A continuously differentiable function -x in 0- satisfying y-1-y2-y0-0-y- is

The correct option is A tanx dydx=1+y^2 ∫dy/1+y^2=∫ dx #160; tan^ 1y=x+c at x=0;y=0 c=0 y=tan x ...

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Variable Separable Method

A continuousl...

Question

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

tan x

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x (x - π)

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(x - π) (1 - e^{x})

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Not possible

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ϕ (x)

(0, π)

y^{'} = 1 + y^{2}, y (0) = 0 = y (π)

ϕ (x)

(0, π)

y^{'} = 1 + y^{2}, y (0) = 0 = y (π)

\frac{d y}{d x} + \frac{y}{2} sec x = \frac{tan x}{2 y},

0 \leq x \leq \frac{π}{2},

y (0) = 1,

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} {(1 - | tan x |)}^{a / | tan x |}, & - \frac{π}{4} < x < 0 \end{matrix} \begin{matrix} b, & x = 0 \end{matrix} \begin{matrix} e^{\frac{sin 3 x}{sin 2 x}}, & 0 < x < \frac{π}{4} \end{matrix} \end{matrix}

x = 0

y = y (x)

\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3},

y (0) = 0,

y (\frac{π}{4})

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