Question Number 67653 by mhmd last updated on 29/Aug/19
Answered by mr W last updated on 29/Aug/19
$${let}\:{t}=\frac{{y}}{{x}} \\ $$$${y}={xt} \\ $$$${y}'={t}+{xt}' \\ $$$$ \\ $$$${xy}'={y}+\mathrm{2}{x}^{\mathrm{3}} \mathrm{sin}^{\mathrm{2}} \:\left(\frac{{y}}{{x}}\right) \\ $$$${y}'=\frac{{y}}{{x}}+\mathrm{2}{x}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\left(\frac{{y}}{{x}}\right) \\ $$$${t}+{xt}'={t}+\mathrm{2}{x}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{t} \\ $$$${t}'=\mathrm{2}{x}\:\mathrm{sin}^{\mathrm{2}} \:{t} \\ $$$$\frac{{dt}}{\mathrm{sin}^{\mathrm{2}} \:{t}}=\mathrm{2}{xdx} \\ $$$$\int\frac{{dt}}{\mathrm{sin}^{\mathrm{2}} \:{t}}=\int\mathrm{2}{xdx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{tan}\:{t}}={x}^{\mathrm{2}} +{C} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} +{C}\right)\mathrm{tan}\:\frac{{y}}{{x}}+\mathrm{1}=\mathrm{0} \\ $$$${or} \\ $$$$\Rightarrow{y}=−{x}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} +{C}} \\ $$