Answer
1) We show that $y(\pi/4)=0$ i.e. that $y$ satisfies given initial conditions.
2) We find $y'$ and then put $y$ and $y'$ into the differential equation and show that left and right sides are equal.
1) and 2) verify the particular solution.
Work Step by Step
Step 1. Show that $y(\pi/4) = 0$.
Really $$y(\pi/4) = \sin\frac{\pi}{4}\cos\frac{\pi}{4} - \cos^2\frac{\pi}{4} = \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}-\left(\frac{1}{\sqrt{2}}\right)^2 = \left(\frac{1}{\sqrt{2}}\right)^2 - \left(\frac{1}{\sqrt{2}}\right)^2 = 0,$$
where we used $\sin(\pi/4) = 1/\sqrt{2}$ and $\cos(\pi/4)=1/\sqrt{2}$.
Step 2. Verify the solution.
$$y'=(\sin x\cos x-\cos^2 x)' = (\sin x\cos x)' - (\cos^2x)'.$$
Using the product rule we get
$$(\sin x\cos x)' = (\sin x)'\cos x + \sin x(\cos x)' = \cos x \cos x -\sin x\sin x = \cos^2 x -\sin^2 x.$$
Using the chain rule we get
$$(\cos^2 x)' = 2\cos x(\cos x)' = -2\sin x\cos x.$$
This gives us
$$y' = \cos^2 x -\sin^2 x +2\sin x\cos x.$$
Putting this into the differential equation we get
The Left side:
$$2y+y' = 2(\sin x\cos x-\cos^2 x) + \cos^2 x - \sin^2 x+2\sin x\cos x= 2\sin x\cos x - 2\cos^2 x + \cos^2 x-\sin^2 x+2\sin x\cos x = 2\times 2\sin x\cos x - (\cos^2 x + \sin^2 x) = 2\sin 2x - 1,$$
where we used $2\sin x\cos x = \sin 2x$ and $\sin^2x+\cos^2 x= 1.$
The Right side is just $2\sin x\cos x -1$. We see thet the sides are equal.
Since in Step 1 we showed that $y$ satisfies the initial condition and in step 2 that it really is a solution to a given differential equation we say that we verified the particular solution.
