### Question: GATE2013 (2 Marks)

The solution of the differential equation \(\displaystyle \frac{d^2u}{dx^2} – k \frac{du}{dx}= 0 \) where k is a constant, subjected to the boundary conditions \(u(0)= 0 \text{ and } u(L) = U \), is

\(\displaystyle (A) u = U \frac{x}{L} \) | \(\displaystyle (B) u = U \bigg(\frac{1 – e^{kx}}{1 – e^{kL}}\bigg) \) |

\(\displaystyle (C) u = U \bigg(\frac{1 – e^{-kx}}{1 – e^{-kL}}\bigg) \) | \(\displaystyle (D)u = U \bigg(\frac{1 + e^{kx}}{1 + e^{kL}}\bigg) \) |

### Solution:

(D^{2} – kD) u = 0

D(D – k) = 0

D = 0, k

u = C_{1}e^{0x} + C_{2}e^{kx}

u = C_{1} + C_{2}e^{kx}

Applying boundary conditions:

u(0) = 0;

0 = C_{1} + C_{2}

C_{1} = –C_{2}

u(L) = U

U = C_{1} + C_{2}e^{kL} = –C_{2} + C_{2}e^{kL} = C_{2}(e^{kL} – 1)

Therefore, \(\displaystyle C_2 = \frac{U}{e^{kL} – 1} \text{ and } C_1 = \frac{-U}{e^{kL} – 1} \)

\(\displaystyle u = \frac{-U}{e^{kL} – 1} + \frac{U}{e^{kL} – 1} e^{kx} \\\displaystyle= \frac{U}{1 – e^{kL}} – \frac{U}{1 – e^{kL}} e^{kx} \)

Solution of the differential equation

\(\displaystyle u = U \bigg(\frac{1 – e^{kx}}{1 – e^{kL}} \bigg) \)