Answer
a)
For the cases $λ=0$ and $λ\lt0$, the solution of the given equation is $y=0$.
b)
value of $λ=\frac{n^{2}\pi^{2}}{L^{2}}$
corresponding solution $y=c_{2}sin\frac{n\pi}{L}x$
Work Step by Step
a)
Given equation is
$y''+λy=0$
If $λ=0$ then the given equation reduces $y''=0$.
By integrating both sides, we get $y'=c_{1}$ where $c_{1}$ is a constant.
Again integrating both sides, we get $y=c_{1}x+c_{2}$ where $c_{2}$ is another constant.
Apply boundary conditions when $x=0$, $y=0$.
Therefore $0=c_{1}(0)+c_{2}$
$c_{2}=0$
when $x=L$, $y=0$
therefore $0=c_{1}L+c_{2}$
$c_{1}L=0$
$c_{1}=0$ since $L\ne0$
Thus
$y=c_{1}x+c_{2}$
$y=0x+0$
$y=0$
When $λ$ is negative,. then the given equation can be written as
$y''-(-λ)y=0$
Auxiliary equation is
$r^{2}-(-λ)y=0$
$r^{2}-(\sqrt (-λ))^{2}=0$
$(r+\sqrt (-λ))(r-\sqrt (-λ))=0$
$r_{1}=-\sqrt (-λ)$
$r_{2}=\sqrt (-λ)$
$y=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$
$y=c_{1}e^{\sqrt (-λ)x}+c_{2}e^{-\sqrt (-λ)x}$
Apply boundary conditions when $x=0$, $y=0$
Therefore
$0=c_{1}e^{0}+c_{2}e^{0}$
$0=c_{1}+c_{2}$
$c_{2}=-c_{1}$
Also when $x=L$, $y=0$
Therefore
$0=c_{1}e^{\sqrt (-λ)L}+c_{2}e^{-\sqrt (-λ)L}$
$c_{1}e^{\sqrt (-λ)L}-c_{1}e^{-\sqrt (λ)L}=0$
$c_{1}(e^{\sqrt (-λ)L}-e^{-\sqrt (λ)L})=0$
$c_{1}=\frac{0}{e^{\sqrt (-λ)L}-e^{-\sqrt (λ)L}}=0$
And $c_{2}=-c_{1}=0$
The solution for the given equation is
$y=c_{1}e^{\sqrt (-λ)x}+c_{2}e^{-\sqrt (-λ)x}$
$y=0$
For the cases $λ=0$ and $λ\lt0$, the solution of the given equation is $y=0$.
b)
$y''+λy=0$, $y(0)=0=y(L)$
Auxiliary equation
$r^{2}+λ=0$
$r^{2}+(\sqrt λ)^{2}=0$
$(r+i\sqrt λ)(r-i\sqrt λ)=0$
$r_{1}=i\sqrt λ$
$r_{2}=-i\sqrt λ$
$y=e^{αx}(c_{1}cosβx+c_{2}sinβx)$
$y=e^{0x}(c_{1}cos(\sqrt λx)+c_{2}sin(\sqrt λx))$
$y=c_{1}cos(\sqrt λx)+c_{2}sin(\sqrt λx)$
Given, when $x=0$, $y=0$
Therefore
$0=c_{1}cos(\sqrt λ(0))+c_{2}sin(\sqrt λ(0))$
$0=c_{1}$
Thus $y=c_{2}sin\sqrt λx$
Also when $x=L, y=0$
Therefore $0=c_{2}sin\sqrt λL$
Trivial solution so $c_{2}\ne0$
Therefore $sin\sqrt λL=0$
$\sqrt λL=n\pi$ where n is an integer
$\sqrt λ=\frac{n\pi}{L}$
$λ=\frac{n^{2}\pi^{2}}{L^{2}}$
Corresponding solution
$y=c_{2}sin\sqrt λx$
$y=c_{2}sin\frac{n\pi}{L}x$
