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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.