From the question it is given that,

(1 + x^{2})dy = xy dx

By cross multiplication,

\( (\dfrac{1}{y}) dy =\dfrac{ x}{(x^2 + 1) }dx\)

Integrating on both side, we get,

\( ∫(\dfrac{1}{y}) dy = ∫ \dfrac{x}{(x^2 + 1)} dx\)

\( ∫(\dfrac{1}{y}) dy = \dfrac{1}{2} ∫ \dfrac{2x}{(x^2 + 1) }dx\)

Log y = \( \dfrac{1}{2}\) log [x^{2} + 1] + log c

\( Log\; y = log\; C(\sqrt{(x^2 + 1)}\)

y = \(C \sqrt{(x^2 + 1)}\)

Therefore, y = \( C \sqrt{(x^2 + 1)}\) is the required solution.

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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