The solution of dy-dx-y tan x - 0 is-

The correct option is B y = a cos x dydx+y tan x=0 ∫dyy= ∫tan x dx+c log y=log c cos x ⇒ y=c cos x ...

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Variable Separable Method

The solution ...

Question

The solution of $\frac{d y}{d x} + y tan x = 0$ is:

y = a cos x

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y = a sin x

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y = log cos x + c

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y = a tan x + c

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y = {tan}^{- 1} [\frac{a cos x + b sin x}{b cos x - a sin x}]

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y = {tan}^{- 1} (\frac{a cos x - b sin x}{b cos x + a sin x})

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y = {tan}^{- 1} (\frac{a sin x + b cos x}{a cos x - b sin x}), - \frac{π}{2} < x < \frac{π}{2}

\frac{b}{a} tan x < 1

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\frac{d y}{d x} = y tan x

y = 1

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\frac{d y}{d x} = sin (x + y) tan (x + y) - 1

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