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Question Number 30421 by abdo imad last updated on 22/Feb/18

integrate y^′ −2ty +ty^2 =0

$${integrate}\:{y}^{'} −\mathrm{2}{ty}\:+{ty}^{\mathrm{2}} =\mathrm{0} \\ $$

Answered by sma3l2996 last updated on 24/Feb/18

y′=t(2y−y^2 )  (dy/(y(2−y)))=tdt  (1/(y(2−y)))=(1/2)((1/y)+(1/(2−y)))  ∫(dy/(y(2−y)))=∫((1/y)+(1/(2−y)))dy=ln∣y∣−ln∣2−y∣=(1/2)t^2 +λ  ln∣(y/(2−y))∣=(1/2)t^2 +λ  (y/(2−y))=e^λ ×e^(t^2 /2)   y=×Ke^(t^2 /2) (2−y)  y(1+Ke^(t^2 /2) )=Ke^(t^2 /2)   y(x)=((Ke^(t^2 /2) )/(1+Ke^(t^2 /2) ))

$${y}'={t}\left(\mathrm{2}{y}−{y}^{\mathrm{2}} \right) \\ $$$$\frac{{dy}}{{y}\left(\mathrm{2}−{y}\right)}={tdt} \\ $$$$\frac{\mathrm{1}}{{y}\left(\mathrm{2}−{y}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{\mathrm{2}−{y}}\right) \\ $$$$\int\frac{{dy}}{{y}\left(\mathrm{2}−{y}\right)}=\int\left(\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{\mathrm{2}−{y}}\right){dy}={ln}\mid{y}\mid−{ln}\mid\mathrm{2}−{y}\mid=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\lambda \\ $$$${ln}\mid\frac{{y}}{\mathrm{2}−{y}}\mid=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\lambda \\ $$$$\frac{{y}}{\mathrm{2}−{y}}={e}^{\lambda} ×{e}^{{t}^{\mathrm{2}} /\mathrm{2}} \\ $$$${y}=×{Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} \left(\mathrm{2}−{y}\right) \\ $$$${y}\left(\mathrm{1}+{Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} \right)={Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} \\ $$$${y}\left({x}\right)=\frac{{Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} }{\mathrm{1}+{Ke}^{{t}^{\mathrm{2}} /\mathrm{2}} } \\ $$$$ \\ $$

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