Question

Consider the PDE \[ a\,\frac{\partial^{2} f(x,y)}{\partial x^{2}} \;+\; b\,\frac{\partial^{2} f(x,y)}{\partial y^{2}} \;=\; f(x,y), \] where $a$ and $b$ are distinct positive real numbers. Select the combinations of real parameters $\xi$ and $\eta$ such that \[ f(x,y)=e^{\,(\xi x+\eta y)} \] is a solution of the PDE.

• A. $\displaystyle \xi=\frac{1}{\sqrt{2a}},\quad \eta=\frac{1}{\sqrt{2b}}$
• B. $\displaystyle \xi=\frac{1}{\sqrt{a}},\quad \eta=0$
• C. $\displaystyle \xi=0,\quad \eta=0$
• D. $\displaystyle \xi=\frac{1}{\sqrt{a}},\quad \eta=\frac{1}{\sqrt{b}}$

Hint:

For exponential trial solutions of linear PDEs, substituting $f=e^{\xi x+\eta y}$ always reduces the equation to an algebraic condition on $\xi$ and $\eta$.

Answer 1

The Correct Option is A, B

Step 1: Substitute the assumed solution \[ f(x,y) = e^{\xi x + \eta y}. \] Compute the required derivatives: \[ \frac{\partial f}{\partial x} = \xi e^{\xi x + \eta y}, \qquad \frac{\partial^{2} f}{\partial x^{2}} = \xi^{2} e^{\xi x + \eta y}, \] \[ \frac{\partial f}{\partial y} = \eta e^{\xi x + \eta y}, \qquad \frac{\partial^{2} f}{\partial y^{2}} = \eta^{2} e^{\xi x + \eta y}. \] Step 2: Substitute into the PDE. \[ a\xi^{2} e^{\xi x+\eta y} + b\eta^{2} e^{\xi x+\eta y} = e^{\xi x+\eta y}. \] Divide both sides by the exponential (never zero): \[ a\xi^{2} + b\eta^{2} = 1. \] Step 3: Check each option.
Option (A): \[ \xi=\frac{1}{\sqrt{2a}},\quad \eta=\frac{1}{\sqrt{2b}}. \] Compute: \[ a\xi^{2}=a\left(\frac{1}{2a}\right)=\frac{1}{2}, \quad b\eta^{2}=b\left(\frac{1}{2b}\right)=\frac{1}{2}. \] Sum: \[ a\xi^{2}+b\eta^{2}=\frac{1}{2}+\frac{1}{2}=1. \] So (A) satisfies the PDE.
Option (B): \[ \xi=\frac{1}{\sqrt{a}},\quad \eta=0. \] Compute: \[ a\xi^{2}=a\left(\frac{1}{a}\right)=1, \quad b\eta^{2}=0. \] So \[ a\xi^{2}+b\eta^{2}=1. \] Thus (B) is valid.
Option (C): \[ \xi=0,\;\eta=0 \quad\Rightarrow\quad a\xi^{2}+b\eta^{2}=0\ne 1. \] Not valid.
Option (D): \[ \xi=\frac{1}{\sqrt{a}},\;\eta=\frac{1}{\sqrt{b}} \] Compute: \[ a\xi^{2}=1,\qquad b\eta^{2}=1. \] Sum: \[ 1+1=2\ne 1. \] Not valid.
Thus the only correct combinations are (A) and (B).
Final Answer: (A), (B)