Question

The solution of $ydx-xdy+logxdx=0$ is:

• A. $(x+1)+\log x=cy$
• B. $y+1+\log x=cx$
• C. $\frac{(y+1{)}^2}{2}+\log x=cy$
• D. $(y+1{)}^2=\log cx$

Answer 1

Answer: (B)

$y+1+\log x=cx$

$ydx=xdy-logxdx$

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

$\Rightarrow \frac{logx}{x}=\frac{dy}{dx}-1/x$

$\Rightarrow \frac{logx}{x^2}=\frac{d}{dx}\left(y/x\right)$

$\int \frac{logx}{x^2}dx=\int d\left(y/x\right)+c_1$

$\int \frac{logx}{x^2}dx=y/x+c_1$

$letlogx=t\Rightarrow 1/xdx=dt$

$x=e^t$

$1/x=e^{-t}$

$\Rightarrow \int t{e}^{-t}dt=-t{e}^{-t}-e^{-t}$

$=-\left(\frac{1+t}{e^t}\right)=-\frac{(1+logx)}{x}$

$\Rightarrow -\left(\frac{1+logx}{x}\right)=y/x+c_1$

$\Rightarrow 1+logx+y+c_{1}x=0$

$put{c}_1=-c$

$\Rightarrow 1+logx+y=cx$