Differential EquationsmediumFree

Largest Interval for a Solution of y′=1+y² | JEE

JEE Maths question with a full step-by-step solution.

Question
The largest value of cc such that there exists a differentiable function h(x)h(x) for −c<x<c-c < x < c that is a solution of y′=1+y2y' = 1 + y^2 with h(0)=0h(0) = 0 is:
A2π2\pi
Bπ\pi
Cπ2\dfrac{\pi}{2}correct
Dπ4\dfrac{\pi}{4}
Solution
Separate and integrate:
dy1+y2=dx⇒tan⁡−1y=x+c1.\frac{dy}{1 + y^2} = dx \Rightarrow \tan^{-1}y = x + c_1.
 Apply h(0)=0h(0) = 0: tan⁡−10=0+c1⇒c1=0\tan^{-1}0 = 0 + c_1 \Rightarrow c_1 = 0, so y=tan⁡xy = \tan x. The function h(x)=tan⁡xh(x) = \tan x is differentiable only where tan⁡−1y=x\tan^{-1}y = x stays in the range (−π2,π2)\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), i.e. −π2<x<π2-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}. Hence the largest such cc is π2\dfrac{\pi}{2}. Correct answer: (3)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Differential Equations · medium
Let f:[1,∞)→Rf:[1,\infty)\to\mathbb{R} be defined as f(x)=∫1xf(t) dt+(1−x)(ln⁡x−1)+ef(x)=\displaystyle\int_1^x f(t)\,dt+(1-x)(\ln x-1)+e. Then the value of f(f(1))f(f(1)) is:
Differential Equations · medium
Let y=y(x)y=y(x) satisfy (tan⁡x)1/2 dy=(sec⁡3x−(tan⁡x)3/2y)dx(\tan x)^{1/2}\,dy=\left(\sec^3x-(\tan x)^{3/2}y\right)dx on (0,π2)\left(0,\frac{\pi}{2}\right), with y ⁣(π4)=625y\!\left(\frac{\pi}{4}\right)=\frac{6\sqrt{2}}{5}. If y ⁣(π3)=45αy\!\left(\frac{\pi}{3}\right)=\frac{4}{5}\alpha, then α4\alpha^4 equals
Differential Equations · easy
The order of the differential equation corresponding to y=c1cos⁡2x+c2cos⁡2x+c3sin⁡2x+c4y = c_1\cos 2x + c_2\cos^2 x + c_3\sin^2 x + c_4 is:
Differential Equations · medium
The degree of the differential equation satisfying the relation 1+x2+1+y2=λ(x1+y2−y1+x2)\sqrt{1+x^2} + \sqrt{1+y^2} = \lambda\big(x\sqrt{1+y^2} - y\sqrt{1+x^2}\big) is:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath