#### Answer

1) We verify that indeed $y(0)= 1$ i.e. that $y$ satisfies the given initial condition.
2) We find $y'$ and then put $y$ and $y'$ into the given differential equation and we see $y$ really is its solution.
Both 1) and 2) verify the particular solution.

#### Work Step by Step

1) Check that $y(0) = 1$.
Indeed
$$y(0) = 6\times 0 - 4\sin 0 +1 = 1$$
where we used $\sin 0=0$.
2) Verify the solution.
Let us first find $y'$.
$$y' = (6x-4\sin x +1)' = (6x)'-(4\sin x)'+(1)' = 6-4\cos x,$$
where we used $(\sin x)'=\cos x$.
Now let us put $y$ and $y'$ into the differential equation:
The Left side is:
$$y'=6-4\cos x.$$
The Right side is just
$$6-4\cos x.$$
We see that the sides are equal which verifies given $y$ as a solution to the given differential equation.
Both 1) and 2) verify $y$ as a particular solution to the given differential equation.