Question Number 127743 by arash sharifi last updated on 01/Jan/21 | ||
$${dx}+{ydy}={x}^{\mathrm{2}} {ydy} \\ $$ | ||
Commented by mr W last updated on 01/Jan/21 | ||
$${you}\:{have}\:{asked}\:{the}\:{same}\:{question}\: \\ $$$${before}\:{and}\:{it}\:{was}\:{answered}! \\ $$ | ||
Answered by Olaf last updated on 01/Jan/21 | ||
$${dx}+{ydx}\:=\:{x}^{\mathrm{2}} {ydy} \\ $$$${dx}\:=\:{y}\left({x}^{\mathrm{2}} −\mathrm{1}\right){dy} \\ $$$$\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{1}}\:=\:{ydy} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right]{dx}\:=\:{ydy} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\mid\:=\:\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{C}_{\mathrm{1}} \\ $$$${y}\:=\:\pm\sqrt{\mathrm{ln}\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\mid+\mathrm{C}_{\mathrm{2}} } \\ $$ | ||
Answered by ZaidMNuri last updated on 01/Jan/21 | ||
$${dx}=\left({x}^{\mathrm{2}} −\mathrm{1}\right){ydy} \\ $$$$\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}={ydy} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}=\int{ydy} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\left[\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right]{dx}=\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\mid=\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{C} \\ $$$${y}=\mp\sqrt{{ln}\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\mid−\mathrm{2}{C}} \\ $$ | ||