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(2xy+y-tan Y)dx+(x^(2)-x Tan^(2)y+sec^(2)y)dy=0

5 min read Jun 16, 2024
(2xy+y-tan Y)dx+(x^(2)-x Tan^(2)y+sec^(2)y)dy=0

Solving the Differential Equation (2xy + y - tan y)dx + (x^2 - x tan^2 y + sec^2 y)dy = 0

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

(2xy + y - tan y)dx + (x^2 - x tan^2 y + sec^2 y)dy = 0

Identifying the Type of Differential Equation

The given equation is a first-order, non-linear, exact differential equation. Here's why:

  • First-order: The highest derivative present is the first derivative (dy/dx).

  • Non-linear: The equation contains terms like xy, tan^2 y, and sec^2 y, which are not linear.

  • Exact: A differential equation of the form M(x, y)dx + N(x, y)dy = 0 is considered exact if ∂M/∂y = ∂N/∂x. Let's verify if our equation satisfies this condition:

    • M(x, y) = 2xy + y - tan y

    • N(x, y) = x^2 - x tan^2 y + sec^2 y

    • ∂M/∂y = 2x + 1 - sec^2 y

    • ∂N/∂x = 2x - tan^2 y

    Since ∂M/∂y = ∂N/∂x, the equation is exact.

Solving the Exact Equation

To solve an exact differential equation, we follow these steps:

  1. Find the potential function (or solution) 'u(x, y)'. This function satisfies the following conditions:

    • ∂u/∂x = M(x, y)
    • ∂u/∂y = N(x, y)

  2. Integrate either equation (∂u/∂x or ∂u/∂y) with respect to its corresponding variable.

  3. Determine the constant of integration by differentiating the result with respect to the other variable and comparing it with the other equation (∂u/∂y or ∂u/∂x).

  4. The final solution is given by u(x, y) = C, where C is an arbitrary constant.

Solution Steps

  1. Find u(x, y):

    • Integrating ∂u/∂x = M(x, y) with respect to x, we get:

      • u(x, y) = ∫(2xy + y - tan y)dx = x^2y + xy - x tan y + g(y)

      Here, g(y) is the constant of integration that could be a function of y.

  2. Determine g(y):

    • Differentiate the obtained u(x, y) with respect to y:

      • ∂u/∂y = x^2 + x - x sec^2 y + g'(y)

    • Compare this with N(x, y):

      • x^2 + x - x sec^2 y + g'(y) = x^2 - x tan^2 y + sec^2 y

    • Solving for g'(y), we get:

      • g'(y) = sec^2 y - x tan^2 y + x sec^2 y

    • Integrating both sides with respect to y:

      • g(y) = tan y - (x/3) tan^3 y + x tan y + C

    • Therefore, the potential function u(x, y) is:

      • u(x, y) = x^2y + xy - x tan y + tan y - (x/3) tan^3 y + x tan y + C
      • u(x, y) = x^2y + xy + tan y - (x/3) tan^3 y + C

  3. The solution:

    • The solution to the differential equation is given by:

      • u(x, y) = C
      • x^2y + xy + tan y - (x/3) tan^3 y = C

Conclusion

By recognizing the given equation as an exact differential equation and applying the appropriate steps, we have successfully solved it to obtain the solution: x^2y + xy + tan y - (x/3) tan^3 y = C. This solution represents a family of curves that satisfy the original differential equation.

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