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Q.
If $x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=\ln x, y(1)=1$ and $y^{\prime}(1)=-1$ then $y^{\prime \prime}\left(\frac{1}{3}\right)$ equals
Continuity and Differentiability
Solution:
Given $\frac{ d }{ dx }\left( x \frac{ dy }{ dx }\right)=\ln x$
$\Rightarrow x \frac{ dy }{ dx }= x \ln x - x + C$
Now $y^{\prime}(1)=-1$, so
$-1=0-1+ C \Rightarrow C =0 $
$\therefore \frac{ dy }{ dx }=\ln x -1$
Hence $\left.\frac{ d ^2 y }{ dx ^2}\right]_{ x =\frac{1}{3}}=\frac{1}{ x }=3$