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

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.

Substitute the value of g(y) in u(x,y).
This gives us:
u(x,y) = (1/2)x² + 2xy + (1/2)y² + C

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.