CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let y=y(x) be a solution of the differential equation, 1x2dydx+1y2=0,|x|<1.
If y(12)=32, then y(12) is equal to:
  1. 12
  2. 32
  3. 32
  4. 12


Solution

The correct option is D 12
1x2dydx+1y2=0
dy1y2+dx1x2=0
sin1y+sin1x=c

If x=12,y=32
then sin132+sin112=c
π3+π6=cc=π2

Therefore, the solution is 
sin1y=π2sin1x=cos1x
sin1y=cos1x

Now, put x=12
sin1y=cos112
sin1y=3π4
y(12)=sin(3π4)=12

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image