The solution of $$ydx-xdy+\log {x}dx=0$$ is
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • A
    $$y-\log {x}-1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • B
    $$x+\log {y}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • C
    $$y+\log {x}+1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$
  • D
    $$y+\log {x}-1=cx$$

Solution

$$ydx - xdy =  - \log x\,dx$$
$$ \Rightarrow \dfrac{{xdy - ydx}}{{{x^2}}} = \dfrac{{\log x\,dx}}{{{x^2}}}$$
$$ \Rightarrow \int {\dfrac{d}{{dx}}\left( {\dfrac{y}{x}} \right) = \,\int {\dfrac{{\log x}}{{{x^2}}}} } .\,dx$$
$$ \Rightarrow Putting\,\,\log x = t \Rightarrow x = {e^t} \Rightarrow \dfrac{1}{x}dx = dt$$
$$ \Rightarrow \dfrac{y}{x} = \int {\mathop t\limits_1 } \,\mathop {{e^{ - t}}}\limits_2 \,dt$$
$$ \Rightarrow \dfrac{y}{x} =  - t{e^{ - t}} + \int {1 \cdot {e^{ - t}}\,dt} $$
$$ \Rightarrow \dfrac{y}{x} =  - t\,{e^{ - t}} - {e^{ - t}} + c$$
$$ \Rightarrow \dfrac{y}{x} =  - {e^{ - t}}\left( {t + 1} \right) + c$$
$$ \Rightarrow \dfrac{y}{x} =  - {e^{ - \log x}}\left( {\log x + 1} \right) + c$$
$$ \Rightarrow \dfrac{y}{x} =  - \dfrac{1}{x}\left( {\log x + 1} \right) + c$$
$$ \Rightarrow y =  - \log x - 1 + cx$$
$$ \Rightarrow y + \log x + 1 = cx$$