(3x^2+6xy^2)dx+(6x^2y+4y^3)dy=0

3 min read Jun 16, 2024
(3x^2+6xy^2)dx+(6x^2y+4y^3)dy=0

Solving the Differential Equation (3x^2+6xy^2)dx+(6x^2y+4y^3)dy=0

This article will guide you through solving the given differential equation:

(3x^2+6xy^2)dx+(6x^2y+4y^3)dy=0

We will employ the method of exact differential equations to find the solution.

1. Identifying the Exact Differential Equation

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

∂M/∂y = ∂N/∂x

Let's apply this to our equation:

  • M(x, y) = 3x^2 + 6xy^2
  • N(x, y) = 6x^2y + 4y^3

Calculate the partial derivatives:

  • ∂M/∂y = 12xy
  • ∂N/∂x = 12xy

Since ∂M/∂y = ∂N/∂x, we can confirm that the given differential equation is indeed exact.

2. Finding the Solution

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

  • ∂u/∂x = M(x, y)
  • ∂u/∂y = N(x, y)

To find u(x, y), we can integrate either ∂u/∂x or ∂u/∂y. Let's integrate ∂u/∂x:

u(x, y) = ∫(3x^2 + 6xy^2) dx = x^3 + 3x^2y^2 + C(y)

Here, C(y) is an arbitrary function of y since we integrated with respect to x.

Now, differentiate u(x, y) with respect to y:

∂u/∂y = 6xy^2 + C'(y)

We know that ∂u/∂y = N(x, y), so:

6xy^2 + C'(y) = 6x^2y + 4y^3

Comparing both sides, we get:

C'(y) = 4y^3

Integrating this equation with respect to y, we obtain:

C(y) = y^4 + K

Where K is a constant of integration.

Therefore, the solution to the given differential equation is:

u(x, y) = x^3 + 3x^2y^2 + y^4 + K = C

Where C is a constant (C = K).

Conclusion

By applying the method of exact differential equations, we have successfully solved the given differential equation. The solution is expressed as an implicit equation representing a family of curves, each corresponding to a specific value of the constant C.

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