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: 7

Answer

We can verify this by differentiating $y=-\cos x \ln|\sec x + \tan x|$ and putting $y$ and $y''$ into the differential equation $y''+4y'=2e^x$. This verifies the solution.

Work Step by Step

We will find the first and the second derivative of $y$: $y' = (-\cos x \ln|\sec x + \tan x|)' = -(\cos x)'\ln|\sec x + \tan x|-\cos x(\ln|\sec x + \tan x|)'$ where we used the product rule. Now we have $$(\cos x)' = -\sin x$$ from the table of derivative as well as $$(\ln|\sec x + \tan x|)'=\frac{1}{|\sec x+\tan x|}(|\sec x+ \tan x|)'$$ where we used the chain rule. Now $$(|\sec x+ \tan x|)'=\text{sign}(\sec x+ \tan x)(\sec x +\tan x)' =\text{sign}(\sec x+ \tan x) ((\sec x)' + (\tan x)') =\text{sign}(\sec x+ \tan x)\left(\frac{\sin x}{\cos^2 x} + \frac{1}{\cos^2 x}\right) = \text{sign}(\sec x+ \tan x) \frac{1+\sin x}{\cos^2 x}.$$ This helps us get $$(\ln|\sec x + \tan x|)' = \frac{\text{sign}(\sec x+ \tan x)}{|\sec x+ \tan x|}\times\frac{1+\sin x}{\cos^2 x}.$$ Since the sign function gives values $\pm 1$ depending of the "sign" of its argulent (whether it is positive or negative) we now that $\text{sign} x = 1/\text{sign} x$ because $1/\pm1=\pm 1$. Putting this into previous expression: $$(\ln|\sec x + \tan x|)' = \frac{\text{sign}(\sec x+ \tan x)}{|\sec x+ \tan x|}\times\frac{1+\sin x}{\cos^2 x} =\frac{1}{\text{sign}(\sec x+ \tan x)|\sec x+ \tan x|}\times\frac{1+\sin x}{\cos^2 x} = \frac{1}{\sec x+ \tan x}\frac{1+\sin x}{\cos ^2 x} = \frac{1+\sin x}{\cos^2 x \frac{1}{\cos x} + \cos^2 x\frac{\sin x}{\cos x}}=\frac{1+\sin x}{\cos x (1+\sin x)} = \frac{1}{\cos x}.$$ Assembling these results into the expression for $y'$ we get $$y' = \sin x \ln|\sec x +\tan x|-\cos x\frac{1}{\cos x} =\sin x \ln|\sec x +\tan x|-1. $$ Now let us find the second derivative: $$y''=(y')' = (\sin x \ln|\sec x +\tan x|-1)' = (\sin x \ln|\sec x +\tan x|)' - (1)'.$$ The derivative of a constrant $(1)'=0$ so we have $$y'' = (\sin x \ln|\sec x +\tan x|)'.$$ Using the product rule we get $$y''= (\sin x)'\ln|\sec x +\tan x|+\sin x (\ln|\sec x +\tan x|)'.$$ From the table of derivatives $$(\sin x)'=\cos x.$$ We already found $$(\ln|\sec x +\tan x|)' = \frac{1}{\cos x}$$ so finally we write $$y''=\cos x \ln|\sec x +\tan x| + \sin x\frac{1}{\cos x} =\cos x \ln|\sec x +\tan x| + \tan x.$$ Putting $y$ and $y''$ into the differential equation from the problem we have The Left side: $$y''+y=\cos x \ln|\sec x +\tan x| + \tan x + (-\cos x \ln|\sec x + \tan x|) = \tan x $$ The Right side is just $\tan x$. We see that the left side is equal to the right side so we can say that we verified that the given $y$ is a solution to the given differential equation.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.