CGPET 2015 Solved Paper

Given differential equation is

dy

dx

=y‌tan‌x−y2‌sec‌x
⇒

dy

dx

−y‌tan‌x=−y2‌sec‌x
⇒

1

y2

dy

dx

−tan‌

x

y

=−sec‌x
Put −

1

y

=u
⇒

1

y2

dy

dx

=

du

dx

...(i)
From Eq. (i),

du

dx

+tan‌x.u=−sec‌x ...(ii)
This is linear differential equation of the form

du

dx

+P.u=Q,where‌P=tan‌x,Q=−sec‌x
∴ IF=e∫tanx‌dx=elog‌sec‌x=sec‌x
Hence, general solution is
u.IF=∫IF.Q‌dx+C1
⇒ u.sec‌x=∫(sec‌x).(−sec‌x)‌dx+C1
⇒u.sec‌x=−∫sec2‌xdx+C1
⇒u‌sec‌x=−tan‌x+c1
⇒

−sec‌x

y

=−tan‌x+C1[∵u=

−1

y

]
⇒sec‌x=y(tan‌x−C1)
⇒sec‌x=y(tan‌x+C)[where,C=−C1]