CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solution of the differential equation
(2xy+2)dx+(4x2y1)dy=0 is

A
2xy=ce(x+2y)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2x+y=ce(2x+y)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x2y=ce(x+2y)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2x+y=ce(x+2y)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2xy=ce(x+2y)
Given equation is dydx={(2xy)+2}2(2xy)1
Put 2xy=v2dydx=dvdx
(dvdx2)=(v+22v1)
dvdx=(v+22v1+2)dvdx=5v2v1
2v15vdv=dx
25v15logv=x+logc
25(2xy)15log(2xy)=x+logc
2(2xy)log(2xy)=5x+5logc
4x2ylog(2xy)=5xlogc
logclog(2xy)=(x+2y)
log(c2xy)=(x+2y)
(c2xy)=e(x+2y)
c.e(x+2y)=(2xy)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon