Question

The solution of the equation $x^2\frac{d^{2}y}{d{x}^2}$ =In ~x, when x=1, y =0 and $\frac{dy}{dx}=-1$ is

• A. $y=\frac{1}{2}(lnx{)}^2+lnx$
  
• B. $y=\frac{1}{2}(lnx{)}^2-lnx$
  
• C. $y=-\frac{1}{2}(lnx{)}^2+lnx$
  
• D. $y=-\frac{1}{2}(lnx{)}^2-lnx$

Answer 1

The correct option is D $y=-\frac{1}{2}(lnx{)}^2-lnx$

$\frac{d^{2}y}{d{x}^2}=\frac{logx}{x^2}\Rightarrow \frac{dy}{dx}=\frac{-(logx+1)}{x}+c$
At $\frac{dy}{dx}=-1,x=1,y=0,\therefore c=0$
$\Rightarrow y=-\int \frac{logx+1}{x}dx=-\frac{1}{2}(logx{)}^2-logx$.