(x+4)(y^2+1)dx+y(x^2+3x+2)dy=0

Jasonbradley

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Jun 16, 2024

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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 first-order differential equation, because it involves only the first derivatives of x and y. Furthermore, it is a non-linear 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:

1. 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.

2. 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.

3. 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)

4. 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.

5. 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.