Question

Solve the partial differential equation \(x^3 \frac{∂u}{∂x} +y^2 \frac{∂u}{∂y} = 0 \) using method of separation of variables if \(u(0,y) = 10 \, e^{\frac{5}{y}}.\)

(a) \(10e^{\frac{5}{2x^2}} e^{\frac{5}{y}} \)

(b) \(10e^{\frac{-5}{2y^2}} e^{\frac{5}{x}} \)

(c) \(10e^{\frac{-5}{2y^2}} e^{\frac{-5}{x}} \)

(d) \(10e^{\frac{-5}{2x^2}} e^{\frac{5}{y}} \)

I had been asked this question during an online exam.

Asked question is from Solution of PDE by Variable Separation Method topic in division Applications of Partial Differential Equations of Engineering Mathematics

Answer 1

Right answer is (d) \(10e^{\frac{-5}{2x^2}} e^{\frac{5}{y}} \)

To elaborate: \(u(x,t) = X(x) T(t) \)

\(x^3 X’Y+y^2 Y’X=0\)

\(\frac{X’}{ X} = \frac{k}{x^3} \) which implies \(X = ce^{\frac{k}{2x^2}}\)

\(\frac{Y’}{Y} = \frac{-k}{y^2} \) which implies \(Y = c’ e^{\frac{k}{y}}\)

\(u(x,t) = cc’e^{\frac{k}{2x^2}} e^{\frac{k}{y}} \)

\(u(0,y) = 10e^{\frac{5}{y}}= cc’e^{\frac{k}{y}} \)

Therefore k = 5 and cc’ = 10

Hence, \(u(x,t) =  10e^{\frac{-5}{2x^2}} e^{\frac{5}{y}}. \)