Solve-the-differential-equation-1-y-2-dx-1-x-2-xydy-0-

Question Number 175670 by Engr_Jidda last updated on 04/Sep/22

$${Solve}\:{the}\:{differential}\:{equation} \\ $$$$\left(\mathrm{1}+{y}^{\mathrm{2}} \right){dx}−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){xydy}=\mathrm{0} \\ $$

Answered by mahdipoor last updated on 04/Sep/22

⇒(1+y^2 )dx=(x+x^3 )ydy ⇒(dx/(x+x^3 ))=((ydy)/(1+y^2 ))⇒ ∫((1/x)−(x/(x^2 +1)))dx=(1/2)∫((2ydy)/((1+y^2 )))⇒ ln∣x∣−(1/2)ln(x^2 +1)=(1/2)ln(y^2 +1)+lnC⇒ (x/( (√(1+x^2 ))))=C(√(1+y^2 ))

$$\Rightarrow\left(\mathrm{1}+{y}^{\mathrm{2}} \right){dx}=\left({x}+{x}^{\mathrm{3}} \right){ydy} \\ $$$$\Rightarrow\frac{{dx}}{{x}+{x}^{\mathrm{3}} }=\frac{{ydy}}{\mathrm{1}+{y}^{\mathrm{2}} }\Rightarrow \\ $$$$\int\left(\frac{\mathrm{1}}{{x}}−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{ydy}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\Rightarrow \\ $$$${ln}\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)+{lnC}\Rightarrow \\ $$$$\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}={C}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$

Commented by Engr_Jidda last updated on 05/Sep/22