Differential EquationsmediumFree

Exact Differential Equation with sin 2y | JEE

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

Question
The solution of the differential equation dydx=sin⁡y+xsin⁡2y−xcos⁡y\dfrac{dy}{dx} = \dfrac{\sin y + x}{\sin 2y - x\cos y} is:
Asin⁡2y=xsin⁡y+x22+C\sin^2 y = x\sin y + \dfrac{x^2}{2} + Ccorrect
Bsin⁡2y=xsin⁡y−x22+C\sin^2 y = x\sin y - \dfrac{x^2}{2} + C
Csin⁡2y=x+sin⁡y+x22+C\sin^2 y = x + \sin y + \dfrac{x^2}{2} + C
Dsin⁡2y=x−sin⁡y+x22+C\sin^2 y = x - \sin y + \dfrac{x^2}{2} + C
Solution
Cross-multiply:
(sin⁡2y−xcos⁡y) dy=(sin⁡y+x) dx⇒sin⁡2y dy−xcos⁡y dy−sin⁡y dx=x dx.(\sin 2y - x\cos y)\,dy = (\sin y + x)\,dx \Rightarrow \sin 2y\,dy - x\cos y\,dy - \sin y\,dx = x\,dx.
 Now sin⁡2y dy=2sin⁡ycos⁡y dy=d(sin⁡2y)\sin 2y\,dy = 2\sin y\cos y\,dy = d(\sin^2 y) and sin⁡y dx+xcos⁡y dy=d(xsin⁡y)\sin y\,dx + x\cos y\,dy = d(x\sin y), so
d(sin⁡2y)−d(xsin⁡y)=x dx⇒sin⁡2y−xsin⁡y=x22+C.d(\sin^2 y) - d(x\sin y) = x\,dx \Rightarrow \sin^2 y - x\sin y = \frac{x^2}{2} + C.
∴sin⁡2y=xsin⁡y+x22+C.\therefore \sin^2 y = x\sin y + \frac{x^2}{2} + C.
 Correct answer: (1)
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