(xy^2+x-2y+3)dx+x^2ydy=2(x+y)dy

4 min read Jun 17, 2024

Solving the Differential Equation (xy^2 + x - 2y + 3)dx + x^2ydy = 2(x + y)dy

This article will explore the solution process for the given differential equation:

(xy^2 + x - 2y + 3)dx + x^2ydy = 2(x + y)dy

This equation is a first-order, non-linear, non-exact differential equation. We can use the following steps to find its solution:

1. Rearrange the Equation

First, we need to rewrite the equation in a more standard form. To do this, we'll group the terms with dx and dy together:

(xy^2 + x - 2y + 3)dx + (x^2y - 2x - 2y)dy = 0

2. Check for Exactness

A differential equation of the form M(x,y)dx + N(x,y)dy = 0 is exact if:

∂M/∂y = ∂N/∂x

In our equation, we have:

M(x,y) = xy^2 + x - 2y + 3
N(x,y) = x^2y - 2x - 2y

Calculating the partial derivatives:

∂M/∂y = 2xy - 2
∂N/∂x = 2xy - 2

Since ∂M/∂y = ∂N/∂x, the equation is exact.

3. Find the Solution

Since the equation is exact, there exists a function u(x, y) such that:

∂u/∂x = M(x,y) and ∂u/∂y = N(x,y)

To find u(x,y), we can integrate M(x,y) with respect to x and N(x,y) with respect to y, and then compare the results:

∫M(x,y) dx = ∫(xy^2 + x - 2y + 3) dx = (1/2)x^2y^2 + (1/2)x^2 - 2xy + 3x + C(y)
∫N(x,y) dy = ∫(x^2y - 2x - 2y) dy = (1/2)x^2y^2 - 2xy - y^2 + D(x)

Comparing the results, we see that C(y) = -y^2 and D(x) = (1/2)x^2 + 3x.

Therefore, the solution to the differential equation is given by:

u(x,y) = (1/2)x^2y^2 + (1/2)x^2 - 2xy + 3x - y^2 = C

where C is an arbitrary constant.

Conclusion

We successfully found the general solution to the given differential equation by first verifying its exactness, and then finding the potential function u(x,y). This process demonstrates a common method for solving exact differential equations, which are frequently encountered in various areas of physics and engineering.