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Question Number 86657 by john santu last updated on 30/Mar/20

$$\mathrm{solve}(1+{\mathrm{x}}^3)\mathrm{dy}-{\mathrm{x}}^2\mathrm{y}\mathrm{dx}=0$$$$\mathrm{y}\left(1\right)=2$$

Answered by jagoll last updated on 30/Mar/20

$$\int \frac{\mathrm{dy}}{\mathrm{y}}=\int \frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^3}\mathrm{dx}$$$$\ln \mathrm{y}=\frac{1}{3}\mathrm{lnC}(1+{\mathrm{x}}^3)=\ln \sqrt[3]{\mathrm{C}(1+{\mathrm{x}}^3)}$$$$\mathrm{y}=\sqrt[3]{\mathrm{C}(1+{\mathrm{x}}^3)}$$$$\mathrm{y}\left(1\right)=2\Rightarrow 2=\sqrt[3]{2\mathrm{C}}$$$$8=2\mathrm{C}\Rightarrow \mathrm{C}=4$$$$\therefore \mathrm{y}=\sqrt[3]{4+{4\mathrm{x}}^3}$$

Commented by john santu last updated on 30/Mar/20

$$\mathrm{joosss}$$

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