Differential EquationsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Differential Equations: Let Solution Curve Differential Equation Curve Passes Throug (JEE Main 2026)

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

Question
Let y=y(x)y=y(x) be the solution curve of the differential equation (1+sin⁡x)dydx+(y+1)cos⁡x=0(1+\sin x)\dfrac{dy}{dx}+(y+1)\cos x=0, y(0)=0y(0)=0. If the curve y=y(x)y=y(x) passes through the point (α,−12)\left(\alpha,-\dfrac12\right), then a value of α\alpha is
Aπ6\dfrac{\pi}{6}
Bπ4\dfrac{\pi}{4}
Cπ3\dfrac{\pi}{3}
Dπ2\dfrac{\pi}{2}correct
Solution
Step 1: Separate variables:
dyy+1=−cos⁡x1+sin⁡x dx.\frac{dy}{y+1}=-\frac{\cos x}{1+\sin x}\,dx.
 Step 2: Integrate both sides (right side has numerator = derivative of denominator):
log⁡e(y+1)=−log⁡e(1+sin⁡x)+C.\log_e(y+1)=-\log_e(1+\sin x)+C.
 Step 3: Apply y(0)=0y(0)=0: log⁡e1=−log⁡e1+C⇒C=0\log_e1=-\log_e1+C\Rightarrow C=0. So
log⁡e(y+1)=−log⁡e(1+sin⁡x) ⇒ y+1=11+sin⁡x.\log_e(y+1)=-\log_e(1+\sin x)\ \Rightarrow\ y+1=\frac{1}{1+\sin x}.
 Step 4: Put y=−12y=-\dfrac12:
11+sin⁡x=12 ⇒ 1+sin⁡x=2 ⇒ sin⁡x=1 ⇒ x=π2.\frac{1}{1+\sin x}=\frac12\ \Rightarrow\ 1+\sin x=2\ \Rightarrow\ \sin x=1\ \Rightarrow\ x=\frac{\pi}{2}.
 Correct answer: (4)
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