Solve
\begin{equation} y^{\prime }+p\left ( x\right ) y=q\left ( x\right ) \left ( y\ln y\right ) \tag {1}\end{equation}
The substitution \(y=e^{u}\) transforms the ode to linear ode.
\[ \frac {dy}{dx}=\frac {du}{dx}e^{u}\]
And the ode becomes
\begin{align*} \frac {du}{dx}e^{u}+pe^{u} & =que^{u}\\ \frac {du}{dx}+p & =qu \end{align*}
Which is linear ode.
\[ \frac {du}{dx}-qu=-p \]
The integrating factor is \(I=e^{\int -qdx}\). Hence the above becomes
\[ d\left ( uI\right ) =-pI \]
Integrating
gives
\begin{align*} uI & =-\int pIdx+c_{1}\\ u & =-I^{-1}\int pIdx+I^{-1}c_{1}\\ u & =-e^{\int qdx}\left ( \int pe^{\int -qdx}dx\right ) +c_{1}e^{\int qdx}\end{align*}
But \(y=e^{u}\) or \(u=\ln y\). Hence the final solution is
\[ \ln \left ( y\right ) =-e^{\int qdx}\left ( \int pe^{\int -qdx}dx\right ) +c_{1}e^{\int qdx}\]
Or
\begin{align} y & =e^{-e^{\int qdx}\left ( \int pe^{\int -qdx}dx\right ) +c_{1}e^{\int qdx}}\nonumber \\ & =e^{-e^{\int qdx}\left ( \int pe^{\int -qdx}dx\right ) }e^{c_{1}e^{\int qdx}}\nonumber \\ & =\frac {e^{c_{1}e^{\int qdx}}}{e^{e^{\int qdx}\left ( \int pe^{\int -qdx}dx\right ) }}\nonumber \\ & =\frac {\exp \left ( c_{1}e^{\int qdx}\right ) }{\exp \left ( e^{\int qdx}\left ( \int pe^{\int -qdx}dx\right ) \right ) } \tag {2}\end{align}
If initial conditions \(y\left ( x_{0}\right ) =y_{0}\) are given then the above becomes
\begin{align} y_{0} & =\frac {\exp \left ( c_{1}e^{\int _{0}^{x_{0}}qd\tau }\right ) }{\exp \left ( e^{\int _{0}^{x_{0}}qd\tau }\left ( \int _{0}^{x_{0}}p\left ( \tau \right ) e^{\int _{0}^{\tau }-q\left ( z\right ) dz}d\tau \right ) \right ) }\nonumber \\ \exp \left ( c_{1}e^{\int _{0}^{x_{0}}qd\tau }\right ) & =y_{0}\exp \left ( e^{\int _{0}^{x_{0}}qd\tau }\left ( \int _{0}^{x_{0}}p\left ( \tau \right ) e^{\int _{0}^{\tau }-q\left ( z\right ) dz}d\tau \right ) \right ) \nonumber \\ c_{1}e^{\int _{0}^{x_{0}}qd\tau } & =\ln \left ( y_{0}\exp \left ( e^{\int _{0}^{x_{0}}qd\tau }\left ( \int _{0}^{x_{0}}p\left ( \tau \right ) e^{\int _{0}^{\tau }-q\left ( z\right ) dz}d\tau \right ) \right ) \right ) \nonumber \\ c_{1} & =\frac {\ln \left ( y_{0}\exp \left ( e^{\int _{0}^{x_{0}}qd\tau }\left ( \int _{0}^{x_{0}}p\left ( \tau \right ) e^{\int _{0}^{\tau }-q\left ( z\right ) dz}d\tau \right ) \right ) \right ) }{e^{\int _{0}^{x_{0}}qd\tau }} \tag {3}\end{align}
Substituting the above in (2) gives
\[ y=\frac {\exp \left ( c_{1}e^{\int qdx}\right ) }{\exp \left ( e^{\int qdx}\left ( \int pe^{\int -qdx}dx\right ) \right ) }\]
Where \(c_{1}\) is given by (3).