A continuously differentiable function $

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

A continuously differentiable function $y=f(x),x∈(\frac{-\pi}{2},\frac{\pi}{2})$ satisfying $y'=1+y^2,y(0)=0$ is:

Options:

y = tan x

y = x(x - π)

y = (x - π)(1 - e^x)

Not possible

Correct Answer:

y = tan x

Explanation:

$\int\frac{dy}{1+y^2}=\int dx⇒tan^{-1}y=x+c$;

y(0) = 0 ⇒ c = 0 ⇒ y = tan x