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