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Question Number 121687 by abdelsalamalmukasabe last updated on 10/Nov/20

Commented by bemath last updated on 11/Nov/20

$$\left(\frac{2y^4+1}{y^2}\right)dx+\left(4xy\right)dy=0$$$$(2y^4+1)dx+\left(4{xy}^3\right)dy=0$$$$\Rightarrow \frac{dy}{dx}=\frac{-(2y^4+1)}{4{xy}^3}$$$$\Rightarrow \frac{y^3}{2y^4+1}dy=-\frac{dx}{4x}$$$$\Rightarrow \int \frac{d(2y^4+1)}{2y^4+1}=-\int \frac{2dx}{x}$$$$\Rightarrow \ln (2y^4+1)=-2\ln \mid x\mid +c$$$$\Rightarrow \ln \left(x^2(2y^4+1)\right)=c$$$$\Rightarrow 2x^2{y}^4+x^2=e^c=C$$$$\therefore 2x^2{y}^4+x^2=C$$

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