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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


Answered 9 months ago
Answered 9 months ago
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Firstly, lets check the initial condition:

y=sinxcosxcos2xy=\sin x \cos x-\cos^2 x

y(π4)=sin(π4)cos(π4)cos2(π4)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(π4)=2222(22)2y\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt 2}{2} \cdot \frac{\sqrt 2}{2} - \bigg(\frac{\sqrt 2}{2}\bigg)^2

y(π4)=2424y\bigg(\frac{\pi}{4}\bigg)=\frac{2}{4} - \frac{2}{4}


We have shown that the initial condition is true!


Now, lets derivate the given function :

y=sinxcosxcos2x/y=\sin x \cos x-\cos^2 x \quad \big/ '

y=cosxcosx+sinx(sinx)2cosx(cosx)y'=\cos x \cos x + \sin x (-\sin x)-2\cos x (\cos x)'

y=cos2xsin2x2cosx(sinx)y'=\cos^2x - \sin^2x -2 cos x (-\sin x)

y=cos2xsin2x+2cosxsinxy'= \cos^2x - \sin^2x +2 \cos x \sin x

y=cos2xsin2x+2cosxsinx\boldsymbol{y'=\cos^2x - \sin^2x +2 \cos x \sin x}

Finally, we can check if the given function yy satisfies the differential equation 2y+y=2sin(2x)12y+y'=2\sin (2x) -1 and in the process we'll be using trigonometric identities sin2x=2sinxcosx\sin 2x=2\sin x\cos x and cos2x+sin2x=1\cos^2x + \sin^2x=1

2y+y=2sin(2x)12y+y'=2\sin (2x) -1

2(sinxcosxcos2x)+cos2xsin2x+2cosxsinx=2sin(2x)12\big(\sin x \cos x-\cos^2 x\big)+\cos^2x - \sin^2x +2 \cos x \sin x=2\sin (2x) -1

2sinxcosx2cos2x+cos2xsin2x+2cosxsinx=2sin(2x)12\sin x \cos x-2\cos^2 x+\cos^2x - \sin^2x +2 \cos x \sin x=2\sin (2x) -1

2sinxcosx+2cosxsinx2cos2x+cos2xsin2x=2sin(2x)12\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)cos2xsin2x=2sin(2x)1\sin (2x)+\sin (2x)-\cos^2x - \sin^2x=2\sin (2x) -1

2sin(2x)1(cos2x+sin2x)=2sin(2x)12\sin (2x)-1(\cos^2x + \sin^2x)=2\sin (2x) -1

2sin(2x)1(1)=2sin(2x)12\sin (2x)-1(1)=2\sin (2x) -1

2sin(2x)1=2sin(2x)12\sin (2x)-1=2\sin (2x) -1

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

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