# How to solve $(y^2 +2x^2y)dx +(2x^3 -xy)dy=0$

I tried solving it by comparing it with $$Mdx+Ndy =0$$ for $$M=y^2 +2x^2y$$ and $$N=2x^3 -xy$$,
$$\frac{\partial M}{\partial y}=2y+2x^2,~~~~\frac{\partial N}{\partial x}=6x^2-y$$

$$\displaystyle\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}$$. I don't know how to procede further. Any help is appreciated

• Maybe It Is my problem, but can you specify what do you mean by "solve". Because i didn't understand what are you looking for Jun 18, 2023 at 12:29
• eliminate the differentials and convert it to simpler x and y terms Jun 18, 2023 at 12:31
• This Is a 1 form. Are you looking for a function $F(x,y)$ such that its differential $dF$ Is your form? Jun 18, 2023 at 12:37

Let the integrating factor be $$\mu=x^ay^b$$,

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

we get

$$(b+2)x^ay^{b+1}+2(b+1)x^{a+2}y^b=2(a+3)x^{a+2}y^b-(a+1)x^ay^{b+1}$$

compare coefficients,

$$b+1=a+3,~~~b+2=-(a+1)$$

hence,

$$a=-\frac32, ~~b=\frac12$$

After multiplying the integral factor $$\mu=x^{-\frac32}y^{\frac12}$$ it becomes exact and you can proceed to solve it.

• thanks for the answer Jun 18, 2023 at 12:49
• you are welcome :) Jun 18, 2023 at 12:53

So your equation can be written as $$\frac{dy}{dx} = \frac{y}{x} \frac{y+2x^2}{y-2x^2} \hspace{3mm} \cdots (1)$$ Now just substitute $$y = vx$$ so $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$

From (1) we get $$v + x\frac{dv}{dx} = v \left( \frac{vx + 2x^2}{vx - 2x^2} \right)$$ $$x\frac{dv}{dx} = \frac{4vx}{v-2x}$$ $$\frac{dv}{dx} = \frac{4v}{v-2x}$$

Now $$\frac{dx}{dv} = \frac{1}{4} \left( 1 - \frac{2x}{v} \right)$$ $$\frac{dx}{dv} + \frac{x}{2v} = \frac{1}{4} \hspace{3mm} \cdots (2)$$

Now do you know about linear differential equations of the form

$$\frac{dy}{dx} + \mathcal{P}y = \mathcal{Q}$$ So solution of this equation is given by $$y (I.F.) = \int \mathcal{Q}(I.F.) dx$$ where $$I.F. = \mathrm{exp}\left( \int \mathcal{P} dx \right)$$

So integrating factor for (2) will be

$$I.F. = \mathrm{exp} \left(\int \frac{1}{2v} dv \right)$$ $$I.F. = \sqrt{v}$$

Now our solution will be $$x (I.F.) = \int \frac{1}{4} (I.F.) dv$$ $$x \sqrt{v} = \frac{1}{6} v^{3/2} + C$$

Now after replacing $$v$$ by $$y/x$$ we get $$\sqrt{xy} = \frac{1}{6} \left(\frac{y}{x} \right)^{3/2} + C$$

$$(y^2 +2x^2y)dx +(2x^3 -xy)dy=0$$ $$y^2dx +2x^2(ydx +xdy) -xydy=0$$ $$y^2dx +2x^2d(xy) -xydy=0$$ $$y(ydx-xdy) +2x^2d(xy) =0$$ $$-yd\left(\dfrac yx \right) +2d(xy) =0$$ Substitute $$v=xy$$ and $$w=\dfrac yx$$: $$-ydw +2dv =0$$ Note that $$y^2=yx\dfrac {y}{x}=vw$$ Then the DE is separable: $$-ydw +2dv =0$$ $$-\sqrt wdw +\dfrac 2{\sqrt v}dv =0$$ Intgrate: $$-\dfrac 23 w^{3/2} +4{\sqrt v} =c$$ $$- \dfrac 13\left( \dfrac yx \right)^{3/2} +2\sqrt {xy} =C$$