Answer:
the method of exact differential equations:
We need to check if this differential equation is exact. To do that, we check if the partial derivative of the first term with respect to y is equal to the partial derivative of the second term with respect to x:
∂/∂y(x(x-y)) = x(-1) = -x
∂/∂x(y^2) = 0
Since these partial derivatives are not equal, the differential equation is not exact. We can try to make it exact by multiplying the entire equation by a suitable integrating factor.
Let's find the integrating factor (IF) by taking the partial derivative of the IF with respect to y and equating it to the partial derivative of the second term with respect to x:
∂/∂y(IF) = -y^2/(x(x-y)^2)
∂/∂x(y^2) = 0
From the first equation, we can see that an integrating factor of IF = x(x-y)^2 should make the equation exact. Multiplying the entire equation by this integrating factor, we get:
x(x-y)^2dy + y^2x(x-y)dx = 0
Now, we just need to find a function φ(x,y) such that:
∂φ/∂x = x(x-y)^2dy
∂φ/∂y = y^2x(x-y)dx
Integrating the first equation with respect to x, we get:
φ(x,y) = ∫x(x-y)^2dy + f(x)
φ(x,y) = -1/3(x-y)^3x + f(x)
Now, we differentiate this equation with respect to x and equate it to the second equation:
∂φ/∂x = -(x-y)^3 + f'(x)
∂φ/∂y = y^2(x-y)^3
Comparing the two, we can see that f'(x) = 0, which means that f(x) is a constant. We can choose this constant to be zero without loss of generality.
Therefore, the solution to the differential equation is given by:
-1/3(x-y)^3x + C = 0
where C is an arbitrary constant.