(x+2y)dx+(2x+y)dy=0 Exact Equation

4 min read Jun 16, 2024
(x+2y)dx+(2x+y)dy=0 Exact Equation

Solving the Exact Differential Equation (x+2y)dx + (2x+y)dy = 0

This article explores the solution to the differential equation (x+2y)dx + (2x+y)dy = 0 using the method of exact equations.

Understanding Exact Equations

A differential equation of the form M(x,y)dx + N(x,y)dy = 0 is considered exact if the following condition holds:

∂M/∂y = ∂N/∂x

where ∂M/∂y represents the partial derivative of M with respect to y, and ∂N/∂x represents the partial derivative of N with respect to x.

Applying the Method to Our Equation

In our case, we have:

  • M(x,y) = x + 2y
  • N(x,y) = 2x + y

Let's check if the condition for exactness is satisfied:

  • ∂M/∂y = 2
  • ∂N/∂x = 2

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

Finding the Solution

  1. Find a function u(x,y) such that ∂u/∂x = M(x,y) and ∂u/∂y = N(x,y).

    Integrating ∂u/∂x = M(x,y) = x + 2y with respect to x, we get:

    u(x,y) = (1/2)x² + 2xy + g(y)

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

    Now, differentiate u(x,y) with respect to y:

    ∂u/∂y = 2x + g'(y)

    We know that ∂u/∂y = N(x,y) = 2x + y. Comparing these two expressions, we get:

    g'(y) = y

    Integrating g'(y) with respect to y, we get:

    g(y) = (1/2)y² + C

    where C is an arbitrary constant.

  2. Substitute the value of g(y) in u(x,y).

    This gives us:

    u(x,y) = (1/2)x² + 2xy + (1/2)y² + C

  3. The solution to the exact differential equation is given by u(x,y) = C.

    Therefore, the solution to the equation (x+2y)dx + (2x+y)dy = 0 is:

    (1/2)x² + 2xy + (1/2)y² = C

Conclusion

By applying the method of exact equations, we successfully solved the differential equation (x+2y)dx + (2x+y)dy = 0. The solution is represented by the equation (1/2)x² + 2xy + (1/2)y² = C, where C is an arbitrary constant. This solution describes a family of curves that represent the integral curves of the given differential equation.

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