Differential EquationHard
Question
The solution of differential equation 
dx + x sec2 
dy = 0 satisfying the initial condition y(1) = 
is
dx + x sec2 dy = 0 satisfying the initial condition y(1) = isOptions
A.x sec 
= √2
= √2B.x tan = 1
= 1C.x tan2 = 1
= 1D.x sec2 = 2
= 2Solution
Put y = vx ⇒ 
= v + x 
(tan v - v sec2v) + sec2v
= 0
⇒ tan v - v sec2 v + v sec2 v + x sec2v
= 0
⇒
⇒ ln tan v = - ln x + ln c ⇒ x tan v = c
⇒ x tan
= c
Now y(1) =
⇒ c = 1
∴ x tan
= 1
= v + x (tan v - v sec2v) + sec2v
= 0⇒ tan v - v sec2 v + v sec2 v + x sec2v
= 0⇒
⇒ ln tan v = - ln x + ln c ⇒ x tan v = c
⇒ x tan
= cNow y(1) =
⇒ c = 1∴ x tan
= 1Create a free account to view solution
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