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Question Number 160903 by blackmamba last updated on 08/Dec/21

$$(1+x^2)y{}^{\prime }=2xy+{(1+x^2)}^2$$

Answered by GuruBelakangPadang last updated on 09/Dec/21

$$(1+x^2)y{}^{\prime }={(1+x^2)}^{\prime }y+{(1+x^2)}^2$$$${(1+x^2)}^{-1}y{}^{\prime }={(1+x^2)}^{-2}{(1+x^2)}^{\prime }y$$$${(1+x^2)}^{-1}y{}^{\prime }-{(1+x^2)}^{-2}{(1+x^2)}^{\prime }y=0$$$${(1+x^2)}^{-1}y{}^{\prime }+{\left({(1+x^2)}^{-1}\right)}^{{}^{\prime }}y=0$$$${(1+x^2)}^{-1}y=\frac{y}{1+x^2}=C$$

Answered by yeti123 last updated on 09/Dec/21

$$(1+x^2)\frac{dy}{dx}=2xy+{(1+x^2)}^2$$$$\frac{dy}{dx}-\frac{2xy}{1+x^2}=1+x^2$$$${(1+x^2)}^{-1}\frac{dy}{dx}-\frac{2xy}{1+x^2}{(1+x^2)}^{-1}=(1+x^2){(1+x^2)}^{-1}$$$$\frac{d}{dx}\left\{{(1+x^2)}^{-1}y\right\}=1$$$${(1+x^2)}^{-1}y=x+c$$$$y=(x+c)(1+x^2)$$

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