Question

The curve satisfying $ydx-xdy+\log xdx=0(x>0)$ passing through $(1,-1)$ is-

• A. $y+\log x+1=0$
• B. $-y^2+\log x+1=0$
• C. $y^3+(\log x{)}^2+1=0$
• D. $Noneofthese$

Answer 1

The correct option is A $y+\log x+1=0$

Given

$ydx-xdy=-\log xdx$

$xdy-ydx=\log xdx$

Now divided by $x^2$ in both sides, we get

$\frac{xdy-ydx}{x^2}=\log x\frac{1}{x^2}dx$

Integrate both side

$\int \frac{xdy-ydx}{x^2}dx=\int \log x\frac{1}{x^2}dx$

We know, differentiation of

$\frac{y}{x}=\frac{(xdy-ydx)}{x^2}$

So, integration of $\frac{(xdy-ydx)}{x^2}=\frac{y}{x}$

Second part

$\int \log x\frac{1}{x^2}dx=-\log x\frac{1}{x}-\frac{1}{x}$

So,

$\frac{y}{x}=\frac{-(\log x+1)}{x}$

Therefore,

$y+\log x+1=0$