Question Number 168613 by LEKOUMA last updated on 14/Apr/22
$${Resolve}\: \\ $$$$\left.\mathrm{1}\right)\:{x}\frac{{dy}}{{dx}}−{y}={y}^{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\:\left({x}−{y}\right){ydx}−{x}^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\left(\mathrm{2}{x}−{y}\right){dx}+\left(\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{3}\right){dy}=\mathrm{0} \\ $$
Commented by Mastermind last updated on 14/Apr/22
$$ \\ $$$${Solution}\:\mathrm{1} \\ $$$${x}\frac{{dy}}{{dx}}−{y}={y}^{\mathrm{3}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\frac{{dy}}{{dx}}−\frac{\mathrm{1}}{{xy}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}}\:………\left(\mathrm{1}\right) \\ $$$${put}\:{z}=−\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:,\:{so}\:{that}\:\frac{{dz}}{{dx}}=\mathrm{2}\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\frac{{dy}}{{dx}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\frac{{dz}}{{dx}}=\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\frac{{dy}}{{dx}} \\ $$$${from}\:\left(\mathrm{1}\right),\:{we}\:{have} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\frac{{dz}}{{dx}}−\frac{{z}}{{x}}=\frac{\mathrm{1}}{{x}} \\ $$$${multiply}\:{all}\:{through}\:{by}\:\mathrm{2},\:{then} \\ $$$$\frac{{dz}}{{dx}}−\frac{\mathrm{2}{z}}{{x}}=\frac{\mathrm{2}}{{x}}\:……..\left(\mathrm{2}\right) \\ $$$${Now},\:{IF}\:=\:{e}^{\int−\frac{\mathrm{2}}{{x}}} \Rightarrow{e}^{{ln}\left({x}^{−\mathrm{2}} \right)} \\ $$$$\Rightarrow\:{x}^{−\mathrm{2}} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${multiply}\:{both}\:{sides}\:{by}\:{IF} \\ $$$${we}\:{have},\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{{dz}}{{dx}}−\frac{\mathrm{2}{z}}{{x}}\right)=\frac{\mathrm{2}}{{x}^{\mathrm{3}} } \\ $$$${Now},\:\frac{{d}}{{dx}}\left({z}\centerdot\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}}{{x}^{\mathrm{3}} } \\ $$$${d}\left({z}\centerdot\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}}{{x}^{\mathrm{3}} }{dx} \\ $$$${On}\:{integrating},\:{we}\:{get} \\ $$$$\frac{{z}}{{x}^{\mathrm{2}} }=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{c} \\ $$$$\Rightarrow{z}=−\mathrm{1}+{cx}^{\mathrm{2}} \\ $$$${putting}\:{z}=−\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$${We}\:{have},\: \\ $$$$−\frac{\mathrm{1}}{{y}^{\mathrm{2}} }=−\mathrm{1}+{cx}^{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} }=\mathrm{1}−{cx}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Resolved} \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by mr W last updated on 15/Apr/22
$$\left(\mathrm{1}\right) \\ $$$$\frac{{dy}}{{y}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}=\frac{{dx}}{{x}} \\ $$$$\left(\frac{\mathrm{1}}{{y}}−\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right){dy}=\frac{{dx}}{{x}} \\ $$$$\mathrm{ln}\:{y}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+{y}^{\mathrm{2}} \right)=\mathrm{ln}\:{x}+{C} \\ $$$$\Rightarrow\frac{{y}^{\mathrm{2}} }{\:\mathrm{1}+{y}^{\mathrm{2}} }={Cx}^{\mathrm{2}} \\ $$$${or} \\ $$$$\Rightarrow\left(\mathrm{1}−{Cx}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$${or} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }=\mathrm{1}−{Cx}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 15/Apr/22
$$\left(\mathrm{2}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{\left({x}−{y}\right){y}}{{x}^{\mathrm{2}} } \\ $$$${y}={xt} \\ $$$$\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}} \\ $$$${t}+{x}\frac{{dt}}{{dx}}={t}\left(\mathrm{1}−{t}\right) \\ $$$${x}\frac{{dt}}{{dx}}=−{t}^{\mathrm{2}} \\ $$$$\frac{{dt}}{{t}^{\mathrm{2}} }=−\frac{{dx}}{{x}} \\ $$$$\int\frac{{dt}}{{t}^{\mathrm{2}} }=−\int\frac{{dx}}{{x}} \\ $$$$−\frac{\mathrm{1}}{{t}}=−\mathrm{ln}\:{x}−{C} \\ $$$$\frac{{x}}{{y}}=\mathrm{ln}\:{x}+{C} \\ $$$$\Rightarrow{y}=\frac{{x}}{\mathrm{ln}\:{x}+{C}} \\ $$
Answered by mr W last updated on 15/Apr/22
$$\left(\mathrm{3}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{−\mathrm{2}{x}+{y}}{\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{3}} \\ $$$${let}\:{u}=−\mathrm{2}{x}+{y} \\ $$$$\frac{{du}}{{dx}}=−\mathrm{2}+\frac{{dy}}{{dx}} \\ $$$$\mathrm{2}+\frac{{du}}{{dx}}=\frac{{u}}{−\mathrm{2}{u}+\mathrm{3}} \\ $$$$\frac{{du}}{{dx}}=\frac{\mathrm{5}{u}−\mathrm{6}}{−\mathrm{2}{u}+\mathrm{3}} \\ $$$$\frac{\left(\mathrm{2}{u}−\mathrm{3}\right){du}}{\mathrm{5}{u}−\mathrm{6}}=−{dx} \\ $$$$\frac{\left(\mathrm{10}{u}−\mathrm{12}−\mathrm{3}\right){du}}{\mathrm{5}{u}−\mathrm{6}}=−\mathrm{5}{dx} \\ $$$$\left(\mathrm{2}−\frac{\mathrm{3}}{\mathrm{5}{u}−\mathrm{6}}\right){du}=−\mathrm{5}{dx} \\ $$$$\mathrm{2}{u}−\frac{\mathrm{3}}{\mathrm{5}}\mathrm{ln}\:\left(\mathrm{5}{u}−\mathrm{6}\right)=−\mathrm{5}{x}+{C} \\ $$$$−\mathrm{4}{x}+\mathrm{2}{y}−\frac{\mathrm{3}}{\mathrm{5}}\mathrm{ln}\:\left(−\mathrm{10}{x}+\mathrm{5}{y}−\mathrm{6}\right)=−\mathrm{5}{x}+{C} \\ $$$$\Rightarrow{x}+\mathrm{2}{y}−\frac{\mathrm{3}}{\mathrm{5}}\mathrm{ln}\:\left(−\mathrm{10}{x}+\mathrm{5}{y}−\mathrm{6}\right)={C} \\ $$