I am given the following ODE : $y'' – 2ay'+by = 0$ for some real constants $a,b$, along with the initial conditions $y(0) = y(1) = 0$. Furthermore I know that some function $y$ is a solution of this ODE.
How do I show that $y(n) = 0$ for all natural numbers $n$?
The idea would be to solve the equation : it is a second order linear differential equation. Hence, (skipping rigour) we solve $r^2 – 2ar + b = 0$ to get solutions $r_1,r_2$ which may or may not be equal (and may be complex).
The solution is now given by :
1 : $y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$ if $r_1 \neq r_2$ (note that if they are complex then by Euler's formula we have a linear combination of trigonometric functions coming in)
2 : $y(t) = c_1 e^{r_1t} + c_2te^{r_1}t$ if $r_1 = r_2$.
In case 1 , substituting $y(0) = 0$ gives $c_1 + c_2 = 0$ and combining with $y(1)=0$ gives $e^{r_1} = e^{r_2}$ and $c_1 = -c_2$, so $\color{red}{e^{r_1 t} = e^{r_2 t} \mbox{ for all } t}$ hence the solution is identically zero.
In case 2, substituting $y(0) = 0$ gives $c_1 = 0$ and then $y(1) = 0$ gives $c_2 = 0$ so the solution is identically zero.
There is definitely something wrong here. I'd like people to point it out, since I think I've blindly used the formula given to me here. This function is supposed to be non-zero at least at the integers, and the fact that no more is provided hints that there are non-trivial solutions.
EDIT : I found the error thanks to the great people below. You may find the error colored in red above.
Best Answer
In 1., for a nontrivial solution with $f(0)=f(1)=0$ has $c_1+c_2=0$ and $c_1e^{r_1}+c_1e^{r_2}=0$. Thus $c_2=-c_1\ne0$ and $c_1(1-e^{r_2-r_1})=0$. This means that $r_2-r_1=2k\pi i$. Then we easily get $f(n)=0$.
There aren't any solutions in case 2.