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dy-dx-2x-2-y-2-y-0-1-

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Question Number 93283 by john santu last updated on 12/May/20
(dy/dx) = 2x^2 +y^2 , y(0) = 1
$$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:,\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\: \\ $$

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