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Solve-the-following-differential-equation-2xy-2-x-y-2-dx-4x-2-ydy-0-




Question Number 4472 by Yozzii last updated on 30/Jan/16
Solve the following differential equation:          (2xy^2 +(x/y^2 ))dx+4x^2 ydy=0
$${Solve}\:{the}\:{following}\:{differential}\:{equation}: \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{2}{xy}^{\mathrm{2}} +\frac{{x}}{{y}^{\mathrm{2}} }\right){dx}+\mathrm{4}{x}^{\mathrm{2}} {ydy}=\mathrm{0} \\ $$
Commented by prakash jain last updated on 01/Feb/16
u=y^4   du=4y^3 dy  (2u+1)dx+xdu=0  (dx/x)=−(du/(2u+1))  ln x=−(1/2)ln(2u+1)+c_1
$${u}={y}^{\mathrm{4}} \\ $$$${du}=\mathrm{4}{y}^{\mathrm{3}} {dy} \\ $$$$\left(\mathrm{2}{u}+\mathrm{1}\right){dx}+{xdu}=\mathrm{0} \\ $$$$\frac{{dx}}{{x}}=−\frac{{du}}{\mathrm{2}{u}+\mathrm{1}} \\ $$$$\mathrm{ln}\:{x}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}{u}+\mathrm{1}\right)+{c}_{\mathrm{1}} \\ $$

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