(3x^2-2xy+2)dx+(6y^2-x^2+3)dy=0

Jasonbradley

3 min read Jun 16, 2024

Solving the Differential Equation: (3x^2 - 2xy + 2)dx + (6y^2 - x^2 + 3)dy = 0

This article explores the solution of the given differential equation:

(3x^2 - 2xy + 2)dx + (6y^2 - x^2 + 3)dy = 0

This equation is a first-order, non-linear differential equation. We can solve it using the following steps:

1. Checking for Exactness

Firstly, we need to check if the equation is exact. A differential equation of the form:

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

is exact if:

∂M/∂y = ∂N/∂x

In our case, we have:

M(x,y) = 3x^2 - 2xy + 2
N(x,y) = 6y^2 - x^2 + 3

Let's calculate the partial derivatives:

∂M/∂y = -2x
∂N/∂x = -2x

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

2. Finding the Potential Function

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

∂φ/∂x = M(x,y)
∂φ/∂y = N(x,y)

To find φ(x,y), we integrate M(x,y) with respect to x:

φ(x,y) = ∫(3x^2 - 2xy + 2)dx = x^3 - x^2y + 2x + g(y)

Here, g(y) is an arbitrary function of y, which arises from the integration.

Now, we differentiate φ(x,y) with respect to y and equate it to N(x,y):

∂φ/∂y = -x^2 + g'(y) = 6y^2 - x^2 + 3

This implies g'(y) = 6y^2 + 3. Integrating with respect to y, we get:

g(y) = 2y^3 + 3y + C

Where C is an arbitrary constant.

Therefore, the potential function is:

φ(x,y) = x^3 - x^2y + 2x + 2y^3 + 3y + C

3. The General Solution

The general solution of the differential equation is given by:

φ(x,y) = C1

where C1 is a constant.

Substituting the expression for φ(x,y), we get:

x^3 - x^2y + 2x + 2y^3 + 3y = C

This is the general solution of the given differential equation.

Conclusion

The given differential equation is exact and we have successfully solved it using the method of exact equations. The general solution of the equation is x^3 - x^2y + 2x + 2y^3 + 3y = C.