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

Concept:
Instead of testing each option by brute force differentiation, use a change of variables. Let $s = x+y$ and $t = x-y$. By the chain rule, differentiating with respect to x and y and combining terms, the operator $\dfrac{\partial^2}{\partial x^2} - \dfrac{\partial^2}{\partial y^2}$ transforms into $4\dfrac{\partial^2}{\partial s\, \partial t}$. So the PDE $u_{xx} - u_{yy}=0$ becomes $u_{st} = 0$ in the new variables.


Step 1: Solve the transformed equation. $u_{st} = 0$ means $\dfrac{\partial u}{\partial s}$ does not depend on t, so integrating with respect to t gives $u = f(s) + g(t)$ for arbitrary functions f, g. Going back to the original variables, the general solution is $u = f(x+y) + g(x-y)$: any function of x+y alone, or any function of x-y alone, is a solution.

Step 2: Sort the four options by which combined variable they use. $(x+y)^5$ is a function of $s = x+y$ only, so it fits the pattern f(s) directly.
$\cos(x+y)$ is also a function of $s = x+y$ only, so it also fits f(s).
$(x-2y)^3$ is a function of x-2y, which is neither $s=x+y$ nor $t=x-y$, so it does not automatically qualify, it must be checked separately.
$\sin(x-2y)$ is also a function of x-2y, same issue as above.

Step 3: Directly test the two suspect options. For $u = (x-2y)^3$: $u_{xx} = 6(x-2y)$ and $u_{yy} = 24(x-2y)$, giving $u_{xx}-u_{yy} = -18(x-2y) \neq 0$.
For $u = \sin(x-2y)$: $u_{xx} = -\sin(x-2y)$ and $u_{yy} = -4\sin(x-2y)$, giving $u_{xx}-u_{yy} = 3\sin(x-2y) \neq 0$.
Neither satisfies the equation, confirming the variable matching argument from Step 2.

Step 4: Conclusion. Only the functions that depend purely on the combination x+y, namely $(x+y)^5$ and $\cos(x+y)$, solve the PDE. \[ \boxed{\text{(A) and (C)}} \]