Calculus 10th Edition

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.
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.