Verify the particular solution of the differential equation. Solution: y = sin x cos x - cos² x Differential Equation and Initial Condition: 2y + y’ = 2 sin(2x) - 1 y(π/4) = 0

Firstly, lets check the initial condition:

$$y=\sin x \cos x-\cos^2 x$$

$$y\bigg(\frac{\pi}{4}\bigg)=\sin \bigg(\frac{\pi}{4}\bigg) \cos \bigg(\frac{\pi}{4}\bigg)-\cos^2 \bigg(\frac{\pi}{4}\bigg)$$

$$y\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt 2}{2} \cdot \frac{\sqrt 2}{2} - \bigg(\frac{\sqrt 2}{2}\bigg)^2$$

$$y\bigg(\frac{\pi}{4}\bigg)=\frac{2}{4} - \frac{2}{4}$$

$$\boldsymbol{y\bigg(\frac{\pi}{4}\bigg)=0}$$

We have shown that the initial condition is true!

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Now, lets derivate the given function :

$$y=\sin x \cos x-\cos^2 x \quad \big/ '$$

$$y'=\cos x \cos x + \sin x (-\sin x)-2\cos x (\cos x)'$$

$$y'=\cos^2x - \sin^2x -2 cos x (-\sin x)$$

$$y'= \cos^2x - \sin^2x +2 \cos x \sin x$$

$$\boldsymbol{y'=\cos^2x - \sin^2x +2 \cos x \sin x}$$

Finally, we can check if the given function $y$ satisfies the differential equation $2y+y'=2\sin (2x) -1$ and in the process we'll be using trigonometric identities $\sin 2x=2\sin x\cos x$ and $\cos^2x + \sin^2x=1$

$$2y+y'=2\sin (2x) -1$$

$$2\big(\sin x \cos x-\cos^2 x\big)+\cos^2x - \sin^2x +2 \cos x \sin x=2\sin (2x) -1$$

$$2\sin x \cos x-2\cos^2 x+\cos^2x - \sin^2x +2 \cos x \sin x=2\sin (2x) -1$$

$$2\sin x \cos x+2 \cos x \sin x-2\cos^2 x+\cos^2x - \sin^2x =2\sin (2x) -1$$

$$\sin (2x)+\sin (2x)-\cos^2x - \sin^2x=2\sin (2x) -1$$

$$2\sin (2x)-1(\cos^2x + \sin^2x)=2\sin (2x) -1$$

$$2\sin (2x)-1(1)=2\sin (2x) -1$$

$$2\sin (2x)-1=2\sin (2x) -1$$

We have shown that the given function $y$ satisfies the differential equation $\boldsymbol{2y+y'=2\sin (2x) -1}$.