# Using integrating factors to solve difficult differential equations

Consider the differential equatioN:

$$(2x^2 y -2y^4)dx+ ( 2x^3 + 3xy^3) dy = 0$$

This equation is of the form:

$$Q dx + P dy=0$$

Now, it's easy to see that this differential is not exact by using the commutativity of second order partial condition. Multiply it by an integrating factor $$\eta(x,y)$$ such that it does:

$$\eta_x P + P_x \eta = \eta_y Q + Q_y \eta$$

Or,

$$\eta_x P - \eta_y Q + ( P_x - Q_y) \eta = 0$$

This becomes:(*)

$$\eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 6x^2 +3 y^3 - 2x^2 -8y^3) \eta =0$$

Or,

$$\eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2 - 11y^3) \eta =0$$

Now... I'm not quite sure what to do.. I don't think I can write $$\eta$$ as function solely of $$x$$ or $$y$$.. did I miss something or... ?

As correctly observed by @Aleksas Domarkas. There is a mistake after the place where I put *. The current working is as follows:

$$\eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 6x^2 +3 y^3 - 2x^2 +8y^3) \eta =0$$

Or,

$$\eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2 + 11y^3) \eta =0$$

$$(2x^2y-2y^4)~dx+(2x^3+3xy^3)~dy=0$$

$$(2x^3+3xy^3)~dy=(2y^4-2x^2y)~dx$$

$$\dfrac{dy}{dx}=\dfrac{2y^4-2x^2y}{2x^3+3xy^3}$$

Let $$r=x^2$$ ,

Then $$\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=2x\dfrac{dy}{dr}$$

$$\therefore2x\dfrac{dy}{dr}=\dfrac{2y^4-2x^2y}{2x^3+3xy^3}$$

$$\dfrac{dy}{dr}=\dfrac{y^4-x^2y}{x^4+3x^2y^3}$$

$$\dfrac{dy}{dr}=\dfrac{y^4-ry}{r^2+3ry^3}$$

Let $$s=y^3$$ ,

Then $$\dfrac{ds}{dr}=3y^2\dfrac{dy}{dr}$$

$$\therefore\dfrac{1}{3y^2}\dfrac{ds}{dr}=\dfrac{y^4-ry}{r^2+3ry^3}$$

$$\dfrac{ds}{dr}=\dfrac{3y^6-3ry^3}{r^2+3ry^3}$$

$$\dfrac{ds}{dr}=\dfrac{3s^2-3rs}{r^2+3rs}$$

I get ("+" in last term). $$\eta_x ( 2x^3 + 3xy^3) - \eta_y ( 2x^2 y - 2y^4) + ( 4x^2 + 11y^3) \eta =0$$ Let $$\eta=x^ay^b$$. Then $$\left( -2 b-3 a-11\right) \, {{x}^{a}}\, {{y}^{b+3}}+\left( 2 b-2 a-4\right) \, {{x}^{a+2}}\, {{y}^{b}}=0$$ Solve system $$-2 b-3 a-11=0,\quad 2 b-2 a-4=0$$.

Then $$a=-3,\quad b=-1$$. Integrating factor is $$\eta=\frac{1}{x^3y}$$ We get general solution of differential equation $$2\ln(xy)+\frac{x^3}{y^2}=C$$