Answer
Proved below.
Work Step by Step
(a) $y(x)=C_{1}\cos x+C_{2}\sin x$
Using $y(0)=0$
$0=C_{1}\cos 0+C_{2}\sin 0=C_{1}$
$\Rightarrow C_{1}=0$
Using $y(π)=1$
$1=C_{1}\cos π+C_{2}\sin π=-C_{1}$
$\Rightarrow C_{1}=-1$
$\Rightarrow$ For any value of $C_{1}$ and $C_{2}$, condition $y(0)=0$ and $y(π)=1$ are not satisfied simultaneously.
Therefore,Boundary-value problem $y''+y=0,y(0)=0,y(π)=1$ has no solution.
(b) $y(x)=C_{1}\cos x+C_{2}\sin x$
Using $y(0)=0$
$0=C_{1}\cos 0+C_{2}\sin 0=C_{1}$
$\Rightarrow C_{1}=0$
Using $y(π)=0$
$0=C_{1}\cos π+C_{2}\sin π=-C_{1}$
$\Rightarrow C_{1}=0$
$\Rightarrow$ Boundary value problem $y''+y=0, y(0)=0, y(π)=0$ has solution.
\[y(x)=C_{2}\sin x\]
and choices of $C_{2}$ are infinite.
Hence, Boundary - value problem $y''+y=0,y(0)=0,y(π)=1$ has infinite number of solutions.
