If the solution of the D.E. (y ln x - 1) y.dx = x.dy. is  $y\left(f\right(x)+cx)=1$
Then f(1)=_________

The given differential equation can be written as   
$\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}={\mathrm{y}}^2\log \mathrm{x}$
Divide by xy².  Hence  $\frac{1}{{\mathrm{y}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{\mathrm{xy}}=\frac{1}{\mathrm{x}}\log \mathrm{x}$
Let $\frac{1}{\mathrm{y}}=\mathrm{v}\Rightarrow -\frac{1}{{\mathrm{y}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}$
so that $\frac{\mathrm{dv}}{\mathrm{dx}}-\frac{1}{\mathrm{x}}\mathrm{v}=-\frac{1}{\mathrm{x}}\log \mathrm{x}$
(3) is the standard linear differential equation
with P $=-\frac{1}{x},Q=-\frac{1}{x}\log x$
If $={\mathrm{e}}^{\int \mathrm{pdx}}={\mathrm{e}}^{\mathrm{j}-1/\mathrm{xdx}}=1/\mathrm{x}$
The solution is given by
$\begin{array}{l}\mathrm{v}\cdot \frac{1}{\mathrm{x}}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\int \frac{1}{\mathrm{x}}\left(-\frac{1}{\mathrm{x}}\log \mathrm{x}\right)\mathrm{dx}=-\int \frac{\log \mathrm{x}}{{\mathrm{x}}^2}\mathrm{dx}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{\log \mathrm{x}}{\mathrm{x}}-\int \frac{1}{\mathrm{x}}\cdot \frac{1}{\mathrm{x}}\mathrm{dx}=\frac{\log \mathrm{x}}{\mathrm{x}}+\frac{1}{\mathrm{x}}+\mathrm{c}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{v}=1+\log \mathrm{x}+\mathrm{cx}=\log \mathrm{ex}+\mathrm{cx}\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{\mathrm{v}}=\log \mathrm{ex}+\mathrm{cx}\text{ or }\mathrm{y}(\log \mathrm{ex}+\mathrm{cx})=1\end{array}$