wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solution of 2ysinxdydx=2sinxcosxy2cosx,x=π2,y=1 is given by

A
y2=sinx
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
y=sin2x
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
y2=cosx+1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A y2=sinx
2ysinxdydx=2sinxcosxy2cosx
2ysinxdydx+y2cosx=2sinxcosx
sinxddx(y2)+y2ddxsinx=2sinxcosx
ddx(y2sinx)=2sinxcosx
d(y2sinx)=sin2xdx
y2sinx=cos2x2+c
At x=π2,y=1
12sinπ2=cosπ2+cc=12
So, y2sinx=cos2x2+12
y2=1cos2x2sinx
y2=sinx

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon