If y(dy)/(dx) = x[ y ^ 2 x ^ 2 + ϕ( y ^ 2 x ^ 2 ) ϕ ^ ' ( y ^ 2 x ^...

Given:\dfrac{y}{x}\dfrac{dy}{dx} = \left[\dfrac{{y}^{2}}{{x}^{2}} + \dfrac{ϕ\left(\dfrac{{y}^{2}}{{x}^{2}}\right)}{{ϕ}^{'}\left(\dfrac{{y}^{2}}{{x}^{2}}\right)}\right] . . . . \left(1\right)$</p><p>Let$\dfrac{y}{x} = t$</p><p>$⇒ y = xt$</p><p>$⇒ \dfrac{dy}{dx} = t + x \cdot \dfrac{dt}{dx}$</p><p>$∴ t\left(t + x\dfrac{dt}{dx}\right) = \left({t}^{2} + \dfrac{ϕ\left({t}^{2}\right)}{{ϕ}^{'}\left({t}^{2}\right)}\right)$</p><p>$⇒ xt\dfrac{dt}{dx} = \dfrac{ϕ\left({t}^{2}\right)}{ϕ ' \left({t}^{2}\right)}$</p><p>$⇒ \dfrac{t \cdot {ϕ}^{'}\left({t}^{2}\right)}{ϕ\left({t}^{2}\right)}dt = \dfrac{1}{x}dx$<br />Integrating both sides</p><p>$∫ \dfrac{t \cdot {ϕ}^{'}\left({t}^{2}\right)}{ϕ\left({t}^{2}\right)}dt = ∫ \dfrac{1}{x}dx$</p><p>Let$ϕ\left({t}^{2}\right) = p$</p><p>$⇒ {ϕ}^{'}\left({t}^{2}\right) . 2t = dp$</p><p>$⇒ \dfrac{1}{2} ∫ \dfrac{1}{p}dp = ∫ \dfrac{1}{x}dx$</p><p>$⇒ \dfrac{1}{2}lnp = \ln x + C$</p><p>$⇒ \dfrac{1}{2}lnϕ\left({t}^{2}\right) = \ln x + C$</p><p>$⇒ \dfrac{1}{2}\ln \left(ϕ\left(\dfrac{{y}^{2}}{{x}^{2}}\right)\right) = \ln x + C . . . \left(2\right)$</p><p>If$x = 1, y = - 1$then$C = \dfrac{1}{2}\ln \left(ϕ\left(1\right)\right)$</p><p>Substituting value of$C$in \left(2\right)$

$\dfrac{1}{2}\ln \left(ϕ\left(\dfrac{{y}^{2}}{{x}^{2}}\right)\right) = \ln x + \dfrac{1}{2}\ln \left(ϕ\left(1\right)\right)$

$⇒$$\ln \left(ϕ\left(\dfrac{{y}^{2}}{{x}^{2}}\right)\right) = ln{x}^{2} + \ln \left(ϕ\left(1\right)\right)$

If $x = 2$ then

$\ln \left(ϕ\left(\dfrac{{y}^{2}}{4}\right)\right) = \ln 4 + \ln \left(ϕ\left(1\right)\right)$

SO, $ϕ\left(\dfrac{{y}^{2}}{4}\right) = 4ϕ\left(1\right)$