Question:
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
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
\[ a\frac{\partial^2 f(x,y)}{\partial x^2} + b\frac{\partial^2 f(x,y)}{\partial y^2} = 8f(x,y) \]
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
Show Hint
Differentiate \(f = e^{2\xi x + \eta y}\) twice in \(x\) and \(y\), substitute into the PDE, and check which pairs satisfy \(4a\xi^2 + b\eta^2 = 8\).
Updated On: Jul 21, 2026
- \(\xi = \dfrac{1}{\sqrt{a}}, \ \eta = \dfrac{2}{\sqrt{b}}\)
- \(\xi = 0, \ \eta = 0\)
- \(\xi = \dfrac{1}{\sqrt{2a}}, \ \eta = \dfrac{1}{\sqrt{2b}}\)
- \(\xi = 0, \ \eta = 2\sqrt{\dfrac{2}{b}}\)
Show Solution
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The Correct Option is A, D
Solution and Explanation
Step 1: Differentiate the trial solution.
We are given \(f(x,y) = e^{2\xi x + \eta y}\). Differentiating with respect to \(x\) twice:
\[ \frac{\partial f}{\partial x} = 2\xi e^{2\xi x+\eta y}, \qquad \frac{\partial^2 f}{\partial x^2} = 4\xi^2 e^{2\xi x+\eta y} = 4\xi^2 f \]
Differentiating with respect to \(y\) twice:
\[ \frac{\partial f}{\partial y} = \eta e^{2\xi x+\eta y}, \qquad \frac{\partial^2 f}{\partial y^2} = \eta^2 e^{2\xi x+\eta y} = \eta^2 f \]
Step 2: Substitute into the PDE.
The PDE is \(a\dfrac{\partial^2 f}{\partial x^2} + b\dfrac{\partial^2 f}{\partial y^2} = 8f\). Plugging in Step 1:
\[ a(4\xi^2 f) + b(\eta^2 f) = 8f \]
Since \(f = e^{2\xi x+\eta y}\) is never zero, we can divide both sides by \(f\), leaving the condition
\[ 4a\xi^2 + b\eta^2 = 8 \]
Any pair \((\xi,\eta)\) that satisfies this equation makes \(f\) a genuine solution of the PDE; any pair that does not satisfy it fails.
Step 3: Test option (A): \(\xi=\dfrac{1}{\sqrt a},\ \eta=\dfrac{2}{\sqrt b}\).
\[ 4a\xi^2 + b\eta^2 = 4a\cdot\frac{1}{a} + b\cdot\frac{4}{b} = 4+4 = 8 \]
This matches, so option (A) is TRUE.
Step 4: Test option (B): \(\xi=0,\ \eta=0\).
\[ 4a(0)^2 + b(0)^2 = 0 \neq 8 \]
This does not match, so option (B) is FALSE.
Step 5: Test option (C): \(\xi=\dfrac{1}{\sqrt{2a}},\ \eta=\dfrac{1}{\sqrt{2b}}\).
\[ 4a\xi^2+b\eta^2 = 4a\cdot\frac{1}{2a} + b\cdot\frac{1}{2b} = 2+\frac12 = \frac52 \]
Since \(\dfrac52 \neq 8\), option (C) is FALSE.
Step 6: Test option (D): \(\xi=0,\ \eta=2\sqrt{\dfrac2b}\).
\[ 4a\xi^2+b\eta^2 = 4a(0)^2 + b\left(4\cdot\frac2b\right) = 0+8 = 8 \]
This matches, so option (D) is TRUE.
Final Answer:
Only the pairs in (A) and (D) satisfy the required condition \(4a\xi^2+b\eta^2=8\).
\[ \boxed{\text{(A) and (D)}} \]
We are given \(f(x,y) = e^{2\xi x + \eta y}\). Differentiating with respect to \(x\) twice:
\[ \frac{\partial f}{\partial x} = 2\xi e^{2\xi x+\eta y}, \qquad \frac{\partial^2 f}{\partial x^2} = 4\xi^2 e^{2\xi x+\eta y} = 4\xi^2 f \]
Differentiating with respect to \(y\) twice:
\[ \frac{\partial f}{\partial y} = \eta e^{2\xi x+\eta y}, \qquad \frac{\partial^2 f}{\partial y^2} = \eta^2 e^{2\xi x+\eta y} = \eta^2 f \]
Step 2: Substitute into the PDE.
The PDE is \(a\dfrac{\partial^2 f}{\partial x^2} + b\dfrac{\partial^2 f}{\partial y^2} = 8f\). Plugging in Step 1:
\[ a(4\xi^2 f) + b(\eta^2 f) = 8f \]
Since \(f = e^{2\xi x+\eta y}\) is never zero, we can divide both sides by \(f\), leaving the condition
\[ 4a\xi^2 + b\eta^2 = 8 \]
Any pair \((\xi,\eta)\) that satisfies this equation makes \(f\) a genuine solution of the PDE; any pair that does not satisfy it fails.
Step 3: Test option (A): \(\xi=\dfrac{1}{\sqrt a},\ \eta=\dfrac{2}{\sqrt b}\).
\[ 4a\xi^2 + b\eta^2 = 4a\cdot\frac{1}{a} + b\cdot\frac{4}{b} = 4+4 = 8 \]
This matches, so option (A) is TRUE.
Step 4: Test option (B): \(\xi=0,\ \eta=0\).
\[ 4a(0)^2 + b(0)^2 = 0 \neq 8 \]
This does not match, so option (B) is FALSE.
Step 5: Test option (C): \(\xi=\dfrac{1}{\sqrt{2a}},\ \eta=\dfrac{1}{\sqrt{2b}}\).
\[ 4a\xi^2+b\eta^2 = 4a\cdot\frac{1}{2a} + b\cdot\frac{1}{2b} = 2+\frac12 = \frac52 \]
Since \(\dfrac52 \neq 8\), option (C) is FALSE.
Step 6: Test option (D): \(\xi=0,\ \eta=2\sqrt{\dfrac2b}\).
\[ 4a\xi^2+b\eta^2 = 4a(0)^2 + b\left(4\cdot\frac2b\right) = 0+8 = 8 \]
This matches, so option (D) is TRUE.
Final Answer:
Only the pairs in (A) and (D) satisfy the required condition \(4a\xi^2+b\eta^2=8\).
\[ \boxed{\text{(A) and (D)}} \]
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