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Solving the Differential Equation (2x^3-xy^2-2y+3)dx-(x^2y+2x)dy=0
This article will guide you through solving the differential equation:
(2x^3 - xy^2 - 2y + 3)dx - (x^2y + 2x)dy = 0
This equation is a non-exact differential equation. To solve it, we'll follow these steps:
1. Identify the form
The given equation is in the form:
M(x, y)dx + N(x, y)dy = 0
Where:
- M(x, y) = 2x^3 - xy^2 - 2y + 3
- N(x, y) = - (x^2y + 2x)
2. Check for exactness
A differential equation is exact if:
∂M/∂y = ∂N/∂x
Let's calculate the partial derivatives:
- ∂M/∂y = -2xy - 2
- ∂N/∂x = -2xy - 2
Since ∂M/∂y = ∂N/∂x, the equation is exact.
3. Find the potential function
An exact differential equation is derived from a potential function, u(x, y), such that:
- ∂u/∂x = M(x, y)
- ∂u/∂y = N(x, y)
To find u(x, y):
-
Integrate M(x, y) with respect to x, treating y as a constant:
u(x, y) = ∫(2x^3 - xy^2 - 2y + 3)dx = (1/2)x^4 - (1/2)x^2y^2 - 2xy + 3x + h(y)
Here, h(y) is an arbitrary function of y, as the integration is done with respect to x.
-
Differentiate u(x, y) with respect to y:
∂u/∂y = -x^2y - 2x + h'(y)
-
Equate this to N(x, y):
-x^2y - 2x + h'(y) = -x^2y - 2x
-
Therefore, h'(y) = 0, which means h(y) = C (a constant).
4. The solution
The solution to the exact differential equation is given by:
u(x, y) = C
Substituting the found potential function:
(1/2)x^4 - (1/2)x^2y^2 - 2xy + 3x + C = 0
This is the implicit solution to the given differential equation.
Conclusion
We successfully solved the non-exact differential equation (2x^3 - xy^2 - 2y + 3)dx - (x^2y + 2x)dy = 0 by identifying it as an exact equation, finding the potential function, and expressing the solution in implicit form. Remember that the process involves checking for exactness, finding the potential function, and expressing the solution in terms of the potential function.