Find the general solution of differential equations $$\dfrac{dy}{dx} = \dfrac{(1 – cos x)}{(1 + cos x)}$$

Asked by Aaryan | 1 year ago |  81

##### Solution :-

From the question it is given that,

$$\dfrac{dy}{dx} = \dfrac{(1 – cos x)}{(1 + cos x)}$$

We know that (1 – cos x)

= 2 sin2 ($$\dfrac{x}{2}$$) and (1 + cos x)

= 2 cos2 ($$\dfrac{x}{2}$$)

So,

$$\dfrac{dy}{dx}=\dfrac{ (2 sin^2 ( \dfrac{x}{2}))}{(2 cos^2 (\dfrac{x}{2})) }$$

Also we know that = $$\dfrac{ sin θ}{cos θ }= tan θ$$

$$\dfrac{dy}{dx}$$ = tan2 ($$\dfrac{x}{2}$$)

By cross multiplication,

dy = tan2 ($$\dfrac{x}{2}$$) dx

Integrating on both side, we get,

∫dy = ∫tan2 ($$\dfrac{x}{2}$$) dx

We know that, sec2 x – 1 = tan2 x

∫dy = ∫sec2 ($$\dfrac{x}{2}$$) – 1 dx

y = 2 tan ($$\dfrac{x}{2}$$) – x + c

Therefore, y = 2 tan ($$\dfrac{x}{2}$$) – x + c

Answered by Aaryan | 1 year ago

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