Question:

For the following partial differential equation, x2fx2+y2fy2=x2+y22 x \frac{\partial^2 f}{\partial x^2} + y \frac{\partial^2 f}{\partial y^2} = \frac{x^2 + y^2}{2} which of the following option(s) is/are CORRECT?

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To classify second-order PDEs, examine the coefficients of the highest-order terms. Elliptic equations have coefficients of the same sign, while hyperbolic equations have coefficients of opposite signs.
Updated On: Jan 24, 2025
  • elliptic for x>0 x>0 and y>0 y>0
  • parabolic for x>0 x>0 and y>0 y>0
  • elliptic for x=0 x = 0 and y>0 y>0
  • hyperbolic for x<0 x<0 and y>0 y>0
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The Correct Option is A

Solution and Explanation

Step 1: The given partial differential equation can be rewritten in the general form: a2fx2+b2fy2=f(x,y) a \frac{\partial^2 f}{\partial x^2} + b \frac{\partial^2 f}{\partial y^2} = f(x, y) where: a=x,b=y a = x, \quad b = y Step 2: The classification of the PDE depends on the sign of the coefficients: If a>0 a>0 and b>0 b>0 , the equation is elliptic. If a<0 a<0 and b>0 b>0 , the equation is hyperbolic. If either a a or b b equals zero, the equation may degenerate to a different type. Step 3: Evaluating the given options: For x>0 x>0 and y>0 y>0 , both coefficients are positive, so the equation is elliptic. Option (A) is correct. For x<0 x<0 and y>0 y>0 , the coefficients have opposite signs, making the equation hyperbolic. Option (D) is correct. Conclusion: The correct answers are (A) and (D), confirming the classification of the given differential equation.
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