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Question Number 84100 by niroj last updated on 09/Mar/20
 Find the differential equations:    x(dy/dx)= y− (√(x^2 +y^2 ))
$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}: \\ $$$$\:\:\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}=\:\mathrm{y}−\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} } \\ $$
Answered by TANMAY PANACEA last updated on 09/Mar/20
xdy−ydx=−(√(x^2 +y^2 )) dx  ((xdy−ydx)/(x^2 ×(√(1+((y/x))^2 ))))=−((xdx)/x^2 )  ∫((d((y/x)))/( (√(1+((y/x))^2 ))))=(−1)∫(dx/x)  ln((y/x)+(√(1+((y/x))^2 )) )=−lnx+lnc  x((y/x)+((√(x^2 +y^2 ))/x))=c  y+(√(x^2 +y^2 )) =c
$${xdy}−{ydx}=−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:{dx} \\ $$$$\frac{{xdy}−{ydx}}{{x}^{\mathrm{2}} ×\sqrt{\mathrm{1}+\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} }}=−\frac{{xdx}}{{x}^{\mathrm{2}} } \\ $$$$\int\frac{{d}\left(\frac{{y}}{{x}}\right)}{\:\sqrt{\mathrm{1}+\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} }}=\left(−\mathrm{1}\right)\int\frac{{dx}}{{x}} \\ $$$${ln}\left(\frac{{y}}{{x}}+\sqrt{\mathrm{1}+\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} }\:\right)=−{lnx}+{lnc} \\ $$$${x}\left(\frac{{y}}{{x}}+\frac{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}{{x}}\right)={c} \\ $$$${y}+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:={c} \\ $$$$ \\ $$
Commented by niroj last updated on 09/Mar/20
its great .
$${its}\:{great}\:. \\ $$
Answered by TANMAY PANACEA last updated on 09/Mar/20
xdy−ydx=−(√(x^2 +y^2 ))  dx  ydx−xdy=(√(x^2 +y^2 ))  dx  ((ydx−xdy)/y^2 )×y^2 =y×(√(1+(x^2 /y^2 )))  dx  ((d((x/y)))/( (√(1+(x^2 /y^2 )))))×(y/x)=(dx/x)  (dp/(p(√(1+p^2 ))))=(dx/x)  p=tanθ  ((sec^2 θ dθ)/(tanθ.secθ))=(dx/x)  ∫((cosθ)/(cosθ×sinθ))dθ=∫(dx/x)  ∫cosecθ dθ=∫(dx/x)  lntan((θ/2))=lnx+lnc  tan((θ/2))=xc  tan(((tan^(−1) p)/2))=xC  tan(((tan^− ((x/y)))/2))=xc
$${xdy}−{ydx}=−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\:{dx} \\ $$$${ydx}−{xdy}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\:{dx} \\ $$$$\frac{{ydx}−{xdy}}{{y}^{\mathrm{2}} }×{y}^{\mathrm{2}} ={y}×\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }}\:\:{dx} \\ $$$$\frac{{d}\left(\frac{{x}}{{y}}\right)}{\:\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }}}×\frac{{y}}{{x}}=\frac{{dx}}{{x}} \\ $$$$\frac{{dp}}{{p}\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}=\frac{{dx}}{{x}} \\ $$$${p}={tan}\theta \\ $$$$\frac{{sec}^{\mathrm{2}} \theta\:{d}\theta}{{tan}\theta.{sec}\theta}=\frac{{dx}}{{x}} \\ $$$$\int\frac{{cos}\theta}{{cos}\theta×{sin}\theta}{d}\theta=\int\frac{{dx}}{{x}} \\ $$$$\int{cosec}\theta\:{d}\theta=\int\frac{{dx}}{{x}} \\ $$$${lntan}\left(\frac{\theta}{\mathrm{2}}\right)={lnx}+{lnc} \\ $$$${tan}\left(\frac{\theta}{\mathrm{2}}\right)={xc} \\ $$$${tan}\left(\frac{{tan}^{−\mathrm{1}} {p}}{\mathrm{2}}\right)={xC} \\ $$$${tan}\left(\frac{{tan}^{−} \left(\frac{{x}}{{y}}\right)}{\mathrm{2}}\right)={xc} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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