Question

The general solution of differential equation $\frac{dy}{dx}=\log x$ is :-

• A. $y=x(\log x+1)+C$
• B. $y+x(\log x+1)=C$
• C. $y=x(\log x-1)+C$
• D. None of these

Answer 1

The correct option is C $y=x(\log x-1)+C$

$\frac{dy}{dx}=logx$

$\Rightarrow dy=logxdx$

$\therefore$ $y=x(logx-1)+c$ [C]