Let  be the solution of the differential equation   such that  , then  is equal to 

Put $y+2\sin x-5=u\Rightarrow \frac{dy}{dx}+2\cos x=\frac{du}{dx}=\frac{dy}{dx}+2\cos x=2(y+2\sin x-5)x$ 
$$\begin{gathered} \frac{du}{dx}=2ux \\ \Rightarrow \frac{1}{u}du=2xdx \\ \Rightarrow \int \frac{1}{u}du=\int 2xdx\Rightarrow \log u=x^2+c \end{gathered}$$ 
$$\begin{gathered} \log (y+2\sin x-5)=x^2+c \\ y\left(0\right)=7\log 2=c \\ \therefore \log (y+2\sin x-5)=x^2+\log 2 \end{gathered}$$ 
$$\begin{gathered} \Rightarrow y+2\sin x-5=2e^{x^2}\text{Now at }x=\pi \\ y-5=2e^{x^2} \\ y=2e^{x^2}+5 \end{gathered}$$