CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
146
You visited us 146 times! Enjoying our articles? Unlock Full Access!
Question

A solution of differential equation (sec2y)dydx+2x tan y=x3, is


A

2 tan y=c.ex2+x21

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

tan y=c.ex2+x21

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

tan y=cex2+x21

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

2 tan y=c.ex2+x21


tan y=zsec2 ydydx=dzdxdzdx+2xz=x3
I.F=e2xdx=ex2
Solution is z.ex2=x3.ex2dx
Let t=x2dt=2x dxtan y.ex2=12[t.etet]+c2 tan y=cex2+x21


flag
Suggest Corrections
thumbs-up
3
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