A) \[y=-1-\log x\]
B) \[y+\log x=0\]
C) \[y={{e}^{x}}-1\]
D) None of the above
Correct Answer: A
Solution :
Given,\[y\,\,dx-x\,\,dy+\log x\,\,dx=0\] \[(y+\log x)dx=x\,\,dy\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{y}{x}+\frac{\log x}{x}\Rightarrow \frac{dy}{dx}-\frac{y}{x}=\frac{\log x}{x}\] \[IF\] \[={{e}^{-\int{\frac{1}{x}dx}}}={{e}^{-\log x}}=\frac{1}{x}\] Solution is \[y\cdot \frac{1}{x}=\int{\frac{\log x}{x}\cdot \frac{1}{x}dx+C}\] Put \[\log x=t\Rightarrow \frac{1}{x}dx=dt\] \[\frac{y}{x}=\int{\frac{t}{{{e}^{t}}}\cdot dt+C=\int{t{{e}^{-t}}}dt+C}\] \[\frac{y}{x}=[-t\,\,{{e}^{-t}}-{{e}^{-t}}]+C\] \[\Rightarrow \] \[y=x\left( \frac{\log x}{x}-\frac{1}{x} \right)+C\] \[\Rightarrow \] \[y=-\log x-1+C\] ... (i) It passes through\[(1,\,\,-1)\] \[\therefore \] \[-1=-\log (1)-1+C=0-1+C\] \[C=0\] \[\therefore \]Required solution is \[y=-\log x-1\]
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