Let y = y ( x ) be the solution of the differential equation (3 y ^...

Given,

$\left(3{y}^{2} - 5{x}^{2}\right)ydx + 2x\left({x}^{2} - {y}^{2}\right)dy = 0$

$⇒ \dfrac{dy}{dx} = \dfrac{y\left(5{x}^{2} - 3{y}^{2}\right)}{2x\left({x}^{2} - {y}^{2}\right)}$

This is homogeneous differential equation

Let $y = tx ⇒ \dfrac{dy}{dx} = t + x\dfrac{dt}{dx}$

$t + x\dfrac{dt}{dx} = \dfrac{( tx ) \left(5{x}^{2} - 3{t}^{2}{x}^{2}\right)}{2x\left({x}^{2} - {t}^{2}{x}^{2}\right)} = \dfrac{t\left(5 - 3{t}^{2}\right)}{2\left(1 - {t}^{2}\right)}$

$⇒ x\dfrac{dt}{dx} = \dfrac{5t - 3{t}^{3}}{2 - 2{t}^{2}} - t = \dfrac{3t - {t}^{3}}{2 - 2{t}^{2}}$

$⇒ \dfrac{2\left(1 - {t}^{2}\right)}{3t - {t}^{3}}dt = \dfrac{1}{x}dx$

Taking integration on both sides we get,

$∫ \dfrac{2\left(1 - {t}^{2}\right)}{3t - {t}^{3}}dt = ∫ \dfrac{1}{x}dx . . . . . . \left(1\right)$

Now let $3t - {t}^{3} = z ⇒ 3\left(1 - {t}^{2}\right)dt = dz$

$⇒ \left(1 - {t}^{2}\right)dt = \dfrac{1}{3}dz$

Substitute in equation \left(1\right)$we get,</p><p>$\dfrac{2}{3} ∫ \dfrac{dz}{z} = ∫ \dfrac{1}{x}dx$</p><p>$⇒ \dfrac{2}{3}\ln ( z ) = \ln x + lnc$</p><p>$⇒ z\dfrac{2}{3} = cx ⇒ z = c\dfrac{3}{2}x\dfrac{3}{2}$</p><p>$⇒ 3t - {t}^{3} = c\dfrac{3}{2}x\dfrac{3}{2}$</p><p>Now put$t = \dfrac{y}{x}$</p><p>$⇒ \dfrac{3y}{x} - \dfrac{{y}^{3}}{{x}^{3}} = c\dfrac{3}{2}x\dfrac{3}{2}$</p><p>Given$y ( 1 ) = 1$when$x = 1, y = 1$</p><p>$⇒ c\dfrac{3}{2} = 2$</p><p>So, equation becomes$3\dfrac{y}{x} - \dfrac{{y}^{3}}{{x}^{3}} = 2x\dfrac{3}{2}$</p><p>$⇒ 3{x}^{2}y - {y}^{3} = 2x\dfrac{9}{2} . . . . . . \left(2\right)$</p><p>Put$x = 2$in equation \left(2\right)$,

$12y ( 2 ) - {\left(y ( 2 )\right)}^{3} = 2 ( 2) \dfrac{9}{2} = 32\sqrt{2}$

$∴ \left|{\left(y ( 2 )\right)}^{3} - 12y ( 2 )\right| = 32\sqrt{2}$