Now differentiating this with respect to x

\(dy/dx= d/dx[2-\int_{0}^{x}(1+y(t))\sin tdt]\)

\(dy/dx= d/dx[2]-d/dx[\int_0^x(1+y(t))\sin tdt]\)

\(dy/dx= 0-[(1+y(t))\sin(x) \cdot d/dx(x)-(1+y(0))\sin(0) \cdot d/dx(0)]\)

\(dy/dx= -[(1+y(x))\sin x \cdot (1)-(1+y(0))(0) \cdot 0]\)

\(dy/dx= -[(1+y(x))\sin x-0]\)

\(dy/dx= -(1+y(x))\sin x\)

\(y'= -(1+y)\sin x\)