Question Number 40399 by rahul 19 last updated on 21/Jul/18
$$\mathrm{Solve}\:: \\ $$$$\left(\mathrm{2}\sqrt{{xy}}\:−{x}\right){dy}\:+\:{ydx}\:=\:\mathrm{0}. \\ $$
Answered by rahul 19 last updated on 21/Jul/18
$$\mathrm{ok}\:\mathrm{so},\:\mathrm{i}\:\mathrm{was}\:\mathrm{able}\:\mathrm{to}\:\mathrm{do}\:\mathrm{by}\:\mathrm{taking}\:\frac{\mathrm{y}}{{x}}={t} \\ $$$${but}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{get}\:\mathrm{this}\:\mathrm{ans}.\:\mathrm{by}\: \\ $$$$\mathrm{method}\:\mathrm{which}\:\mathrm{ajfour}\:\mathrm{sir}\:\mathrm{did}\:\mathrm{in}\:\mathrm{my}\:\mathrm{last} \\ $$$$\mathrm{doubt}. \\ $$$$\mathrm{My}\:\mathrm{try}: \\ $$$$\frac{\mathrm{2}\sqrt{{xy}}\:}{{x}^{\mathrm{2}} }{dy}\:\:=\:\frac{{xdy}−{ydx}}{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{d}\left(\frac{\mathrm{y}}{{x}}\right)=\:\frac{\mathrm{2}\sqrt{{xy}}}{{x}^{\mathrm{2}} }\:{dy}\: \\ $$$$…………………… \\ $$$${H}\mathrm{ow}\:\mathrm{to}\:\mathrm{proceed}\:\mathrm{further}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jul/18
$${x}^{\frac{\mathrm{3}}{\mathrm{2}}} {d}\left(\frac{{y}}{{x}}\right)=\mathrm{2}{y}^{\frac{\mathrm{1}}{\mathrm{2}}} {dy} \\ $$$$\frac{{d}\left(\frac{{y}}{{x}}\right)}{\left(\frac{{y}}{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\mathrm{2}\frac{{dy}}{{y}}\:\:\:{devide}\:{both}\:{side}\:{by}\:{y}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${intregating} \\ $$$$\int\frac{{d}\left(\frac{{y}}{{x}}\right)}{\left(\frac{{y}}{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\mathrm{2}\int\frac{{dy}}{{y}} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\left(\frac{{y}}{{x}}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }{−\frac{\mathrm{1}}{\mathrm{2}}}={lny}+{lnc} \\ $$$$−\left(\frac{{y}}{{x}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} ={ln}\left({yc}\right) \\ $$$${yc}={e}^{−\frac{\sqrt{{x}}}{\:\sqrt{{y}}}} \: \\ $$
Commented by rahul 19 last updated on 21/Jul/18
thank you sir ��
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jul/18
$${ydx}={dy}\left({x}−\mathrm{2}\sqrt{{xy}}\:\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{{y}}{{x}−\mathrm{2}\sqrt{{xy}}\:} \\ $$$$\frac{{dy}}{{dx}}=\frac{\frac{{y}}{{x}}}{\mathrm{1}−\mathrm{2}\sqrt{\frac{{y}}{{x}}}} \\ $$$${t}=\frac{{y}}{{x}}\:\:\:{put}\:\:{y}={tx}\:\:\:{so}\:\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}} \\ $$$${t}+{x}\frac{{dt}}{{dx}}=\frac{{t}}{\mathrm{1}−\mathrm{2}\sqrt{{t}}\:} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{{t}}{\mathrm{1}−\mathrm{2}\sqrt{{t}}\:}−{t} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{{t}−{t}+\mathrm{2}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{1}−\mathrm{2}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$\frac{\mathrm{1}−\mathrm{2}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt}=\frac{{dx}}{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{t}^{\frac{−\mathrm{3}}{\mathrm{2}}} {dt}−\int\frac{{dt}}{{t}}=\int\frac{{dx}}{{x}} \\ $$$$\:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{−\mathrm{1}}{\mathrm{2}}}{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} −{lnt}={lnx}+{lnc} \\ $$$$\:\:\:−\frac{\mathrm{1}}{\:\sqrt{{t}}\:}={ln}\left({xtc}\right) \\ $$$$\:\:\:\:−\frac{\sqrt{{x}}\:}{\:\sqrt{{y}}}={ln}\left({yc}\right) \\ $$$${yc}={e}^{\frac{−\sqrt{{x}}}{\:\sqrt{{y}}}} \\ $$
Answered by ajfour last updated on 21/Jul/18
$$\mathrm{2}\sqrt{{xy}}{dy}\:=\:{xdy}−{ydx} \\ $$$$\Rightarrow\:\frac{\mathrm{2}\sqrt{{xy}}{dy}}{{x}^{\mathrm{2}} }\:=\:{d}\left(\frac{{y}}{{x}}\right) \\ $$$${or}\:\:\:\:\mathrm{2}\sqrt{{y}}{dy}\:=\:{x}\sqrt{{x}}{d}\left(\frac{{y}}{{x}}\right) \\ $$$${or}\:\:\:\:\frac{\mathrm{2}\sqrt{{y}}{dy}}{{y}\sqrt{{y}}}\:=\:\left(\frac{{x}}{{y}}\right)^{\mathrm{3}/\mathrm{2}} {d}\left(\frac{{y}}{{x}}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{2ln}\:{y}\:=\:−\mathrm{2}\sqrt{\frac{{x}}{{y}}}\:+{c} \\ $$$$\:\:\:{or}\:\:\:{y}\:=\:{ke}^{−\sqrt{\frac{{x}}{{y}}}} \:. \\ $$
Commented by rahul 19 last updated on 21/Jul/18
thank you sir ��