4 min read Jun 17, 2024

Solving the Partial Differential Equation (x^2+y^2)p+2xyq=(x+y)z

This article explores the solution of the given partial differential equation (PDE):


where 'p' represents the partial derivative of 'z' with respect to 'x' (∂z/∂x), and 'q' represents the partial derivative of 'z' with respect to 'y' (∂z/∂y). This equation falls under the category of non-linear first-order PDEs.

Identifying the Type

This PDE is non-linear due to the presence of the product term '2xyq'. It's also a first-order PDE because the highest order derivative present is the first derivative (p and q).

Solution Approach

A common approach to solving non-linear first-order PDEs is using the Lagrange's method or the Charpit's method.

1. Lagrange's Method:

This method involves finding two independent solutions (u and v) of the auxiliary equations:

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

Solving the first two equations, we get:

(x^2 + y^2)dy = 2xydx

This can be solved by separation of variables, leading to:

y^2 / x^2 = C1

where C1 is an arbitrary constant.

Similarly, solving the second and third equations, we get:

(x+y)dz = z(2xydy)

This also can be solved by separation of variables, resulting in:

z / (x^2 * y) = C2

where C2 is another arbitrary constant.

Finally, we can express the general solution of the PDE as:

F(C1, C2) = 0 

where F is an arbitrary function. This means the solution can be expressed as:

F( y^2 / x^2,  z / (x^2 * y)) = 0

2. Charpit's Method:

This method involves finding a complete integral of the PDE. This can be achieved by introducing auxiliary equations:

dp = (x+y)z * ds
dq = -(x^2+y^2)z * ds
dr = p(x^2+y^2) * ds + q(2xy) * ds

where 's' is an auxiliary variable.

Solving this system of equations can be challenging and may require specific techniques to find a complete integral.


Solving non-linear first-order PDEs like (x^2+y^2)p+2xyq=(x+y)z can be complex. Lagrange's method and Charpit's method provide frameworks to approach the solution, although finding a complete integral might require advanced techniques. Remember, the solution will usually involve an arbitrary function, leading to a family of solutions.

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