(x^3+3xy^2)p+(y^3+3x^2y)q=2z(x^2+y^2)

Jasonbradley

4 min read Jun 17, 2024

Solving the Partial Differential Equation: (x^3+3xy^2)p+(y^3+3x^2y)q=2z(x^2+y^2)

This article will delve into the solution of the given partial differential equation (PDE):

(x^3+3xy^2)p+(y^3+3x^2y)q=2z(x^2+y^2)

where p = ∂z/∂x and q = ∂z/∂y.

Understanding the Equation

This PDE is a non-linear first-order PDE which is non-homogeneous. We can see this by observing the following:

Non-linear: The equation contains terms like x^3, y^3, x^2y, and xy^2, indicating that the equation is not linear in z, p, or q.
First-order: The highest order derivative is the first derivative (p and q).
Non-homogeneous: The right-hand side of the equation is not zero.

Method of Solution

This specific equation can be solved using the Lagrange's Method which involves finding the characteristics of the PDE.

Step 1: Writing the Auxiliary Equations

Lagrange's method utilizes the following auxiliary equations:

dx/(x^3 + 3xy^2) = dy/(y^3 + 3x^2y) = dz/(2z(x^2 + y^2))

Step 2: Solving the Auxiliary Equations

We need to find two independent solutions from these equations. We can simplify the equations by factoring out common terms:

dx/x(x^2 + 3y^2) = dy/y(y^2 + 3x^2) = dz/(2z(x^2 + y^2))

Now, we can solve the equations in pairs:

Pair 1: dx/x(x^2 + 3y^2) = dy/y(y^2 + 3x^2)

We can rearrange this as:
```
y(y^2 + 3x^2)dx = x(x^2 + 3y^2)dy
```
Integrating both sides, we get:
```
y^4/4 + x^2y^2 = C1
```
Pair 2: dx/x(x^2 + 3y^2) = dz/(2z(x^2 + y^2))

We can rearrange this as:
```
2(x^2 + y^2)dx = x(x^2 + 3y^2)dz/z
```
Integrating both sides, we get:
```
x^3/3 + x^2y^2/2 = C2z 
```

Step 3: Obtaining the General Solution

We have obtained two independent solutions, C1 and C2z. The general solution is obtained by eliminating C1 and C2, resulting in an equation of the form:

F(C1, C2z) = 0

where F is an arbitrary function.

Therefore, the general solution of the given PDE is:

F(y^4/4 + x^2y^2, x^3/3 + x^2y^2/2 - z) = 0

Conclusion

The given PDE is solved using Lagrange's method. The general solution obtained is an implicit equation involving an arbitrary function F, representing a family of solutions. This solution provides a framework for finding specific solutions depending on initial or boundary conditions.