\[ y^{\prime }\sin \left ( y^{\prime }\right ) +\cos \left ( y^{\prime }\right ) =y \] Since \(x\) is missing then this is of the form \(y^{\prime }=f\left ( y\right ) \) we just need to solve for \(y^{\prime }\). The solution is in terms of RootOf\[ y^{\prime }=\operatorname {RootOf}\left ( \_Z\sin \left ( \_Z\right ) +\cos \left ( \_Z\right ) -y\right ) \] Integrating gives\begin {align*} \int \frac {dy}{\operatorname {RootOf}\left ( \_Z\sin \left ( \_Z\right ) +\cos \left ( \_Z\right ) -y\right ) } & =\int dx\\ \int ^{y\left ( x\right ) }\frac {d\tau }{\operatorname {RootOf}\left ( \_Z\sin \left ( \_Z\right ) +\cos \left ( \_Z\right ) -\tau \right ) } & =x+c \end {align*}
Hence the solution is implicit\[ x-\int ^{y\left ( x\right ) }\frac {d\tau }{\operatorname {RootOf}\left ( \_Z\sin \left ( \_Z\right ) +\cos \left ( \_Z\right ) -\tau \right ) }+c=0 \] We should also find the singular solution since we divided by \(\operatorname {RootOf}\left ( \_Z\sin \left ( \_Z\right ) +\cos \left ( \_Z\right ) -y\right ) \). i.e. ask what is \(y\) which will make this zero? Solving \[ \operatorname {RootOf}\left ( \_Z\sin \left ( \_Z\right ) +\cos \left ( \_Z\right ) -y\right ) =0 \] For \(y\) gives\[ y=1 \] Hence this is solution also. We see that if we plug in \(y=1\) in the ode, this is correct solution.