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: 12

Answer

1) We show that $y(\pi/2) = 1$ i.e. that given $y$ satisfies given initial conditions. 2) We find $y'$ and put $y$ and $y'$ into the differential equation to show that $y$ really is its solution. By 1) and 2) we verified $y$ as a particular solution to this differential equation.

Work Step by Step

1) Show that $y(\pi/2)=1$. Indeed $$y(\pi/2) = e^{-\cos \pi/2} = e^0= 1,$$ where we used $\cos\pi/2 = 0.$ 2) Show that $y$ is a solution to this differential equation. We find $y'$ $$y' = (e^{-\cos x})' = e^{-\cos x}(-\cos x)' = e^{-\cos x}\sin x,$$ where we used the chain rule in the first step and then the fact that $(\cos x)'=-\sin x.$ Now we put $y$ and $y'$ into the differential equation. The Left side is $$y'=e^{-\cos x}\sin x$$ The Right side is $$y\sin x= e^{-\cos x}\sin x.$$ We see that the sides are equal to eachother so $y$ satisfies the given differential equation. 1) and 2) tell us that we verified $y$ as a particular solution to this differential equation.
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