The general solution of the differential equation dydx=y tan x−

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Question

The general solution of the differential equation

\frac{d y}{d x} = y t a n x - y^{2} s e c x

is

Options:

A . tan x = (c + sec x)y

B . sec y = (c + tan y )x

C . sec x = (c + tan x)y

D . None of these

Answer: Option C
:
C
We have

\frac{d y}{d x} = y t a n x - y^{2} s e c x \Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} t a n x = - s e c x

Putting

\frac{1}{y} = v \Rightarrow \frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d v}{d x}

, we obtain

\frac{d v}{d x} + t a n x . v = s e c x

which is linear

I . F = e^{\int t a n x d x} = e^{l o g s e c x} = s e c x

∴

The solution is

v s e c x = \int s e c^{2} x d x + c \Rightarrow \frac{1}{y} s e c x = t a n x + c

\Rightarrow s e c x = y (c + t a n x)

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