(2xy+3)dx+(x^2-1)dy=0

3 min read Jun 16, 2024

Solving the Differential Equation (2xy + 3)dx + (x^2 - 1)dy = 0

This article will guide you through the process of solving the differential equation (2xy + 3)dx + (x^2 - 1)dy = 0. We will determine if the equation is exact, and if not, find an integrating factor to make it exact. Finally, we will solve the resulting equation.

Identifying the Equation Type

The given equation is a first-order differential equation. It is in the form:

M(x, y) dx + N(x, y) dy = 0

Where:

M(x, y) = 2xy + 3
N(x, y) = x^2 - 1

Checking for Exactness

A differential equation is considered exact if:

∂M/∂y = ∂N/∂x

Let's calculate the partial derivatives:

∂M/∂y = 2x
∂N/∂x = 2x

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

Solving the Exact Equation

To solve an exact differential equation, we follow these steps:

Find a potential function Ψ(x, y):
- ∂Ψ/∂x = M(x, y)
- ∂Ψ/∂y = N(x, y)
Integrate ∂Ψ/∂x with respect to x, treating y as a constant: ∫ (2xy + 3) dx = x^2y + 3x + g(y)

Here, g(y) is an arbitrary function of y that arises from the integration.
Differentiate the result with respect to y and equate it to N(x, y): ∂/∂y (x^2y + 3x + g(y)) = x^2 + g'(y) = x^2 - 1
Solve for g'(y) and integrate to find g(y): g'(y) = -1 g(y) = -y + C
Substitute the value of g(y) back into the potential function Ψ(x, y): Ψ(x, y) = x^2y + 3x - y + C
The general solution to the differential equation is given by Ψ(x, y) = C': x^2y + 3x - y = C'

Where C' is an arbitrary constant.

Conclusion

Therefore, the general solution to the differential equation (2xy + 3)dx + (x^2 - 1)dy = 0 is x^2y + 3x - y = C', where C' is an arbitrary constant.