The general solution of the differential equation d y d x = y t a n x − y 2 s e c x is Options: A . &nbsp tan x = (c + sec x)y B . &nbsp sec y = (c + tan y )x C . &nbsp sec x = (c + tan x)y D . &nbsp None of these

Answer: Option C : C We have d y d x = y t a n x − y 2 s e c x ⇒ 1 y 2 d y d x − 1 y t a n x = − s e c x Putting 1 y = v ⇒ − 1 y 2 d y d x = d v d x , we obtain d v d x + t a n x . v = s e c x which is linear I . F = e ∫ 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 = ∫ s e c 2 x d x + c ⇒ 1 y s e c x = t a n x + c ⇒ s e c x = y ( c + t a n x )

The general solution of the differential equation dydx=y tan xȡ

Discussion Forum : Differential Equations

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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