y"+6y'+9=0 can someone help me to explain this ODE?

Question

anonymous · Answer

you can assume the solution is in the form of $$y=e^{\lambda x}$$then when you sub that in for y, you just need to solve for the lambda values.

anonymous · Answer

so you get labda = -3 twice right, so y = $$e^{-3x}$$ +  x*$$e^{-3x}$$ right?

anonymous · Answer

Well, what you have in front of you is a very normal differential equation, I will assume you want the solution in R and not in C... So, here we go:
The characteristic equation to this equation is
$$r^2+6r+9=0 $$
the discriminant is therefore
$$\Delta=6^2-4*9=36-36=0$$
So we have one solution
$$r=\frac{-6}{2*9}=\frac{-1}{3}$$
and the canonical solution for a differential equation in real values with a characteristic equation with only one solution is, finally
$$y:t \rightarrow (A*t+B)*e^{\frac{-1}{3}*t} $$
where A and B are constants determined by the initial conditions

anonymous · Answer

r should be -6/(2*1) no?

anonymous · Answer

not (2*9)?

anonymous · Answer

oh yes, small mistake... so you end up with r=-3 instead, my bad

anonymous · Answer

great thanks :)