Solving the Differential Equation (x^2y^2)dx + (x^2  2xy)dy = 0
This article will guide you through the process of solving the given differential equation:
(x^2  y^2)dx + (x^2  2xy)dy = 0
We will utilize the method of exact differential equations.
1. Identifying the Equation as Exact
A differential equation of the form M(x,y)dx + N(x,y)dy = 0 is called exact if the following condition holds:
∂M/∂y = ∂N/∂x
Let's check if our equation fits this condition:
 M(x, y) = x^2  y^2
 N(x, y) = x^2  2xy
Calculating the partial derivatives:
 ∂M/∂y = 2y
 ∂N/∂x = 2x  2y
Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact in its current form.
2. Finding an Integrating Factor
To make the equation exact, we need to find an integrating factor, denoted by μ(x,y). An integrating factor is a function that, when multiplied by the original equation, will make it exact.
There are two common methods for finding integrating factors:
 If (∂M/∂y  ∂N/∂x)/N is a function of x only, then μ(x) = exp[∫(∂M/∂y  ∂N/∂x)/N dx]
 If (∂N/∂x  ∂M/∂y)/M is a function of y only, then μ(y) = exp[∫(∂N/∂x  ∂M/∂y)/M dy]
Let's try the first method:
 (∂M/∂y  ∂N/∂x)/N = (2y  (2x  2y))/(x^2  2xy) = 2x/(x^2  2xy)
This expression is a function of x and y, not just x. Therefore, this method does not work.
Let's try the second method:
 (∂N/∂x  ∂M/∂y)/M = (2x  2y  (2y))/(x^2  y^2) = 2x/(x^2  y^2)
This expression is a function of x and y, not just y. Therefore, this method also doesn't work.
Since neither method yielded an integrating factor directly, we need to look for an integrating factor that is a function of both x and y. In this case, we can observe that:
 (∂N/∂x  ∂M/∂y)/M = 2x/(x^2  y^2) = (2/x) * (x^2/(x^2  y^2))
 (∂M/∂y  ∂N/∂x)/N = 2x/(x^2  2xy) = (2/x) * (x^2/(x^2  2xy))
These expressions suggest that the integrating factor could be of the form μ(x, y) = 1/x.
3. Multiplying by the Integrating Factor
Let's multiply the original equation by μ(x, y) = 1/x:
(1/x)(x^2  y^2)dx + (1/x)(x^2  2xy)dy = 0
Simplifying:
(x  y^2/x)dx + (x  2y)dy = 0
Now, let's check if this equation is exact:

M(x, y) = x  y^2/x

N(x, y) = x  2y

∂M/∂y = 2y/x

∂N/∂x = 1
Since ∂M/∂y ≠ ∂N/∂x, the equation is still not exact.
We need to adjust the integrating factor slightly. We can multiply the original equation by μ(x, y) = 1/x^2 instead:
(1/x^2)(x^2  y^2)dx + (1/x^2)(x^2  2xy)dy = 0
Simplifying:
(1  y^2/x^2)dx + (1  2y/x)dy = 0
Now, let's check if this equation is exact:

M(x, y) = 1  y^2/x^2

N(x, y) = 1  2y/x

∂M/∂y = 2y/x^2

∂N/∂x = 2y/x^2
Since ∂M/∂y = ∂N/∂x, the equation is now exact.
4. Solving the Exact Equation
To solve the exact differential equation, we need to 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) = 1  y^2/x^2 with respect to x, we get:
u(x, y) = x + y^2/x + g(y)
where g(y) is an arbitrary function of y.
Now, differentiate this expression for u(x, y) with respect to y and equate it to N(x, y):
∂u/∂y = 2y/x + g'(y) = 1  2y/x
Solving for g'(y):
g'(y) = 1
Integrating this expression with respect to y:
g(y) = y + C
where C is an arbitrary constant.
Therefore, the general solution of the original differential equation is:
u(x, y) = x + y^2/x + y + C = 0
This can be rearranged to:
x^2 + xy + y^2 + Cx = 0
Conclusion
We successfully solved the differential equation (x^2  y^2)dx + (x^2  2xy)dy = 0 by applying the method of exact differential equations. We found an integrating factor that made the equation exact and then solved for the general solution, which is x^2 + xy + y^2 + Cx = 0.