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1-x-2-dy-dx-2xy-0-




Question Number 85498 by Roland Mbunwe last updated on 22/Mar/20
(1−x^2 )(dy/(dx ))−2xy=0
$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}\:}−\mathrm{2}{xy}=\mathrm{0} \\ $$
Commented by niroj last updated on 22/Mar/20
 (1−x^2 )(dy/dx)−2xy=0      (1−x^2 )(dy/dx)= 2xy    (1/y)dy= ((2x)/(1−x^2 ))dx    On integrating both side,    ∫(1/y)dy=2∫(( xdx)/(1−x^2 ))    put,  1−x^2 =t     −2xdx=dt         xdx=−(dt/2)   logy =−2∫(dt/(t.2))   logy= −log t+log c    log y= −log(1−x^2 )+log c   log y = log(c/(1−x^2 ))        y=(c/(1−x^2 ))     y−x^2 y=c     y=x^2 y+C  //.
$$\:\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{2xy}=\mathrm{0} \\ $$$$\:\:\:\:\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\frac{\mathrm{dy}}{\mathrm{dx}}=\:\mathrm{2xy} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{y}}\mathrm{dy}=\:\frac{\mathrm{2x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:\mathrm{On}\:\mathrm{integrating}\:\mathrm{both}\:\mathrm{side}, \\ $$$$\:\:\int\frac{\mathrm{1}}{\mathrm{y}}\mathrm{dy}=\mathrm{2}\int\frac{\:\mathrm{xdx}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\mathrm{put},\:\:\mathrm{1}−\mathrm{x}^{\mathrm{2}} =\mathrm{t} \\ $$$$\:\:\:−\mathrm{2xdx}=\mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\mathrm{xdx}=−\frac{\mathrm{dt}}{\mathrm{2}} \\ $$$$\:\mathrm{logy}\:=−\mathrm{2}\int\frac{\mathrm{dt}}{\mathrm{t}.\mathrm{2}} \\ $$$$\:\mathrm{logy}=\:−\mathrm{log}\:\mathrm{t}+\mathrm{log}\:\mathrm{c} \\ $$$$\:\:\mathrm{log}\:\mathrm{y}=\:−\mathrm{log}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)+\mathrm{log}\:\mathrm{c} \\ $$$$\:\mathrm{log}\:\mathrm{y}\:=\:\mathrm{log}\frac{\mathrm{c}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\mathrm{y}=\frac{\mathrm{c}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\:\mathrm{y}−\mathrm{x}^{\mathrm{2}} \mathrm{y}=\mathrm{c} \\ $$$$\:\:\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{C}\:\://. \\ $$$$ \\ $$$$ \\ $$
Answered by mind is power last updated on 22/Mar/20
⇔(dy/y)=((2x)/(1−x^2 ))  ⇒ln∣y∣=−ln∣x^2 −1∣  ∣y∣=(c/(∣x^2 −1∣))  ⇒y=(c/(∣x^2 −1∣))
$$\Leftrightarrow\frac{{dy}}{{y}}=\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{ln}\mid{y}\mid=−{ln}\mid{x}^{\mathrm{2}} −\mathrm{1}\mid \\ $$$$\mid{y}\mid=\frac{{c}}{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid} \\ $$$$\Rightarrow{y}=\frac{{c}}{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid} \\ $$

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