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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

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  • $\begingroup$ 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 $\endgroup$
    – Marco
    Jun 18, 2023 at 12:29
  • $\begingroup$ eliminate the differentials and convert it to simpler x and y terms $\endgroup$
    – Gowhar1998
    Jun 18, 2023 at 12:31
  • $\begingroup$ This Is a 1 form. Are you looking for a function $F(x,y)$ such that its differential $dF$ Is your form? $\endgroup$
    – Marco
    Jun 18, 2023 at 12:37

3 Answers 3

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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.

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  • $\begingroup$ thanks for the answer $\endgroup$
    – Gowhar1998
    Jun 18, 2023 at 12:49
  • $\begingroup$ you are welcome :) $\endgroup$
    – MathFail
    Jun 18, 2023 at 12:53
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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 $$ \sqrt{xy} = \frac{1}{6} \left(\frac{y}{x} \right)^{3/2} + C $$

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    $\begingroup$ thanks for your answer $\endgroup$
    – Gowhar1998
    Jun 18, 2023 at 12:55
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    $\begingroup$ @Gowhar1998 is my answer correct? just asking out of curiosity because I make lots of mistake :') $\endgroup$ Jun 18, 2023 at 12:56
  • $\begingroup$ I did't verified the answer ,I just what to know how to procede, but your contribution is worth it $\endgroup$
    – Gowhar1998
    Jun 18, 2023 at 13:00
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    $\begingroup$ @Gowhar1998 If you want to express your thanks, please upvote the answer. Writing a comment just to say thanks is a nice gesture, but on MathSE it is recommended not to. $\endgroup$
    – IraeVid
    Jun 18, 2023 at 13:00
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$$(y^2 +2x^2y)dx +(2x^3 -xy)dy=0$$ $$y^2dx +2x^2(ydx +xdy) -xydy=0$$ $$y^2dx +2x^2d(xy) -xydy=0$$ $$y(ydx-xdy) +2x^2d(xy) =0$$ $$-yd\left(\dfrac yx \right) +2d(xy) =0$$ 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: $$-\dfrac 23 w^{3/2} +4{\sqrt v} =c$$ $$- \dfrac 13\left( \dfrac yx \right)^{3/2} +2\sqrt {xy} =C$$

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