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Let , $a$ and $b$ be two real numbers. If the function $x(t)$ is a solution of the differential equation $\displaystyle \frac{d^2 x}{dt^2}-2a\frac{dx}{dt}+bx=0$ with $x(0)=x(1)=0$ then show that $x(n)=0$ for all $n\in \Bbb N$.

I got that when $a^2-b\ge 0$ then $x(t)\equiv 0$ and the proof is trivial. But when $a^2-b<0$ then the solution becomes $x(t)=Be^{at}\sin\sqrt{a^2-b}t$. From here how I can show that $x(n)=0$ for all $n\in \Bbb N$ ?

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    Make a change $y = t+1$ and note that $x(y) = 0 $ if $t = 0$. Now replace $x(y)$ on the equation and note thar $x(y) = x(t)$, i.e, $x(1+t) = x(t)$. Now $x(t)$ is periodic with period $1$, thus $x(n) = x(n-1) = ... =x(1) = 0$. – 2017-02-08
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    Changing $y=t+1$ , I got , $x(y-1)=Be^{a(y-1)}\sin{\sqrt{a^2-b}(y-1)}$. When $t=0$ then how $x(y)=0$ ? – 2017-02-08
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    Work on the ODE, not in the solution. – 2017-02-08

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$$ \frac{d^2 x}{dt^2}-2a\frac{dx}{dt}+bx=0$$ The general solution is $$\quad x(t)=C_1e^{(a+\sqrt{a^2-b})\:t}+C_2e^{(a-\sqrt{a^2-b})\:t}$$ Looking for real solutions with $a$ , $b$ , $C_1$ and $C_2$ reals , the solutions can be expressed as : $$\begin{cases} \text{If} \quad a^2-b< 0 \quad \text{ then } \quad x(t)=c_1e^{at}\sin(\sqrt{b-a^2}\:t)+c_2e^{at}\cos(\sqrt{b-a^2}\:t) \\ \text{If}\quad b=a^2 \quad \text{ then } \quad x(t)=c_1e^{at}t+c_2e^{at}\\ \text{If} \quad a^2-b> 0 \quad \text{ then } \quad x(t)=c_1e^{at}\sinh(\sqrt{a^2-b}\:t)+c_2e^{at}\cosh(\sqrt{a^2-b}\:t) \end{cases}$$ We see that, in all casses, the condition $\quad x(0)=0 \quad$ implies $\quad c_2=0\quad$ thus : $$x(0)=0\quad\to\quad \begin{cases} \text{If} \quad a^2-b< 0 \quad \text{ then } \quad x(t)=c_1e^{at}\sin(\sqrt{b-a^2}\:t) \\ \text{If}\quad b=a^2 \quad \text{ then } \quad x(t)=c_1e^{at}t\\ \text{If} \quad a^2-b> 0 \quad \text{ then } \quad x(t)=c_1e^{at}\sinh(\sqrt{a^2-b}\:t)) \end{cases}$$ And the condition $\quad x(1)=0 \quad$ implies $\quad c_1=0\quad$ except in the case of $\sqrt{a^2-b}=k\pi\quad$ with $k$ integer.

$$x(0)=x(1)=0\quad\to\quad \begin{cases} \text{If} \quad a^2-b< 0 \text{ and } \sqrt{b-a^2}=k\pi \quad \text{ then } \quad x(t)=c_1e^{at}\sin(k\pi\:t) \\ \text{If} \quad a^2-b< 0 \text{ and } \sqrt{b-a^2}\neq k\pi \quad \text{ then } \quad x(t)=0 \\ \text{If} \quad a^2-b\ge 0 \quad \text{ then } \quad x(t)=0 \end{cases}$$ Even in the non-trivial case $\quad b=a^2+k^2\pi^2 \quad\to\quad x(n)=0\quad$ because $\quad \sin(k\pi n)=0$ .