## Calculus: Early Transcendentals 8th Edition

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 $y=c_{2}sin\frac{n\pi}{L}x$
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 $y=c_{2}sin\sqrt λx$ $y=c_{2}sin\frac{n\pi}{L}x$