## anonymous 4 years ago y"+6y'+9=0 can someone help me to explain this ODE?

1. anonymous

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.

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

not (2*9)?

6. anonymous

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

7. anonymous

great thanks :)