A continuously differentiable function ϕ(x),x∈(0,π/2) satisfying y′=1+y2,y(0)=0, is

dydx=1+y2⇒11+y2dy=dx⇒ ∫11+y2dy=∫dx+C⇒ tan−1⁡y=x+C⇒y=tan⁡(x+C)⇒ϕ(x)=tan⁡(x+C)Now, y(0)=0⇒C=0∴ ϕ(x)=tan⁡x

A continuously differentiable function ϕ(x),x∈(0,π/2) satisfying y′=1+y2,y(0)=0, is

dydx=1+y2⇒11+y2dy=dx⇒ ∫11+y2dy=∫dx+C⇒ tan−1⁡y=x+C⇒y=tan⁡(x+C)⇒ϕ(x)=tan⁡(x+C)Now, y(0)=0⇒C=0∴ ϕ(x)=tan⁡x

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A continuously differentiable function $ϕ (x), x \in (0, π / 2)$ satisfying $y^{'} = 1 + y^{2}, y (0) = 0$ , is

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$\tan x$

$x (x - π)$

$x \log y = y y^{'}$

answer is A.

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$\begin{aligned} \frac{d y}{d x} = 1 + y^{2} \Rightarrow \frac{1}{1 + y^{2}} d y = d x \\ \Rightarrow \int \frac{1}{1 + y^{2}} d y = \int d x + C \end{aligned}$

$\Rightarrow \tan^{- 1} y = x + C \Rightarrow y = \tan (x + C) \Rightarrow ϕ (x) = \tan (x + C)$

Now, $y (0) = 0 \Rightarrow C = 0$

$∴ ϕ (x) = \tan x$

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A continuously differentiable function ϕ(x),x∈(0,π/2) satisfying y′=1+y2,y(0)=0, is

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A continuously differentiable function $ϕ (x), x \in (0, π / 2)$ satisfying $y^{'} = 1 + y^{2}, y (0) = 0$ , is