Question Number 62869 by aliesam last updated on 26/Jun/19
$${y}\frac{{dy}}{{dx}}\:−\:\frac{{y}}{\frac{{dy}}{{dx}}}\:=\:\mathrm{2}{a} \\ $$$$ \\ $$$${a}\:{is}\:{a}\:{real}\:{number} \\ $$
Answered by Hope last updated on 26/Jun/19
![y((dy/dx))^2 −y=2a((dy/dx)) yp^2 −2ap−y=0 p=((2a±(√(4a^2 +4y^2 )))/(2y)) (dy/dx)=((2a±(√(4a^2 +4y^2 )))/(2y)) ((ydy)/(a±(√(a^2 +y^2 ))))=dx t^2 =a^2 +y^2 →tdt=ydy now ∫((ydy)/(a+(√(a^2 +y^2 ))))=∫dx →[considering + sign] ∫((tdt)/(a+t))=∫dx ∫((a+t−a)/(a+t))dt=∫dx ∫dt−a∫(dt/(a+t))=∫dx t−aln(a+t)=x+c (√(a^2 +y^2 )) −aln(a+(√(a^2 +y^2 )) )=x+c if cosider −ve sign ∫((ydy)/(a−(√(a^2 +y^2 ))))=∫dx ∫((tdt)/(a−t))=∫dx ∫((a−t−a)/(a−t))dt=−∫dx ∫dt−a∫(dt/(a−t))=−∫dx ∫dt+∫((adt)/(t−a))=−∫dx t+aln(t−a)=−x+c_1 (√(a^2 +y^2 )) +aln((√(a^2 +y^2 )) −a)=−x+c_1](Q62888.png)
$${y}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} −{y}=\mathrm{2}{a}\left(\frac{{dy}}{{dx}}\right) \\ $$$${yp}^{\mathrm{2}} −\mathrm{2}{ap}−{y}=\mathrm{0} \\ $$$${p}=\frac{\mathrm{2}{a}\pm\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }}{\mathrm{2}{y}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}{a}\pm\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }}{\mathrm{2}{y}} \\ $$$$\frac{{ydy}}{{a}\pm\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }}={dx} \\ $$$${t}^{\mathrm{2}} ={a}^{\mathrm{2}} +{y}^{\mathrm{2}} \rightarrow{tdt}={ydy} \\ $$$${now}\:\int\frac{{ydy}}{{a}+\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }}=\int{dx}\:\rightarrow\left[{considering}\:+\:{sign}\right] \\ $$$$\int\frac{{tdt}}{{a}+{t}}=\int{dx} \\ $$$$\int\frac{{a}+{t}−{a}}{{a}+{t}}{dt}=\int{dx} \\ $$$$\int{dt}−{a}\int\frac{{dt}}{{a}+{t}}=\int{dx} \\ $$$${t}−{aln}\left({a}+{t}\right)={x}+{c} \\ $$$$\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{aln}\left({a}+\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\right)={x}+{c} \\ $$$$ \\ $$$${if}\:{cosider}\:−{ve}\:{sign} \\ $$$$\int\frac{{ydy}}{{a}−\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }}=\int{dx} \\ $$$$\int\frac{{tdt}}{{a}−{t}}=\int{dx} \\ $$$$\int\frac{{a}−{t}−{a}}{{a}−{t}}{dt}=−\int{dx} \\ $$$$\int{dt}−{a}\int\frac{{dt}}{{a}−{t}}=−\int{dx} \\ $$$$\int{dt}+\int\frac{{adt}}{{t}−{a}}=−\int{dx} \\ $$$${t}+{aln}\left({t}−{a}\right)=−{x}+{c}_{\mathrm{1}} \\ $$$$\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:+{aln}\left(\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}\right)=−{x}+{c}_{\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$
Commented by aliesam last updated on 26/Jun/19
$${thank}\:{you}\:{sir}\:{brilliant}\:{sol} \\ $$
Commented by Hope last updated on 26/Jun/19
$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }}×\mathrm{2}{y}\frac{{dy}}{{dx}}+\frac{{a}}{\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }}×\mathrm{2}{y}\frac{{dy}}{{dx}}=−\mathrm{1} \\ $$$$\frac{{y}\frac{{dy}}{{dx}}}{\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:}\left(\mathrm{1}+\frac{{a}}{\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}}\right)=−\mathrm{1} \\ $$$$\frac{{y}\frac{{dy}}{{dx}}}{\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\left(\frac{\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:}{\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}}\right)=−\mathrm{1} \\ $$$${y}\frac{{dy}}{{dx}}=−\left(\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{−\left(\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}\right)}{{y}} \\ $$$${y}\frac{{dy}}{{dx}}−\frac{{y}}{\frac{{dy}}{{dx}}} \\ $$$$\frac{−\left(\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}\right)}{\mathrm{1}}+\frac{{y}^{\mathrm{2}} }{\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}} \\ $$$$\frac{{y}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{y}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\right)}{\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}} \\ $$$$=\frac{\mathrm{2}{a}\left(\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}\right)}{\sqrt{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:−{a}} \\ $$$$=\mathrm{2}{a}\:{proved} \\ $$$${so}\:{we}\:{can}\:{get}... \\ $$$$ \\ $$
Commented by mr W last updated on 26/Jun/19
$${very}\:{nice}\:{sir}! \\ $$
Commented by Hope last updated on 26/Jun/19
$${thank}\:{you}\:{sir}... \\ $$