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:
(D2 – kD) u = 0
D(D – k) = 0
D = 0, k
u = C1e0x + C2ekx
u = C1 + C2ekx
Applying boundary conditions:
u(0) = 0;
0 = C1 + C2
C1 = –C2
u(L) = U
U = C1 + C2ekL = –C2 + C2ekL = C2(ekL – 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
\(\displaystyle u = U \bigg(\frac{1 – e^{kx}}{1 – e^{kL}} \bigg) \)