Find the particular solution satisfying the given condition (1 + x2)dy = xy dx

Asked by Sakshi | 1 year ago |  44

Solution :-

From the question it is given that,

(1 + x2)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 [x2 + 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 ago

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