Question

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

Answer 1

$\bold{Answer :}$

Now,

(3xy² - y³) dx - (2x²y - xy²) dy = 0

⇒ dy/dx = (3xy² - y³)/(2x²y - xy²) ...(i)

Let us take,

y = vx

Then,

dy/dx = v + x (dv/dx)

From (i), we get

v + x (dv/dx) = (3x v²x² - v³x³)/(2x² vx - x v²x²)

⇒ v + x (dv/dx) = (3v² - v³)/(2v - v²)

⇒ x (dv/dx) = (3v - v²)/(2 - v) - v

⇒ x (dv/dx) = (3v - v² - 2v + v²)/(2 - v)

⇒ x (dv/dx) = v/(2 - v)

⇒ {(2 - v)/v} dv = (dx)/x

⇒ 2 (dv)/v - dv = (dx)/x

∴ integrating we get

2 ∫ (dv)/v - ∫ dv = ∫ (dx)/x

⇒ 2 logv - v = logx + c, where c is intergral constant

⇒ 2 log (y/x) - (y/x) = logx + c [ ∵ y = vx ]

⇒ 2 logy - 2 logx - (y/x) = logx + c

⇒ 2 logy - 3 logx = (y/x) + c,

which is the required primitive

#$\bold{MarkAsBrainliest}$

Answer 2

Solve the differential equation $(3xy^2-y^3)dx-(2x^2y-xy^2)dy=0$

Explanation:

1. given the differential equation: $(3xy^2-y^3)dx-(2x^2y-xy^2)dy=0$                              

                                                             $\frac{3xy^2-y^3}{2x^2y-xy^2}=\frac{dy}{dx}$      ----(a)

    2. let $y=ux$          then , $\frac{dy}{dx}=u+x\frac{du}{dx}$     ---(b)

    3. substituting (b) in (a) we get,                  

                    [tex]->\frac{3x^3u^2-u^3x^3}{2x^3u-x^3u^2}=u+x\frac{du}{dx} \\ -> \frac{3x^3u^2-u^3x^3-2x^3u^2+x^3u^3}{2x^3u-x^3u^2}= \frac{du}{dx}\\ ->\frac{x^3u^2}{x^3u(2-u)}=\frac{du}{dx}\\ ->\frac{u}{2-u} =\frac{du}{dx}\\\\ -> dx=\frac{2-u}{u} du[/tex]   ------(c)

      4. integrating (c) on both sides we get,                        

                    [tex]->\int\ dx =\int\ {\frac{2-u}{u}} \, du\\ ->x=2ln(|u|)-u+c[/tex]     (here 'c' is the integration constant)

      5. now from (b) we have, $u=\frac{y}{x}$   putting this back in above equation

          we get,  (here 'c' is the integration constant)                                                                      

         [tex]->x=2ln(|\frac{y}{x} |)-\frac{y}{x}+c\\\\ ->2xln(|y|)-2xln(|x|)-y+cx-x^2=0\\[/tex]        ------>ANSWER