Step 1

Given,

\(y=2 \sin x+3\)

Differentiate the given equation with respect to x we get,

\(y'=2 \cos x\)

Now again differentiate with respect to x we get,

\(y":=−2\sin x\)

Step 2

Now,

\(L.H.S=y"+y\)

\(L.H.S=−2\sin x+2\sin x+3\)

\(L.H.S=0+3\)

\(L.H.S=3\)

\(L.H.S=R.H.S\)