Question

Find the solution of \(\frac{∂u}{∂x}=36 \frac{∂u}{∂t}+10u \) if \( \frac{∂u}{∂x} (t=0)=3e^{-2x} \) using the method of separation of variables.

(a) \(\frac{-3}{2} e^{-2x} e^{-t/3}\)

(b) \(3e^x e^{-t/3} \)

(c) \(\frac{3}{2} e^{2x} e^{-t/3} \)

(d) \(3e^{-x} e^{-t/3} \)

The question was posed to me in exam.

Question is taken from Solution of PDE by Variable Separation Method topic in chapter Applications of Partial Differential Equations of Engineering Mathematics

Answer 1

Right answer is (a) \(\frac{-3}{2} e^{-2x} e^{-t/3}\)

Easiest explanation: \(u(x,t) = X(x) T(t) \)

Substituting in the given equation, \(X’T = 36T’X + 10XT \)

\(\frac{X’}{X} = k \) which implies \(X = c e^{kx} \)

\(\frac{T’}{T} = \frac{(k-10)}{36} \) which implies \(T = c’e^{\frac{k-10}{36} t} \)

\(\frac{∂u}{∂x} (t=0)=3e^{-2x}= cc’ke^{kx} \)

Therefore k = -2 and cc’ = \(\frac{-3}{2} \)

Hence, \(u(x,t) = \frac{-3}{2} e^{-2x} e^{-t/3}. \)