Consider the following partial differential equation:
\[ a\frac{\partial^2 f(x,y)}{\partial x^2} + b\frac{\partial^2 f(x,y)}{\partial y^2} = 8f(x,y) \]
where $a$ and $b$ are distinct positive real numbers.
The combination(s) of the values of the real parameters $\xi$ and $\eta$ for which $f(x,y) = e^{2\xi x+\eta y}$ is a solution of the given partial differential equation, is/are

Question

Let $A$ be a square matrix. If $A^2 = A$, then the matrix $A$ is called:

Answer 1

Rather than deriving one general algebraic condition first, we can plug each candidate pair $(\xi,\eta)$ straight into the original equation $af_{xx} + bf_{yy} = 8f$ and check the arithmetic directly, using $f_{xx} = 4\xi^2 f$ and $f_{yy} = \eta^2 f$ from differentiating $f = e^{2\xi x+\eta y}$ twice.

(A) $\xi = 1/\sqrt{a}$, $\eta = 2/\sqrt{b}$: Here $\xi^2 = 1/a$ and $\eta^2 = 4/b$. The left side becomes $a \cdot 4 \cdot (1/a) f + b \cdot (4/b) f = 4f + 4f = 8f$, which equals the right side exactly. This pair works.
(B) $\xi = 0$, $\eta = 0$: Both second derivative terms vanish since $\xi = \eta = 0$, so the left side is $0$, but the right side is $8f$, which is never zero for real $x,y$ (an exponential is always positive). This pair fails.
(C) $\xi = 1/\sqrt{2a}$, $\eta = 1/\sqrt{2b}$: Here $\xi^2 = 1/(2a)$ and $\eta^2 = 1/(2b)$. The left side is $a\cdot 4 \cdot \frac{1}{2a}f + b\cdot\frac{1}{2b}f = 2f + 0.5f = 2.5f$, which is not equal to $8f$. This pair fails.
(D) $\xi = 0$, $\eta = 2\sqrt{2/b}$: Here $\xi = 0$ kills the $x$-term entirely, and $\eta^2 = 4 \cdot (2/b) = 8/b$. The left side is $0 + b \cdot (8/b) f = 8f$, matching the right side exactly. This pair works.

Since $a$ and $b$ are distinct positive numbers, none of these divisions run into trouble (we never divide by zero), so the arithmetic above holds for every valid choice of $a,b$.

Let's summarize:

The pair in (A) makes both terms on the left equal to $4f$ each, adding to $8f$.
The pair in (D) sends the $x$-term to zero and makes the $y$-term alone equal to $8f$.
(B) gives $0$ and (C) gives $2.5f$, neither of which reaches $8f$.

The correct options are (A) and (D).

\[ \boxed{\text{(A) and (D)}} \]

Answer 2

Rather than deriving one general algebraic condition first, we can plug each candidate pair $(\xi,\eta)$ straight into the original equation $af_{xx} + bf_{yy} = 8f$ and check the arithmetic directly, using $f_{xx} = 4\xi^2 f$ and $f_{yy} = \eta^2 f$ from differentiating $f = e^{2\xi x+\eta y}$ twice.

(A) $\xi = 1/\sqrt{a}$, $\eta = 2/\sqrt{b}$: Here $\xi^2 = 1/a$ and $\eta^2 = 4/b$. The left side becomes $a \cdot 4 \cdot (1/a) f + b \cdot (4/b) f = 4f + 4f = 8f$, which equals the right side exactly. This pair works.
(B) $\xi = 0$, $\eta = 0$: Both second derivative terms vanish since $\xi = \eta = 0$, so the left side is $0$, but the right side is $8f$, which is never zero for real $x,y$ (an exponential is always positive). This pair fails.
(C) $\xi = 1/\sqrt{2a}$, $\eta = 1/\sqrt{2b}$: Here $\xi^2 = 1/(2a)$ and $\eta^2 = 1/(2b)$. The left side is $a\cdot 4 \cdot \frac{1}{2a}f + b\cdot\frac{1}{2b}f = 2f + 0.5f = 2.5f$, which is not equal to $8f$. This pair fails.
(D) $\xi = 0$, $\eta = 2\sqrt{2/b}$: Here $\xi = 0$ kills the $x$-term entirely, and $\eta^2 = 4 \cdot (2/b) = 8/b$. The left side is $0 + b \cdot (8/b) f = 8f$, matching the right side exactly. This pair works.

Since $a$ and $b$ are distinct positive numbers, none of these divisions run into trouble (we never divide by zero), so the arithmetic above holds for every valid choice of $a,b$.

Let's summarize:

The pair in (A) makes both terms on the left equal to $4f$ each, adding to $8f$.
The pair in (D) sends the $x$-term to zero and makes the $y$-term alone equal to $8f$.
(B) gives $0$ and (C) gives $2.5f$, neither of which reaches $8f$.

The correct options are (A) and (D).

\[ \boxed{\text{(A) and (D)}} \]

Show Hint

The Correct Option is A, D

Solution and Explanation

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