Answer
\[y=\frac{x^3}{3}\ln x-\frac{x^3}{9}+\frac{19}{9}\]
Work Step by Step
\[y'=x^2\ln x\]
\[\frac{dy}{dx}=x^2\ln x\]
Seperating the variables
\[dy=x^2\ln x dx\]
Integrating
\begin{eqnarray*}
y&=&\ln x\int x^2dx-\int \left(\frac{d}{dx}(\ln x)\int x^2dx\right)dx+c\\
&=&\frac{x^3}{3}\ln x-\int \frac{x^3}{3x}dx+c\\
&=&\frac{x^3}{3}\ln x-\frac{1}{3}\int x^2dx+c\\
&=&\frac{x^3}{3}\ln x-\frac{x^3}{9}+c
\end{eqnarray*}
Using initial condition $y(1)=2$
$2=\frac{1}{3}\ln 1-\frac{1}{9}+c \Rightarrow c=2+\frac{1}{9}=\frac{19}{9}$
Hence solution is $y=\frac{x^3}{3}\ln x-\frac{x^3}{9}+\frac{19}{9}$
