\[ y'(x)-y(x) (a+\sin (\log (x))+\cos (\log (x)))=0 \] ✓ Mathematica : cpu = 0.0278232 (sec), leaf count = 19
\[\left \{\left \{y(x)\to c_1 e^{a x+x \sin (\log (x))}\right \}\right \}\] ✓ Maple : cpu = 0.025 (sec), leaf count = 14
\[y \relax (x ) = c_{1} {\mathrm e}^{x \left (\sin \left (\ln \relax (x )\right )+a \right )}\]
Hand solution
\begin {equation} \frac {dy}{dx}-y\relax (x) \left [ a+\sin \left (\log \relax (x) \right ) +\cos \left (\log \relax (x) \right ) \right ] =0\tag {1} \end {equation}
Integrating factor \(\mu =e^{-\int a-\sin \left (\log \relax (x) \right ) -\cos \left (\log \relax (x) \right ) dx}=e^{-ax}e^{-\int \sin \left ( \log \relax (x) \right ) +\cos \left (\log \relax (x) \right ) dx}\). To integrate \(\int \sin \left (\log \relax (x) \right ) +\cos \left (\log \relax (x) \right ) dx\), let \(r=\log \relax (x) \), \(\frac {dr}{dx}=\frac {1}{x}\), then \(dx=xdr\), But \(x=e^{r}\), hence the integral becomes
\begin {align} \int \sin \left (\log \relax (x) \right ) +\cos \left (\log \left ( x\right ) \right ) dx & =\int \left [ \sin \relax (r) +\cos \left ( r\right ) \right ] e^{r}dr\nonumber \\ & =\int e^{r}\sin \relax (r) dr+\int e^{r}\cos \relax (r) dr\tag {2} \end {align}
Integrating by parts \(\int e^{r}\cos \relax (r) dr,\) \(\int udv=uv-\int vdu\), Let \(u=e^{r}\rightarrow du=e^{r}\) and \(dv=\cos \relax (r) \rightarrow v=\sin \relax (r) \), hence (2) becomes
\begin {align*} \int e^{r}\sin \relax (r) dr+\int e^{r}\cos \relax (r) dr & =\int e^{r}\sin \relax (r) dr+e^{r}\sin \relax (r) -\int \sin \left ( r\right ) e^{r}dr\\ & =e^{r}\sin \relax (r) \end {align*}
Therefore, substituting back \(r=\log \relax (x) \) gives
\begin {align*} \int \sin \left (\log \relax (x) \right ) +\cos \left (\log \left ( x\right ) \right ) dx & =e^{\log \relax (x) }\sin \left (\log \left ( x\right ) \right ) \\ & =x\sin \left (\log \relax (x) \right ) \end {align*}
Hence the integration factor is
\begin {align*} \mu & =e^{-ax}e^{-\int \sin \left (\log \relax (x) \right ) +\cos \left ( \log \relax (x) \right ) dx}\\ & =e^{-ax}e^{-x\sin \left (\log \relax (x) \right ) } \end {align*}
Therefore (1) becomes
\[ \frac {d}{dx}\left (\mu y\relax (x) \right ) =0 \]
Integrating
\begin {align*} y\relax (x) e^{-ax}e^{-x\sin \left (\log \relax (x) \right ) } & =C\\ y\relax (x) & =Ce^{ax}e^{x\sin \left (\log \relax (x) \right ) }\\ & =Ce^{ax+x\sin \left (\log \relax (x) \right ) }\\ & =Ce^{x\left (a+\sin \left (\log \relax (x) \right ) \right ) } \end {align*}