Question

For all real values of \( x \) and \( y \), the partial differential equation in terms of \( \psi(x, y) \) given by \[ \frac{\partial^2 \psi}{\partial x^2} + 2 \frac{\partial^2 \psi}{\partial x \partial y} - 3 \frac{\partial^2 \psi}{\partial y^2} = 0 \] is ________

• A. hyperbolic
• B. parabolic
• C. elliptic
• D. elliptic within the region, \( x^2 - y<0 \)

Hint:

To classify a second-order linear PDE, calculate the discriminant \( D = B^2 - AC \). - \( D>0 \): hyperbolic - \( D = 0 \): parabolic - \( D<0 \): elliptic

Answer 1

The Correct Option is A

Step 1: Standard form of a second-order PDE.
The given second-order partial differential equation is: \[ \frac{\partial^2 \psi}{\partial x^2} + 2 \frac{\partial^2 \psi}{\partial x \partial y} - 3 \frac{\partial^2 \psi}{\partial y^2} = 0. \] This is a second-order linear partial differential equation with constant coefficients. 
Step 2: Classification of the PDE.
The general form of a second-order linear partial differential equation is: \[ A \frac{\partial^2 \psi}{\partial x^2} + 2B \frac{\partial^2 \psi}{\partial x \partial y} + C \frac{\partial^2 \psi}{\partial y^2} = 0, \] where \( A = 1 \), \( B = 1 \), and \( C = -3 \) in the given equation. The discriminant \( D \) is given by: \[ D = B^2 - AC = 1^2 - (1)(-3) = 1 + 3 = 4. \] Step 3: Classifying the equation.
If \( D>0 \), the equation is hyperbolic.
If \( D = 0 \), the equation is parabolic.
If \( D<0 \), the equation is elliptic.
Here, \( D = 4>0 \), so the given equation is hyperbolic. 
Step 4: Conclusion.
Therefore, the correct classification of the partial differential equation is: \[ \boxed{{hyperbolic}}. \]