(1+y^2)(1+logx)dx+xdy=0

Jasonbradley

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Jun 16, 2024

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Solving the Differential Equation (1+y^2)(1+logx)dx + xdy = 0

This article will guide you through the process of solving the given differential equation:

(1+y^2)(1+logx)dx + xdy = 0

Identifying the Type of Differential Equation

Firstly, we need to determine the type of differential equation we are dealing with. Observing the equation, we can see that it is a first-order, non-linear, and exact differential equation.

• First-order because the highest derivative present is the first derivative of y (dy/dx).
• Non-linear because the equation contains a term with y^2.
• Exact because it can be expressed in the form M(x,y)dx + N(x,y)dy = 0, where ∂M/∂y = ∂N/∂x.

Solving the Exact Differential Equation

1. Verification of Exactness:

   • M(x,y) = (1+y^2)(1+logx)
   • N(x,y) = x
   • ∂M/∂y = 2y
   • ∂N/∂x = 1

   Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact in its current form.

2. Finding the Integrating Factor:

   To make the equation exact, we need to find an integrating factor. Let's assume the integrating factor is a function of x only, denoted by μ(x). Multiplying the entire equation by μ(x), we get:

   μ(x)(1+y^2)(1+logx)dx + μ(x)xdy = 0

   For this equation to be exact, we need:

   ∂[μ(x)(1+y^2)(1+logx)]/∂y = ∂[μ(x)x]/∂x

   Simplifying the above equation, we get:

   2yμ(x)(1+logx) = μ'(x)x + μ(x)

   Solving this differential equation for μ(x), we find the integrating factor to be μ(x) = x.

3. Multiplying by the Integrating Factor:

   Multiplying the original equation by x, we get:

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

   Now, this equation is exact, as ∂[x(1+y^2)(1+logx)]/∂y = ∂[x^2]/∂x = 2x.

4. Solving the Exact Equation:

   Since the equation is exact, there exists a function φ(x,y) such that:

   ∂φ/∂x = x(1+y^2)(1+logx) ∂φ/∂y = x^2

   Integrating the first equation with respect to x, we get:

   φ(x,y) = (1/2)x^2(1+y^2)(1+logx) + g(y)

   where g(y) is an arbitrary function of y.

   Differentiating this expression with respect to y and equating it to the second equation, we get:

   x^2y + g'(y) = x^2

   Solving for g'(y), we get g'(y) = 0. Integrating g'(y), we get g(y) = C, where C is an arbitrary constant.

   Therefore, the general solution of the given differential equation is:

   (1/2)x^2(1+y^2)(1+logx) + C = 0

Conclusion

By following these steps, we have successfully solved the given differential equation. The solution is a function that implicitly defines y in terms of x, and it involves an arbitrary constant C. This constant represents the family of solutions to the original differential equation.