Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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∣))

$$\Leftrightarrow\frac{{dy}}{{y}}=\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{ln}\mid{y}\mid=−{ln}\mid{x}^{\mathrm{2}} −\mathrm{1}\mid \\ $$$$\mid{y}\mid=\frac{{c}}{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid} \\ $$$$\Rightarrow{y}=\frac{{c}}{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com