The solution of the differential equation dydx-ytanx=exsecx is
y=excosx+c
ycosx=ex+c
y=exsinx+c
ysinx=ex+c
Explanation of the correct option.
Compute the required value.
Given : dydx-ytanx=exsecx
Compare the equation with standard L.D.E.dydx+Py=Q,
P=-tanx and Q=exsecx
I.F.=e∫Pdx=e∫-tanxdx=e-logsecx=elogcosx=cosx
Since the solution of L.D.E. is given by y(I.F.)=∫QI.F.dx+c.
ycosx=∫exsecxcosxdx+c⇒ycosx=∫exdx+c⇒ycosx=ex+c
Hence option B is the correct option.