Question

Solution of the PDE \(u_x+yu_y=0\) with the initial condition \(u(0,y)=y^3\) is ____.

• A. \(u(x,y)=y^3e^{-5x}\)
• B. \(u(x,y)=y^3e^{-4x}\)
• C. \(u(x,y)=y^3e^{-3x}\)
• D. \(u(x,y)=y^3e^{-2x}\)

Hint:

Use Lagrange's auxiliary equations \(\frac{dx}{1}=\frac{dy}{y}=\frac{du}{0}\) to find the invariant along each characteristic.

Answer 1

The Correct Option is C

Step 1: Since the initial data is \(u(0,y) = y^3\), try a solution of the separable form \(u(x,y) = y^3 g(x)\) with \(g(0) = 1\).

Step 2: Compute the partial derivatives: \(u_x = y^3 g'(x)\) and \(u_y = 3y^2 g(x)\).

Step 3: Substitute into the PDE \(u_x + y u_y = 0\):

\[y^3 g'(x) + y \cdot 3y^2 g(x) = y^3\left[g'(x) + 3g(x)\right] = 0\]

Step 4: Since this must hold for all \(y\), the bracket must vanish:

\[g'(x) + 3g(x) = 0\]

Step 5: This is a first order linear ODE in \(g\). Its solution is \(g(x) = C e^{-3x}\).

Step 6: Use \(g(0) = 1\) to find \(C = 1\), so \(g(x) = e^{-3x}\).

Step 7: Therefore

\[u(x,y) = y^3 g(x) = y^3 e^{-3x}\] \[\boxed{u(x,y) = y^3 e^{-3x}}\]