If dydx+xsin⁡2y=x3cos2⁡y,y(0)=0 then tan⁡y(1)=

sec2⁡ydydx+2xtan⁡y=x3 I. F=e∫2xdx=ex2 Solution is tan⁡yex2=∫x3⋅ex2dx=12x2−1ex2+c⇒tan⁡y(x)=12x2−1+c⋅e−x2x=0,y=0⇒c=12⇒tan⁡y(x)=12x2−1+e−x2tan⁡y(y(1))=12e

If dydx+xsin⁡2y=x3cos2⁡y,y(0)=0 then tan⁡y(1)=

sec2⁡ydydx+2xtan⁡y=x3 I. F=e∫2xdx=ex2 Solution is tan⁡yex2=∫x3⋅ex2dx=12x2−1ex2+c⇒tan⁡y(x)=12x2−1+c⋅e−x2x=0,y=0⇒c=12⇒tan⁡y(x)=12x2−1+e−x2tan⁡y(y(1))=12e

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$If \frac{d y}{d x} + x \sin 2 y = x^{3} \cos^{2} y, y (0) = 0 then \tan y (1) =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{e}$

$\frac{1}{2 e}$

$2 e$

$e$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \sec^{2} y \frac{d y}{d x} + 2 x \tan y = x^{3} \\ I. F = e^{\int 2 x d x} = e^{x^{2}} \\ Solution is \tan y e^{x^{2}} = \int x^{3} \cdot e^{x^{2}} d x = \frac{1}{2} (x^{2} - 1) e^{x^{2}} + c \\ \Rightarrow \tan y (x) = \frac{1}{2} (x^{2} - 1) + c \cdot e^{- x^{2}} \\ x = 0, y = 0 \Rightarrow c = \frac{1}{2} \\ \Rightarrow \tan y (x) = \frac{1}{2} [x^{2} - 1 + e^{- x^{2}}] \\ \tan y (y (1)) = \frac{1}{2 e} \end{array}$