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

Solve the differential equation cos2xdydx+y=tanx

Open in App
Solution

cos2xdydx+y=tanx
dydx+ycos2x=tanxcos2x (1)
This is of the form :dydx+P(x)y=Q(x)
where P(x)=1cos2x=sec2x
Q(x)=tanxsec2x

Integrating factor: esec2xdx=etanx
Mutiplying (1) by Integrating Factor:
etanxdydx+ysec2etanx=tanxsec2etanx
d(yetanx)dx=tanxsec2etanx
Integerating both sides,
ytanx=tanxsec2etanxdx
Taking RHS
Let tanx=t
sec2xdx=dt
tanxsec2etanxdx=tetdt=tet+et=(t+1)et
(t+1)et=(tanx+1)etanx
Hence
ytanx=(tanx+1)etanx+c where c is constant of integration
y=(tanx+1)etanxtanx+ctanx

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
General and Particular Solutions of a DE
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon