Question Number 30421 by abdo imad last updated on 22/Feb/18
$${integrate}\:{y}^{'} −\mathrm{2}{ty}\:+{ty}^{\mathrm{2}} =\mathrm{0} \\ $$
Answered by sma3l2996 last updated on 24/Feb/18
$${y}'={t}\left(\mathrm{2}{y}−{y}^{\mathrm{2}} \right) \\ $$$$\frac{{dy}}{{y}\left(\mathrm{2}−{y}\right)}={tdt} \\ $$$$\frac{\mathrm{1}}{{y}\left(\mathrm{2}−{y}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{\mathrm{2}−{y}}\right) \\ $$$$\int\frac{{dy}}{{y}\left(\mathrm{2}−{y}\right)}=\int\left(\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{\mathrm{2}−{y}}\right){dy}={ln}\mid{y}\mid−{ln}\mid\mathrm{2}−{y}\mid=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\lambda \\ $$$${ln}\mid\frac{{y}}{\mathrm{2}−{y}}\mid=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\lambda \\ $$$$\frac{{y}}{\mathrm{2}−{y}}={e}^{\lambda} ×{e}^{{t}^{\mathrm{2}} /\mathrm{2}} \\ $$$${y}=×{Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} \left(\mathrm{2}−{y}\right) \\ $$$${y}\left(\mathrm{1}+{Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} \right)={Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} \\ $$$${y}\left({x}\right)=\frac{{Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} }{\mathrm{1}+{Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} } \\ $$$$ \\ $$