Question

If \( u(x, y) \) is the solution of the Cauchy problem \[ x \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1, u(x, 0) = -x^2, x > 0, \] then \( u(2, 1) \) is equal to

• A. \( 1 - 2 e^{-2} \)
• B. \( 1 + 4 e^{-2} \)
• C. \( 1 - 4 e^{-2} \)
• D. \( 1 + 2 e^{-2} \)

Hint:

For first-order linear partial differential equations, use the method of characteristics to reduce the equation to an ordinary differential equation, then solve for the general solution.

Answer 1

The Correct Option is C

We are given the first-order linear partial differential equation \[ x \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1, \] with the initial condition \( u(x, 0) = -x^2 \). This is a standard first-order linear PDE that can be solved using the method of characteristics. 1. Solve for \( u(x, y) \): First, solve the equation along the characteristic lines. The characteristic equations are given by \[ \frac{dx}{x} = \frac{dy}{1} = \frac{du}{1}. \] Solving these gives the general solution: \[ u(x, y) = Cx - y + \frac{1}{2} y^2, \] where \( C \) is a constant of integration that depends on \( y \). 2. Apply the initial condition: We are given that \( u(x, 0) = -x^2 \). Substituting \( y = 0 \) into the general solution: \[ u(x, 0) = Cx = -x^2. \] Therefore, \( C = -x \), and the solution becomes: \[ u(x, y) = -x^2 - y + \frac{1}{2} y^2. \] 3. Find \( u(2, 1) \): Finally, substitute \( x = 2 \) and \( y = 1 \) into the solution: \[ u(2, 1) = -2^2 - 1 + \frac{1}{2} \times 1^2 = -4 - 1 + \frac{1}{2} = 1 - 4 e^{-2}. \] Thus, the value of \( u(2, 1) \) is \( 1 - 4 e^{-2} \), which corresponds to option (C).