# Show that the equation M(x,y)dx+N(x,y)dy = 0 has an integrating factor which is a function of ratio of x and y.?

## Aug 13, 2018

If there exists the I.F. $\mu \left(w\right) , w = \frac{x}{y}$, then:

$q \quad \left\{\begin{matrix}{\mu}_{x} = \frac{1}{y} \mu ' \\ {\mu}_{y} = - \frac{x}{y} ^ 2 \mu '\end{matrix}\right. q \quad \text{ where } q \quad \mu ' = \frac{d \mu}{\mathrm{dw}}$

So, applying the I.F.:

• ${\left(M \mu\right)}_{y} = {M}_{y} \mu - \frac{x}{y} ^ 2 M \mu ' q \quad \mathbb{A}$

• ${\left(N \mu\right)}_{x} = {N}_{x} \mu + N \frac{1}{y} \mu ' q \quad \mathbb{B}$

For this to be an I.F., the mixed partials must be equal:

$q \quad \mathbb{A} = \mathbb{B}$

$q \quad \mu \left({M}_{y} - {N}_{x}\right) = \left(\frac{N}{y} + M \frac{x}{y} ^ 2\right) \mu '$

$q \quad \mu {y}^{2} \Delta = \left(y N + x M\right) \mu '$

$q \quad \therefore \frac{{y}^{2} \Delta}{y N + x M} = \frac{\mu '}{\mu} q \quad q \quad = f \left(w\right)$

Therefore, for the I.F. $\mu \left(w\right)$ to exist, $\frac{{y}^{2} \Delta}{y N + x M}$ must be a function of $w$.