Answer
1) We show that $y(\pi/2) = 1$ i.e. that given $y$ satisfies given initial conditions.
2) We find $y'$ and put $y$ and $y'$ into the differential equation to show that $y$ really is its solution.
By 1) and 2) we verified $y$ as a particular solution to this differential equation.
Work Step by Step
1) Show that $y(\pi/2)=1$.
Indeed
$$y(\pi/2) = e^{-\cos \pi/2} = e^0= 1,$$
where we used $\cos\pi/2 = 0.$
2) Show that $y$ is a solution to this differential equation.
We find $y'$
$$y' = (e^{-\cos x})' = e^{-\cos x}(-\cos x)' = e^{-\cos x}\sin x,$$
where we used the chain rule in the first step and then the fact that $(\cos x)'=-\sin x.$
Now we put $y$ and $y'$ into the differential equation.
The Left side is
$$y'=e^{-\cos x}\sin x$$
The Right side is
$$y\sin x= e^{-\cos x}\sin x.$$
We see that the sides are equal to eachother so $y$ satisfies the given differential equation.
1) and 2) tell us that we verified $y$ as a particular solution to this differential equation.
