Question

The solution of differential equation $x\frac{dy}{dx}+y=y^2$ is:

• A. $y=1+cxy$
• B. $y=\ln \left(cxy\right)$
• C. $y+1=cxy$
• D. $y=c+xy$

Answer 1

Answer: (A)

$y=1+cxy$

$x\frac{dy}{dx}+y=y^2$

$\therefore \frac{dy}{y^2-y}=\frac{dx}{x}$

$\therefore \frac{dy}{y-1}-\frac{dy}{y}=\frac{dx}{x}$

$\therefore \ln (y-1)-\ln y=\ln x+C^{{}^{\prime }}$

$\therefore \ln \left(\frac{y-1}{yx}\right)=C^{{}^{\prime }}$

$\therefore \frac{y-1}{yx}=e^{C^{{}^{\prime }}}$

$\therefore \frac{y-1}{yx}=c$ ........ $[e^{C^{{}^{\prime }}}=c]$

$\therefore y=1+cxy$