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Question Number 122281 by benjo_mathlover last updated on 15/Nov/20

$$\frac{dy}{dx}=\frac{-y}{4y^3+3y^2+x}withy\left(1\right)=-1$$

Answered by liberty last updated on 15/Nov/20

$$({4\mathrm{y}}^3+{3\mathrm{y}}^2+\mathrm{x})\mathrm{dy}+\mathrm{ydx}=0$$$$\Rightarrow ({4\mathrm{y}}^3+{3\mathrm{y}}^2)\mathrm{dy}+\mathrm{d}\left(\mathrm{xy}\right)=0$$$$\Rightarrow \int ({4\mathrm{y}}^3+{3\mathrm{y}}^2)\mathrm{dy}+\int \mathrm{d}\left(\mathrm{xy}\right)=\mathrm{C}$$$$\Rightarrow {\mathrm{y}}^4+{\mathrm{y}}^3+\mathrm{xy}=\mathrm{C}$$$$\mathrm{where}\mathrm{y}\left(1\right)=-1\mathrm{give}{(-1)}^4+{(-1)}^3+\left(1\right)(-1)=\mathrm{C}$$$$\Rightarrow 1-1-1=\mathrm{C};\mathrm{C}=-1$$$$\because {\mathrm{y}}^4+{\mathrm{y}}^3+\mathrm{xy}+1=0.$$

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