Question

The solution of the differential equation $k^2\frac{d^{2}y}{d{x}^2}=y-y_2$ under the boundary conditions
(i) $y=y_1$ at $x=0$ and
(ii) $y=y_2$ at $x=\infty$, where $k,y_1$ and $y_2$ are constnats, is

• A. $y=(y_1-y_2)\text{exp}(-x/k^2)+y_2$
• B. $y=(y_2-y_1)\text{exp}(-x/k)+y_1$
• C. $y=(y_1-y_2)\text{sinh}\left(x/k\right)+y_1$
• D. $y=(y_1-y_2)\text{exp}(-x/k)+y_2$

Answer 1

The correct option is D $y=(y_1-y_2)\text{exp}(-x/k)+y_2$

$k^2\frac{d^{2}y}{d{x}^2}=y-y_2$
$\frac{d^{2}y}{d{x}^2}-\frac{1}{k^2}y=-\frac{y_2}{k^2}$
$A.E$ is $D^2-\frac{1}{k^2}=0$
$D=\pm \frac{1}{k}$
$C.F.=c_1{e}^{x/k}+c_2{e}^{-x/k}$
$P.I.=\frac{1}{D^2-\frac{1}{k^2}}(-\frac{y_2}{k^2})=y_2$
$\therefore y=c_1{e}^{x/k}+c_2{e}^{-x/k}+y_2$
At $x=0,y=y_1$
$\Rightarrow y_1=c_1+c_2+y_2$
$c_1+c_2=y_1-y_2$ ... (i)
At $x=\infty ,y=y_2$
$\Rightarrow y_2=c_1{e}^{\infty }+0+y_2$
$\Rightarrow c_1=0$
$\left(1\right)\Rightarrow c_2=y_1-y_2$
$y=(y_1-y_2)e^{-x/k}+y_2$