2nd order diff. eqn: y''+ 12y'+ 36y=0, y(0)=1, y'(0)=-1

mathstresser

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Jan 28, 2006
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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?
 
Re: 2nd order differential equations

Hello, mathstresser!

Find the solution to: \(\displaystyle y''\,+\,12y'\,+\,36y\:=\:0,\;y(0)\,=\,1,\;y'(0)\,=\,-1\)

The characteristic solution is: \(\displaystyle \:y_c\:=\:r^2\,+\,12r\,+\,36\:=\:0\;\;\Rightarrow\;\;r\,=\,-6,\,-6\)

So the general solution is: \(\displaystyle \:y\:=\:Ae^{-6t}\,+\,Bte^{-6t}\;\) Right!

Now how do I find the solution?

We are told that \(\displaystyle y(0)\,=\,1\)

So we have: \(\displaystyle \:A\cdot e^0\,+\,B\cdot0\cdot e^0 \:=\:1\;\;\Rightarrow\;\;\fbox{A\,=\,1}\)


We are told that \(\displaystyle y'(0)\,=\,-1\)
. . We find that: \(\displaystyle y' \:=\:-6Ae^{-6t}\,+\,Be^{-6t}\,-\,6Bte^{-6t}\)
So we have: \(\displaystyle \:-6\cdot1\cdot e^0\,+\,B\cdot e^0\,-\,6B\cdot0\cdot e^0 \:=\:-1\;\;\Rightarrow\;\;-6\,+\,B\:=\:-1\;\;\Rightarrow\;\;\fbox{B\,=\,5}\)


Therefore, the solution is: \(\displaystyle \L\:y\:=\:e^{-6t}\,+\,5te^{-6t}\)

 
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