Question

The solution to the differential equation $\frac{d^{2}u}{d{x}^2}-k\frac{du}{dx}=0$ where k is constant, subjected to the boundary conditions $u\left(0\right)=0$ and $u\left(L\right)=U$, is

• A. $u=U\frac{x}{L}$
• B. $u=U\left(\frac{1-e^{kx}}{1-e^{kL}}\right)$
• C. $u=U\left(\frac{1-e^{-kx}}{1-e^{-kL}}\right)$
• D. $u=U\left(\frac{1+e^{kx}}{1+e^{kL}}\right)$

Answer 1

The correct option is B $u=U\left(\frac{1-e^{kx}}{1-e^{kL}}\right)$

Given differential equation is $\frac{d^{2}u}{d{x}^2}-k\frac{du}{dx}=0....\left(1\right)$
$(D^2-KD)u=0$
Auxiliary equation become
$m(m-k)=0$
$m=0,k$
$\therefore$ $u\left(x\right)=c_1{e}^{0.x}+c_2{e}^{Kx}=c_1+c_2{e}^{Kx}$
$u\left(0\right)=0$
$\Rightarrow 0=c_1+c_2$ ...(2)
$u\left(L\right)=U$
$\Rightarrow U=C_1+C_2{e}^{KL}$ ... $\left(3\right)$
Solving equation $\left(2\right)$ and $\left(3\right)$ we get
$\Rightarrow (1-e^{KL})c_2=-U$
$\Rightarrow c_2=\left(\frac{U}{e^{KL}-1}\right)$
Putting value of $c_2$ in equation $\left(2\right)$ we get
$c_1=-c_2=\frac{-U}{e^{KL}-1}$
$\therefore$ $u\left(x\right)=\frac{U}{1-e^{KL}}+\left(\frac{-U}{1-e^{KL}}\right)e^{Kx}$
$u=U\left(\frac{1-e^{Kx}}{1-e^{KL}}\right)$