Question

A partial differential equation is given below.
\[ \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = 0 \]
Possible solution(s) is/are:

• A. \( (x+y)^5 \)
• B. \( (x-2y)^3 \)
• C. \( \cos(x+y) \)
• D. \( \sin(x-2y) \)

Hint:

The PDE \( u_{xx}-u_{yy}=0 \) is solved by any function of x+y alone or x-y alone, test each option against this pattern.

Answer 1

The Correct Option is A, C

Step 1: Identify the general form of solution.
The given PDE is \( \dfrac{\partial^2 u}{\partial x^2} - \dfrac{\partial^2 u}{\partial y^2} = 0 \). This is a second order PDE of the wave equation type. Its general solution is known to be \( u = F(x+y) + G(x-y) \), where F and G are any twice differentiable functions.
This works because if \( u = F(x+y) \), then by the chain rule \( u_x = F'(x+y) \) and \( u_{xx} = F''(x+y) \), and also \( u_y = F'(x+y) \) and \( u_{yy} = F''(x+y) \), so \( u_{xx} - u_{yy} = F''(x+y) - F''(x+y) = 0 \). The same check works for any function of (x-y) alone.

Step 2: Check option (A), \( (x+y)^5 \).
This is of the exact form F(x+y) with \( F(s) = s^5 \). \( u_x = 5(x+y)^4 \), so \( u_{xx} = 20(x+y)^3 \). \( u_y = 5(x+y)^4 \), so \( u_{yy} = 20(x+y)^3 \). \( u_{xx} - u_{yy} = 20(x+y)^3 - 20(x+y)^3 = 0 \). This satisfies the equation.

Step 3: Check option (B), \( (x-2y)^3 \).
This function mixes x and y in the ratio 1:-2, not 1:1 or 1:-1, so it is not automatically a solution, it needs to be tested directly. \( u_x = 3(x-2y)^2 \), so \( u_{xx} = 6(x-2y) \). \( u_y = -6(x-2y)^2 \), so \( u_{yy} = 24(x-2y) \). \( u_{xx} - u_{yy} = 6(x-2y) - 24(x-2y) = -18(x-2y) \), which is not zero in general. This option fails.

Step 4: Check option (C), \( \cos(x+y) \).
This is F(x+y) with \( F(s) = \cos s \). \( u_x = -\sin(x+y) \), so \( u_{xx} = -\cos(x+y) \). \( u_y = -\sin(x+y) \), so \( u_{yy} = -\cos(x+y) \). \( u_{xx} - u_{yy} = -\cos(x+y) - (-\cos(x+y)) = 0 \). This satisfies the equation.

Step 5: Check option (D), \( \sin(x-2y) \).
\( u_x = \cos(x-2y) \), so \( u_{xx} = -\sin(x-2y) \). \( u_y = -2\cos(x-2y) \), so \( u_{yy} = -4\sin(x-2y) \). \( u_{xx} - u_{yy} = -\sin(x-2y) - (-4\sin(x-2y)) = 3\sin(x-2y) \), not zero in general. This option fails.

Final Answer:
Only \( (x+y)^5 \) and \( \cos(x+y) \) satisfy the PDE, since both depend on x and y only through the combination (x+y).
\[ \boxed{\text{Options (A) and (C)}} \]