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,

\( 1 – (\dfrac{dv}{dx}) = \dfrac{(v + 3)}{(2v + 3)}\)

Transposing we get,

\(\dfrac{dv}{dx} = \dfrac{1 – (v + 3)}{(2v + 5)}\)

\(\dfrac{dv}{dx} = \dfrac{(2v + 5 – v – 3)}{(2v + 5)}\)

\(\dfrac{dv}{dx} =\dfrac{ (v + 2)}{(2v + 5)}\)

Now, taking like variables on same side,

\(\dfrac{2v + 5}{(v + 2)}dv = dx\)

\( \dfrac{(2v + 4) + 1)}{(v + 2)} dv = dx\)

On dividing we get,

\( 2 + \dfrac{1}{(v + 2)}dv = dx\)

Integrating on both side we get,

\(∫2 + \dfrac{1}{(v + 2)} dv = ∫dx\)

We know that,

\( ∫\dfrac{dx}{x} = log x + c \;and\; ∫adx = ax + c\)

2v + log [v + 2] = x + c

2(x – y) + log [x – y – 2] = x + c

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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