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Solving the Differential Equation (1 + y^2 + xy^2)dx + (x^2y + y + 2xy)dy = 0
This article will guide you through the process of solving the given differential equation:
(1 + y^2 + xy^2)dx + (x^2y + y + 2xy)dy = 0
This equation is classified as an exact differential equation because it can be expressed as the total differential of a function.
Identifying the Exactness
To verify if the equation is exact, we need to check if the following condition holds:
∂M/∂y = ∂N/∂x
Where:
- M = 1 + y^2 + xy^2
- N = x^2y + y + 2xy
Calculating the partial derivatives:
- ∂M/∂y = 2y + 2xy
- ∂N/∂x = 2xy + 2y
Since ∂M/∂y = ∂N/∂x, we confirm that the given differential equation is exact.
Finding the Solution
Now, let's find the solution. We know that there exists a function u(x, y) such that:
du = Mdx + Ndy
Therefore, we can obtain u by integrating M with respect to x and N with respect to y.
1. Integrating M with respect to x:
∫(1 + y^2 + xy^2)dx = x + xy^2 + (1/2)x^2y^2 + g(y)
Where g(y) is an arbitrary function of y.
2. Integrating N with respect to y:
∫(x^2y + y + 2xy)dy = (1/2)x^2y^2 + (1/2)y^2 + xy^2 + h(x)
Where h(x) is an arbitrary function of x.
3. Combining the results:
Comparing both results, we can see that g(y) = (1/2)y^2 and h(x) = x. Therefore, the general solution of the differential equation is:
u(x, y) = x + xy^2 + (1/2)x^2y^2 + (1/2)y^2 = C
Where C is an arbitrary constant.
Conclusion
We have successfully solved the given differential equation, proving it to be exact and obtaining its general solution. Remember, this method can be applied to other similar exact differential equations.