Question

The solution of $ydx-xdy+\log xdx=0$ is

• A. $y-\log x-1=Cx$
• B. $x+\log y+1=Cx$
• C. $y+\log x+1=Cx$
• D. $y+\log x-1=Cx$

Answer 1

The correct option is C $y+\log x+1=Cx$

$-ydx+xdy-\log xdx=0$

$\frac{dy}{dx}-\frac{y}{x}=\frac{\log x}{x}$

$I.F=e^{\int -1/x.dx}=e^{-\ln x}=\frac{1}{x}$

$\frac{y}{x}=\int \frac{1}{x}.\frac{\ln x}{x}.dx$

$\ln x=tx=e^t$

$\frac{y}{x}=\int t.e^{-t}.dt$

$=-t.e^{-t}+\int e^{-t}.dt$

$=-t.e^{-t}-e^{-t}+C$

$y=x[\frac{-\ln 1}{x}-\frac{1}{x}]+Cx$

$Cx=y+\log x+1$