If ydydx - x - y2x2 - -y2x2- -y2x2 - x - 0- - - 0 and y1 - -1- then -y24 is equal to

Let y^2x^2 = u ⇒ y^2 = ux^2 ⇒ 2yy' = u'x^2 + 2ux …(1) Given, nbsp; yy' = x [ nbsp; y^2x^2 + ϕ(y^2x^2)ϕ ' (y^2x^2) ] ⇒ nbsp; yy' = x [u + ϕ(u)ϕ ' (u) ...

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Byju's Answer

Standard XII

Mathematics

Condition for Strictly Increasing

If ydydx = x ...

Question

If $y \frac{d y}{d x} = x ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{y^{2}}{x^{2}} + \frac{ϕ (\frac{y^{2}}{x^{2}})}{ϕ^{'} (\frac{y^{2}}{x^{2}})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦,$ $x > 0,$ $ϕ > 0$ and $y (1) = - 1,$ then $ϕ (\frac{y^{2}}{4})$ is equal to

4 ϕ (2)

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4 ϕ (1)

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2 ϕ (1)

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ϕ (1)

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Solution

The correct option is B $4 ϕ (1)$
Let $\frac{y^{2}}{x^{2}} = u$
$\Rightarrow y^{2} = u x^{2}$
$\Rightarrow 2 y y^{'} = u^{'} x^{2} + 2 u x \dots (1)$

Given,
$y y^{'} = x ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{y^{2}}{x^{2}} + \frac{ϕ (\frac{y^{2}}{x^{2}})}{ϕ^{'} (\frac{y^{2}}{x^{2}})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow y y^{'} = x [u + \frac{ϕ (u)}{ϕ^{'} (u)}]$
Using equation $(1)$
$\Rightarrow \frac{1}{2} [u^{'} x^{2} + 2 u x] = x [u + \frac{ϕ (u)}{ϕ^{'} (u)}]$
$\Rightarrow \frac{1}{2} u^{'} x^{2} = x \frac{ϕ (u)}{ϕ^{'} (u)}$
$\Rightarrow x u^{'} = \frac{2 ϕ (u)}{ϕ^{'} (u)}$
$\Rightarrow x \frac{d u}{d x} = \frac{2 ϕ (u)}{ϕ^{'} (u)}$
$\int \frac{ϕ^{'} (u) d u}{ϕ (u)} = \int \frac{2 d x}{x}$
$\Rightarrow ln ϕ (u) = 2 ln x + ln c$
$\Rightarrow ϕ (u) = c x^{2}$
$ϕ (\frac{y^{2}}{x^{2}}) = c x^{2}$
$∵ y (1) = - 1$
$∴ ϕ (1) = c$
$\Rightarrow ϕ (\frac{y^{2}}{x^{2}}) = ϕ (1) x^{2}$
Put $x = 2$
$ϕ (\frac{y^{2}}{4}) = 4 ϕ (1)$

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If ydydx=x⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣y2x2+ϕ(y2x2)ϕ′(y2x2)⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦, x>0, ϕ>0 and y(1)=−1, then ϕ(y24) is equal to

If $y \frac{d y}{d x} = x ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{y^{2}}{x^{2}} + \frac{ϕ (\frac{y^{2}}{x^{2}})}{ϕ^{'} (\frac{y^{2}}{x^{2}})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦,$ $x > 0,$ $ϕ > 0$ and $y (1) = - 1,$ then $ϕ (\frac{y^{2}}{4})$ is equal to