dx%2B(2xy%5E2-x%5E3)dy%3D0.jpeg "(y^3-2x^2y)dx+(2xy^2-x^3)dy=0")
Solving the Differential Equation: (y^3 - 2x^2y)dx + (2xy^2 - x^3)dy = 0
This article will guide you through the process of solving the given differential equation:
(y^3 - 2x^2y)dx + (2xy^2 - x^3)dy = 0
1. Identifying the Type of Differential Equation
Firstly, we need to identify the type of differential equation we are dealing with. This equation is an exact differential equation.
To verify this, we can check if the following condition holds:
∂M/∂y = ∂N/∂x
Where:
- M = y^3 - 2x^2y
- N = 2xy^2 - x^3
Calculating the partial derivatives:
- ∂M/∂y = 3y^2 - 2x^2
- ∂N/∂x = 2y^2 - 3x^2
Since ∂M/∂y = ∂N/∂x, the equation is indeed an exact differential equation.
2. Finding the Solution
To solve an exact differential equation, we need to find a function u(x, y) such that:
- ∂u/∂x = M
- ∂u/∂y = N
Integrating the first equation with respect to x, we get:
u(x, y) = ∫(y^3 - 2x^2y)dx = xy^3 - (2/3)x^3y + h(y)
Where h(y) is an arbitrary function of y.
Now, we differentiate u(x, y) with respect to y:
∂u/∂y = 3xy^2 - (2/3)x^3 + h'(y)
Equating this to N, we get:
3xy^2 - (2/3)x^3 + h'(y) = 2xy^2 - x^3
Solving for h'(y):
h'(y) = (1/3)x^3
Integrating this with respect to y, we get:
h(y) = (1/9)x^3y + C
Where C is an arbitrary constant.
Substituting h(y) back into the expression for u(x, y), we obtain the general solution:
u(x, y) = xy^3 - (2/3)x^3y + (1/9)x^3y + C = xy^3 - (5/9)x^3y + C
Therefore, the general solution to the given differential equation is xy^3 - (5/9)x^3y = C.
Conclusion
By recognizing the equation as an exact differential equation, we were able to find the general solution through a straightforward process of finding a potential function u(x, y) and incorporating the arbitrary constant C.