Question

solve the differential equation
dy/dx= tanx-y/1+x^2​

Answer 1

Question:-The general solution of the differential equation:-dxdy=ytanx−y 2 secx is

Answer:-

We have, dxdy =ytanx−y 2 secx⇒ y21 dxdy − y1tanx=−secx

Put − y1 =v⇒ y 21 dy=dv

∴ dxdv +vtanx=secx ...(1)

Here P=tanx⇒∫tanxdx=logsecx

∴e logsecx =secx

Multiplying (1) by I.F. we get

secx dxdv +vsecx=sec 2 x

Integrating both sides w.r.t x we get

vsecx=∫sec 2 xdx⇒ y1

secx=tanx+c

secx=y(tanx+c)