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Question Number 40399 by rahul 19 last updated on 21/Jul/18

Solve : (2(√(xy)) −x)dy + ydx = 0.

$$\mathrm{Solve}\:: \\ $$$$\left(\mathrm{2}\sqrt{{xy}}\:−{x}\right){dy}\:+\:{ydx}\:=\:\mathrm{0}. \\ $$

Answered by rahul 19 last updated on 21/Jul/18

ok so, i was able to do by taking (y/x)=t but i want to get this ans. by method which ajfour sir did in my last doubt. My try: ((2(√(xy)) )/x^2 )dy = ((xdy−ydx)/x^2 ) ⇒ d((y/x))= ((2(√(xy)))/x^2 ) dy ........................ How to proceed further?

$$\mathrm{ok}\:\mathrm{so},\:\mathrm{i}\:\mathrm{was}\:\mathrm{able}\:\mathrm{to}\:\mathrm{do}\:\mathrm{by}\:\mathrm{taking}\:\frac{\mathrm{y}}{{x}}={t} \\ $$$${but}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{get}\:\mathrm{this}\:\mathrm{ans}.\:\mathrm{by}\: \\ $$$$\mathrm{method}\:\mathrm{which}\:\mathrm{ajfour}\:\mathrm{sir}\:\mathrm{did}\:\mathrm{in}\:\mathrm{my}\:\mathrm{last} \\ $$$$\mathrm{doubt}. \\ $$$$\mathrm{My}\:\mathrm{try}: \\ $$$$\frac{\mathrm{2}\sqrt{{xy}}\:}{{x}^{\mathrm{2}} }{dy}\:\:=\:\frac{{xdy}−{ydx}}{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{d}\left(\frac{\mathrm{y}}{{x}}\right)=\:\frac{\mathrm{2}\sqrt{{xy}}}{{x}^{\mathrm{2}} }\:{dy}\: \\ $$$$........................ \\ $$$${H}\mathrm{ow}\:\mathrm{to}\:\mathrm{proceed}\:\mathrm{further}? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jul/18

x^(3/2) d((y/x))=2y^(1/2) dy ((d((y/x)))/(((y/x))^(3/2) ))=2(dy/y) devide both side by y^(3/2) intregating ∫((d((y/x)))/(((y/x))^(3/2) ))=2∫(dy/y) (1/2).((((y/x))^((−1)/2) )/(−(1/2)))=lny+lnc −((y/x))^(−(1/2)) =ln(yc) yc=e^(−((√x)/(√y)))

$${x}^{\frac{\mathrm{3}}{\mathrm{2}}} {d}\left(\frac{{y}}{{x}}\right)=\mathrm{2}{y}^{\frac{\mathrm{1}}{\mathrm{2}}} {dy} \\ $$$$\frac{{d}\left(\frac{{y}}{{x}}\right)}{\left(\frac{{y}}{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\mathrm{2}\frac{{dy}}{{y}}\:\:\:{devide}\:{both}\:{side}\:{by}\:{y}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${intregating} \\ $$$$\int\frac{{d}\left(\frac{{y}}{{x}}\right)}{\left(\frac{{y}}{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\mathrm{2}\int\frac{{dy}}{{y}} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\left(\frac{{y}}{{x}}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }{−\frac{\mathrm{1}}{\mathrm{2}}}={lny}+{lnc} \\ $$$$−\left(\frac{{y}}{{x}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} ={ln}\left({yc}\right) \\ $$$${yc}={e}^{−\frac{\sqrt{{x}}}{\sqrt{{y}}}} \: \\ $$

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(x−2(√(xy)) ) (dy/dx)=(y/(x−2(√(xy)) )) (dy/dx)=((y/x)/(1−2(√(y/x)))) t=(y/x) put y=tx so (dy/dx)=t+x(dt/dx) t+x(dt/dx)=(t/(1−2(√t) )) x(dt/dx)=(t/(1−2(√t) ))−t x(dt/dx)=((t−t+2t^(3/2) )/(1−2t^(1/2) )) ((1−2t^(1/2) )/(2t^(3/2) ))dt=(dx/x) (1/2)∫t^((−3)/2) dt−∫(dt/t)=∫(dx/x) ((1/2)/((−1)/2))t^((−1)/2) −lnt=lnx+lnc −(1/((√t) ))=ln(xtc) −(((√x) )/(√y))=ln(yc) yc=e^((−(√x))/(√y))

$${ydx}={dy}\left({x}−\mathrm{2}\sqrt{{xy}}\:\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{{y}}{{x}−\mathrm{2}\sqrt{{xy}}\:} \\ $$$$\frac{{dy}}{{dx}}=\frac{\frac{{y}}{{x}}}{\mathrm{1}−\mathrm{2}\sqrt{\frac{{y}}{{x}}}} \\ $$$${t}=\frac{{y}}{{x}}\:\:\:{put}\:\:{y}={tx}\:\:\:{so}\:\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}} \\ $$$${t}+{x}\frac{{dt}}{{dx}}=\frac{{t}}{\mathrm{1}−\mathrm{2}\sqrt{{t}}\:} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{{t}}{\mathrm{1}−\mathrm{2}\sqrt{{t}}\:}−{t} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{{t}−{t}+\mathrm{2}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{1}−\mathrm{2}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$\frac{\mathrm{1}−\mathrm{2}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt}=\frac{{dx}}{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{t}^{\frac{−\mathrm{3}}{\mathrm{2}}} {dt}−\int\frac{{dt}}{{t}}=\int\frac{{dx}}{{x}} \\ $$$$\:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{−\mathrm{1}}{\mathrm{2}}}{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} −{lnt}={lnx}+{lnc} \\ $$$$\:\:\:−\frac{\mathrm{1}}{\sqrt{{t}}\:}={ln}\left({xtc}\right) \\ $$$$\:\:\:\:−\frac{\sqrt{{x}}\:}{\sqrt{{y}}}={ln}\left({yc}\right) \\ $$$${yc}={e}^{\frac{−\sqrt{{x}}}{\sqrt{{y}}}} \\ $$

Answered by ajfour last updated on 21/Jul/18

2(√(xy))dy = xdy−ydx ⇒ ((2(√(xy))dy)/x^2 ) = d((y/x)) or 2(√y)dy = x(√x)d((y/x)) or ((2(√y)dy)/(y(√y))) = ((x/y))^(3/2) d((y/x)) ⇒ 2ln y = −2(√(x/y)) +c or y = ke^(−(√(x/y))) .

$$\mathrm{2}\sqrt{{xy}}{dy}\:=\:{xdy}−{ydx} \\ $$$$\Rightarrow\:\frac{\mathrm{2}\sqrt{{xy}}{dy}}{{x}^{\mathrm{2}} }\:=\:{d}\left(\frac{{y}}{{x}}\right) \\ $$$${or}\:\:\:\:\mathrm{2}\sqrt{{y}}{dy}\:=\:{x}\sqrt{{x}}{d}\left(\frac{{y}}{{x}}\right) \\ $$$${or}\:\:\:\:\frac{\mathrm{2}\sqrt{{y}}{dy}}{{y}\sqrt{{y}}}\:=\:\left(\frac{{x}}{{y}}\right)^{\mathrm{3}/\mathrm{2}} {d}\left(\frac{{y}}{{x}}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{2ln}\:{y}\:=\:−\mathrm{2}\sqrt{\frac{{x}}{{y}}}\:+{c} \\ $$$$\:\:\:{or}\:\:\:{y}\:=\:{ke}^{−\sqrt{\frac{{x}}{{y}}}} \:. \\ $$

Commented by rahul 19 last updated on 21/Jul/18

thank you sir ��

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