If $y \frac{d y}{d x}=x\left[\frac{y^{2}}{x^{2}}+\frac{\phi\left(\frac{y^{2}}{x^{2}}\right)}{\phi^{\prime}\left(\frac{y^{2}}{x^{2}}\right)}\right], x>0, \phi>0$, and $y(1)=-1$, then $\phi\left(\frac{y^{2}}{4}\right)$ is equal to :

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Differential Equations

y \frac{d y}{d x} = x \frac{y ^{2}}{x ^{2}} + \frac{ϕ ( \frac{y ^{2}}{x ^{2}} )}{ϕ ^{'} ( \frac{y ^{2}}{x ^{2}} )}, x > 0, ϕ > 0

, and

y (1) = - 1

, then

ϕ (\frac{y ^{2}}{4})

is equal to :

Answer: (C)

With

v = \frac{y ^{2}}{x ^{2}}

, differentiating and substituting the equation gives

\frac{d v}{d x} = \frac{2}{x} \cdot \frac{ϕ ( v )}{ϕ ^{'} ( v )}

. Separating,

\frac{ϕ ^{'} ( v )}{ϕ ( v )} d v = \frac{2}{x} d x

, so

ln ϕ (v) = 2 ln x + const

, i.e.

ϕ (v) = ϕ (1) x^{2}

once

y (1) = - 1

(where

v = 1

) fixes the constant. The quantity

ϕ (\frac{y ^{2}}{4})

corresponds to

x = 2

, hence

ϕ (\frac{y ^{2}}{4}) = 4 ϕ (1)

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