Question

The type of partial differential equation \(\frac{{{\partial ^2}P}}{{\partial {x^2}}} + \frac{{{\partial ^2}P}}{{\partial {y^2}}} + 3\frac{{{\partial ^2}P}}{{\partial x\partial y}} + 2\frac{{\partial P}}{{\partial x}} - \frac{{\partial P}}{{\partial y}} = 0\) is 
1. Elliptic
2. Parabolic
3. Hyperbolic
4. None of these

Answer 1

Correct Answer - Option 3 : Hyperbolic

Concept:

The general form of 2^nd order linear partial differential equation is given by

\(A\frac{{{\partial ^2}U}}{{\partial {X^2}}} + B\frac{{{\partial ^2}U}}{{\partial X\partial Y}} + C\frac{{{\partial ^2}U}}{{\partial {Y^2}}} + f\left( {X,Y,Z\frac{{\partial U}}{{\partial X}},\frac{{\partial U}}{{\partial Y}}\;} \right) = 0\)

The above equation is said to be parabolic, elliptic, and hyperbolic based on the following,

1. Parabolic     = B² – 4AC = 0
2. Elliptic          = B² – 4AC < 0
3. Hyperbolic   = B² – 4AC > 0

Calculation:

Given equation,

\(\frac{{{\partial ^2}P}}{{\partial {X^2}}} + \frac{{{\partial ^2}P}}{{\partial {Y^2}}} + 3\frac{{{\partial ^2}P}}{{\partial X\partial Y}} + 2\frac{{\partial P}}{{\partial X}} - \frac{{\partial P}}{{\partial Y}} = 0\)

Compare the given equation with the following general equation

\(A\frac{{{\partial ^2}U}}{{\partial {X^2}}} + B\frac{{{\partial ^2}U}}{{\partial X\partial Y}} + C\frac{{{\partial ^2}U}}{{\partial {Y^2}}} + f\left( {X,Y,Z\frac{{\partial U}}{{\partial X}},\frac{{\partial U}}{{\partial Y}}\;} \right) = 0\)

Hence we get,

A = 1, B = 3, C = 1

Put value of A,B, and C in the following equation

B² – 4AC

∴ B² – 4AC = 3² – (4× 1 × 1) = 5

We can see B² – 4AC is greater than Zero, i.e

B² – 4AC > 0

Hence the given partial differential equation is Hyperbolic