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 (\( \dfrac{v}{2}\)) + c_{1}

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

\( c + y = \dfrac{cot (x – y)}{2}\)

Answered by Aaryan | 1 year agoFind the particular solution satisfying the given condition \( (\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y\)

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