Find the particular solution satisfying the given condition $$\dfrac{dy}{dx} =\dfrac{1+y^2}{y^3}$$

Asked by Sakshi | 1 year ago |  59

##### Solution :-

From the question it is given that,

$$\dfrac{dy}{dx}= \dfrac{(1 + y^2)}{y^3}$$

By cross multiplication,

$$\dfrac{y^3}{1 + y^2} dy = dx$$

Integrating on both side, we get,

$$∫y – \dfrac{y}{1 + y^2} dy = ∫dx$$

$$∫ydy – ∫\dfrac{y}{(1 + y^2)}dy = ∫ dx$$

$$∫ydy – \dfrac{1}{2} ∫\dfrac{2y}{(1 + y^2)} dy = ∫ dx$$

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

Answered by Aaryan | 1 year ago

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