Solving the Differential Equation (x+4)(y^2+1)dx + y(x^2+3x+2)dy = 0
This article will guide you through solving the given differential equation:
(x+4)(y^2+1)dx + y(x^2+3x+2)dy = 0
Identifying the Type of Differential Equation
First, we need to identify the type of differential equation we are dealing with. Looking at the equation, we can see that it is a firstorder differential equation, because it involves only the first derivatives of x and y. Furthermore, it is a nonlinear equation due to the presence of the terms y^2 and x^2.
This equation is also exact, which means it can be written in the form:
M(x, y)dx + N(x, y)dy = 0
where ∂M/∂y = ∂N/∂x.
Solving the Exact Differential Equation
To solve an exact differential equation, we follow these steps:

Check for exactness: Verify that ∂M/∂y = ∂N/∂x. In our case,
 M(x, y) = (x+4)(y^2+1)
 N(x, y) = y(x^2+3x+2)
Calculating the partial derivatives:
 ∂M/∂y = 2y(x+4)
 ∂N/∂x = 2xy + 3y
We can see that ∂M/∂y = ∂N/∂x, confirming that the equation is exact.

Find the potential function: Integrate M(x, y) with respect to x, treating y as a constant:
∫M(x, y)dx = ∫(x+4)(y^2+1)dx = (1/2)x^2(y^2+1) + 4x(y^2+1) + C(y)
Here, C(y) is an arbitrary function of y that arises from the integration.

Differentiate the potential function with respect to y:
∂/∂y [(1/2)x^2(y^2+1) + 4x(y^2+1) + C(y)] = x^2y + 8xy + C'(y)

Equate this to N(x, y) and solve for C'(y):
x^2y + 8xy + C'(y) = y(x^2+3x+2) C'(y) = 5xy + 2y
Integrating both sides with respect to y:
C(y) =  (5/2)y^2 + y^2 + K
Where K is an arbitrary constant.

The solution is given by:
(1/2)x^2(y^2+1) + 4x(y^2+1)  (5/2)y^2 + y^2 + K = 0
Simplifying:
x^2(y^2+1) + 8x(y^2+1)  3y^2 + 2K = 0
Conclusion
The solution to the differential equation (x+4)(y^2+1)dx + y(x^2+3x+2)dy = 0 is x^2(y^2+1) + 8x(y^2+1)  3y^2 + 2K = 0, where K is an arbitrary constant.