\[ y^{\prime \prime }+\left ( 3+x\right ) y^{\prime }+y\left [ y^{\prime }\right ] ^{2}=0 \] Comparing to \[ y^{\prime \prime }\left ( x\right ) +p\left ( x\right ) y^{\prime }\left ( x\right ) +q\left ( y\right ) \left [ y^{\prime }\left ( x\right ) \right ] ^{2}=0 \] Show that \(p=\left ( 3+x\right ) \) and \(q\left ( y\right ) =y\). Hence the conditions are satisfied to use this method. Therefore equation (4) becomes\begin {align*} \ln y^{\prime }+\int q\left ( y\right ) dy & =-\int p\left ( x\right ) dx\\ \ln y^{\prime }+\int ydy & =-\int \left ( 3+x\right ) dx\\ \ln y^{\prime }+\frac {y^{2}}{2} & =-\frac {\left ( 3+x\right ) ^{2}}{2}+c\\ \ln y^{\prime } & =-\frac {\left ( 3+x\right ) ^{2}}{2}-\frac {y^{2}}{2}+c \end {align*}
Hence\[ y^{\prime }=c_{1}e^{-\frac {\left ( 3+x\right ) ^{2}}{2}-\frac {y^{2}}{2}}\] This is separable. \begin {align*} \frac {dy}{dx} & =c_{1}e^{-\frac {\left ( 3+x\right ) ^{2}}{2}}e^{-\frac {y^{2}}{2}}\\ e^{\frac {y^{2}}{2}}dy & =c_{1}e^{-\frac {\left ( 3+x\right ) ^{2}}{2}}dx \end {align*}
Integrating gives\begin {align*} \int e^{\frac {y^{2}}{2}}dy & =\int c_{1}e^{-\frac {\left ( 3+x\right ) ^{2}}{2}}dx+c_{2}\\ -\frac {i}{2}\sqrt {2\pi }\operatorname {erf}\left ( \frac {i}{\sqrt {2}}y\right ) & =-\frac {c_{1}}{2}\sqrt {2\pi }\operatorname {erf}\left ( \frac {x}{\sqrt {2}}+\frac {3}{\sqrt {2}}\right ) +c_{2} \end {align*}
And the above is the implicit solution for \(y\).