and B.... But I find out that A and B are both 0. I don't really think this is right... What else should I be doing?
y(0)=1, y'(0)=-1
the characteristic solution is
yc= r^2 + 12r +36=0
(r+6)(r+6)=0
r= -6, -6
So, the general solution is
y=Ae^(rt)+Bte^(rt)
and the general solution for this equation is
y=Ae^(-6y) + Bte^(-6t)
Now how do I find the solution?
I find the particular solution and get
yp= Ae^(-6t)+ Bte^(-6t)
yp'= -6Ae^(-6t)- 6Be^(-6t) - 6Bte^(-6t)
yp''= 36Ae(-6t)- 12Be^(-6t)+ 36Bte^(-6t)
So I sub thos back into the original equation to find A and B.... But I find out that A and B are both 0. I don't really think this is right... What else should I be doing?