Solve the differential equations condition $$\dfrac{dy}{dx} cos (x – y) = 1$$

Asked by Sakshi | 1 year ago |  67

##### Solution :-

Let us assume that, x – y = v

Differentiating with respect to x on both side, we get,

$$1 – (\dfrac{dy}{dx}) = \dfrac{dv}{dx}$$

Transposing,

$$\dfrac{dy}{dx} = 1 – (\dfrac{dv}{dx})$$ … [equation (ii)]

Substituting equation (ii) in equation (i),

Then,

$$1 – (\dfrac{dv}{dx}) cos v = 1$$

$$1 – (\dfrac{dv}{dx}) = sec v$$

$$1 – sec v = \dfrac{dv}{dx}$$

Now, taking like variables on same side,

$$dx = \dfrac{dv}{(1 – sec v)}$$

$$dx = \dfrac{cos}{(1 – cos v)} dv$$

Integrating on both side we get,

$$∫dx = ∫cos^2 (\dfrac{v}{2}) – \dfrac{sin^2 (\dfrac{v}{2})}{2 sin^2( \dfrac{v}{2}) dv}$$

We know that,

$$∫cosec^2 x = – cot x + c$$

$$∫dx = ∫\dfrac{1}{2} cot (\dfrac{v}{2}) dv – ∫\dfrac{1}{2}dv$$

$$2∫dx = ∫cosec^2 (\dfrac{v}{2} – 1) dv – ∫ dv$$

$$2x = – 2 cot (\dfrac{v}{2}) dv – v – v + c_1$$

2(x + v) = – 2 cot ($$\dfrac{v}{2}$$) + c1

$$x + x – y = – cot \dfrac{(x – y)}{2} + c$$

$$c + y = \dfrac{cot (x – y)}{2}$$

Answered by Aaryan | 1 year ago

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