$$( \frac { \operatorname { sin } ( y ) } { y } - 2 e ^ { - x } \operatorname { sin } ( x ) ) + ( \frac { \operatorname { cos } ( y ) + 2 e ^ { - x } \operatorname { cos } ( x ) } { y } ) y ^ { \prime } = 0 , \quad \mu ( x , y ) = y e ^ { x };$$
mutiply both sides by $\mu$:
Then $( \operatorname { sin } ( y ) e ^ { x } - 2 y \operatorname { sin } ( x ) ) + ( \operatorname { cos } ( y ) e ^ { x } + 2 \operatorname { cos } ( x ) ) y ^ { \prime } = 0$
Let $M(x,y)=\sin(y)e^x-2y\sin(x)$ and $N(x,y)=\cos(y)e^x+2\cos(x)$
Then $M_y=\cos(y)e^x-2\sin(x)$ $\quad N_x=\cos(y)e^x-\sin(x)$
Then the equation is exact.
Therefore, there exists a function $\varphi(x,y)$ such that
$$\left. \begin{array} { l } { \varphi _ { x } ( x , y ) = \operatorname { sin } ( y ) e ^ { x } - 2 y \operatorname { sin } ( x ) = M } \\ { \varphi _ { y } ( x , y ) = \operatorname { cos } ( y ) e ^ { x } + 2 \operatorname { cos } ( x ) = N } \end{array} \right.$$
Then $$\left. \begin{array} { l } { \varphi ( x , y ) = \int N d y = \operatorname { sin } ( y ) e ^ { x } + h ( x ) } \\ { \varphi _ { x } ( x , y ) = \operatorname { sin } ( y ) e ^ { x } + h ^ { \prime } ( x ) } \end{array} \right.$$
Then $$\left. \begin{array}{l}{ h ^ { \prime } ( x ) = - 2 y \operatorname { sin } ( x ) }\\{ h ( x ) = \int - 2 y \operatorname { sin } ( x ) }\\{ \quad\,\quad= 2 y \operatorname { cos } ( x ) + c }\end{array} \right.$$
Therefore $\sin(y)e^x+2y\cos(x)=c$
