# Non-exact D.E with Integration Factor

$DE : (y^2)dx + (x^2-xy)dy = 0$

$\frac{ \delta M } {\delta y} = 2y$ and $\frac{ \delta N } {\delta x} = 2x-y$

I've found the Integrating factor to be:

$e^\int\ \frac{((\delta M/\delta y) - (\delta N/\delta x))}{N} \$ = $\frac {x-y}{x^3}\$

However after multiplying the Integrating factor with the initial equations , they don't seem to be becoming exact.

$\delta M * \frac {x-y}{x^3} = y^2 * \frac {x-y}{x^3}$ and $\delta N * \frac {x-y}{x^3} = x^2-xy * \frac {x-y}{x^3}$

After differentiating the new $\delta M$ and $\delta N$

$\frac{\delta M } {\delta y} = - \frac {y(3y-2x)}{x^3} = \frac {-3y^2+2xy}{x^3}$ and $\frac{\delta N } {\delta x} = \frac {2y(x-y)}{x^3} = \frac {-2y^2 + 2xy}{x^3}$

As you can see they are not exact, even though they should be. I don't know where I've made an error exactly but it seems that it's most likely in the calculation of the Integrating factor? Any help will be greatly appreciated.

• I do unfortunately. It's part of my differential equations coursework. – Saloni Mude Mar 27 '17 at 15:26

The integrating factor only works if $\dfrac{\dfrac{\partial M }{\partial y } - \dfrac{\partial N }{\partial x }}{N}$ is only in terms of $x$.
Fortunately, the equation is homogeneous. Homogeneous equations can be solved using the substitution $y = u x$ and $\dfrac{dy}{dx} = x \dfrac{du}{dx} + u$, from the Product Rule.
$y^2dx + (x^2 - xy)dy = 0$ can be rewritten as $\dfrac{dy}{dx} = \dfrac{y^2}{xy - x^2}$, which simplifies to the following separable equation:
$$x \dfrac{du}{dx} = \dfrac{1}{u-1}$$