\[ y'(x)-y(x)^2+y(x) \sin (x)-\cos (x)=0 \] ✓ Mathematica : cpu = 7.28133 (sec), leaf count = 57
\[\left \{\left \{y(x)\to \frac {c_1 \sin (x) \left (\int _1^x e^{-\cos (K[1])} \, dK[1]\right )+c_1 \left (-e^{-\cos (x)}\right )+\sin (x)}{c_1 \int _1^x e^{-\cos (K[1])} \, dK[1]+1}\right \}\right \}\]
✓ Maple : cpu = 0.21 (sec), leaf count = 25
\[ \left \{ y \left ( x \right ) =-{\frac {{{\rm e}^{-\cos \left ( x \right ) }}}{{\it \_C1}+\int \!{{\rm e}^{-\cos \left ( x \right ) }}\,{\rm d}x}}+\sin \left ( x \right ) \right \} \]
\begin {align} y^{\prime }-y^{2}+y\sin \left ( x\right ) -\cos \left ( x\right ) & =0\nonumber \\ y^{\prime } & =y^{2}-y\sin \left ( x\right ) +\cos \left ( x\right ) \tag {1} \end {align}
This is Riccati first order non-linear ODE of the form of the general form \(y^{\prime }=P\left ( x\right ) +Q\left ( x\right ) y+R\left ( x\right ) y^{2}\) where \(P\left ( x\right ) =\cos \left ( x\right ) ,Q\left ( x\right ) =-\sin \left ( x\right ) ,R\left ( x\right ) =1\). It is best to first try to spot a particular solution \(y_{p}\) and use the transformation \(y=y_{p}+\frac {1}{u}\) otherwise we use \(y=-\frac {u^{\prime }}{yR\left ( x\right ) }\) transformation. For this problem \[ y_{p}=\sin \left ( x\right ) \] Therefore
\begin {align*} y & =\sin x+\frac {1}{u}\\ y^{\prime } & =\cos x-\frac {u^{\prime }}{u^{2}} \end {align*}
Equating this to (1) gives
\begin {align*} y^{2}-y\sin \left ( x\right ) +\cos \left ( x\right ) & =\cos x-\frac {u^{\prime }}{u^{2}}\\ \left ( \sin x+\frac {1}{u}\right ) ^{2}-\left ( \sin x+\frac {1}{u}\right ) \sin x+\cos x & =\cos x-\frac {u^{\prime }}{u^{2}}\\ \sin ^{2}x+\frac {1}{u^{2}}+\frac {2}{u}\sin x-\sin ^{2}x-\frac {1}{u}\sin x & =-\frac {u^{\prime }}{u^{2}}\\ \frac {1}{u^{2}}+\frac {1}{u}\sin x & =-\frac {u^{\prime }}{u^{2}}\\ 1+u\sin x & =-u^{\prime }\\ u^{\prime }+u\sin x & =-1 \end {align*}
Integrating factor is \(e^{\int \sin x}=e^{-\cos x}\), hence
\[ d\left ( e^{-\cos x}u\right ) =-e^{-\cos x}\]
Integrating both sides
\begin {align*} e^{-\cos x}u & =-{\displaystyle \int } e^{-\cos x}dx+C\\ u & =e^{\cos x}\left ( C-{\displaystyle \int } e^{-\cos x}dx\right ) \end {align*}
Since \(y=\sin x+\frac {1}{u}\) then
\[ y=\sin x+\frac {e^{-\cos x}}{C-{\displaystyle \int } e^{-\cos x}dx}\]
Or letting \(C_{1}=-C\) to make match Maple form, we obtain
\[ y=-\frac {e^{-\cos x}}{C_{1}+{\displaystyle \int } e^{-\cos x}dx}+\sin x \]