Solving the Differential Equation (x+y)^2dx + (2xy + x^2 - 1)dy = 0
This article will explore the solution to the given differential equation:
(x+y)^2dx + (2xy + x^2 - 1)dy = 0
This equation is a first-order, non-linear differential equation. We can solve this equation using the following steps:
1. Identifying the Type of Equation
The given equation is an exact differential equation. This means that it can be written in the form:
M(x, y)dx + N(x, y)dy = 0
where the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x:
∂M/∂y = ∂N/∂x
In our case:
- M(x, y) = (x+y)^2
- N(x, y) = 2xy + x^2 - 1
Let's verify if it's exact:
- ∂M/∂y = 2(x+y)
- ∂N/∂x = 2y + 2x
We can see that ∂M/∂y = ∂N/∂x, confirming that the equation is exact.
2. Finding the Solution
Since the equation is exact, we know 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 M(x, y) with respect to x or N(x, y) with respect to y. Let's integrate M(x, y):
∫M(x, y)dx = ∫(x+y)^2 dx = (1/3)(x+y)^3 + g(y)
Here, g(y) is an arbitrary function of y that arises from the integration.
Now, we can differentiate this result with respect to y and equate it to N(x, y):
∂/∂y [(1/3)(x+y)^3 + g(y)] = 2xy + x^2 - 1
** (x+y)^2 + g'(y) = 2xy + x^2 - 1**
This implies that g'(y) = -1 and therefore, g(y) = -y + C, where C is a constant.
Finally, we can substitute the value of g(y) back into the expression for u(x, y):
u(x, y) = (1/3)(x+y)^3 - y + C
The general solution to the given differential equation is therefore:
(1/3)(x+y)^3 - y = C
Conclusion
We have successfully solved the given differential equation using the method for exact differential equations. The solution is an implicit function of x and y, representing a family of curves defined by the equation:
(1/3)(x+y)^3 - y = C
This solution highlights the importance of recognizing the type of differential equation and employing the appropriate solution techniques.