Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.1 Exercises - Page 403: 9


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.
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