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: 8

Answer

To verify the solution we find $y'$, $y''$ and then put $y$, $y'$ and $y''$ into the differential equation. Since we see that the left and the right side are equal we verified the solution.

Work Step by Step

Let us find $y'$ and $y''$. $$y' = \left(\frac{2}{5}(e^{-4x}+e^x)\right)' = \frac{2}{5}(e^{-4x}+e^x)' = \frac{2}{5}((e^{-4x})'+(e^x)').$$ From table of derivatives $$(e^x)'=e^x.$$ Now to find $(e^{-4x})'$ use the chain rule: $$(e^{-4x})'=e^{-4x}(-4x)'=-4e^{-4x}.$$ Putting this into the expression for $y'$ we have $$y'=\frac{2}{5}e^x -\frac{8}{5}e^{-4x}.$$ Now lets find the second derivative $y''$: $$y''=(y')' = \left(\frac{2}{5}e^x -\frac{8}{5}e^{-4x}\right)' = \frac{2}{5}(e^x)'-\frac{8}{5}(e^{-4x})'.$$ We already found $(e^x)' = e^x$ and $(e^{-4x})'=-4e^{-4x}$ so putting this into previous expression gives us $$y'' =\frac{2}{5}e^x -(-4) \frac{8}{5}e^{-4x} =\frac{2}{5}e^x + \frac{32}{5}e^{-4x}.$$ Puttiong $y$, $y'$ and $y''$ into the differential equation we obtain: The Left side: $$y''+4y' = \frac{2}{5}e^x + \frac{32}{5}e^{-4x} + 4(\frac{2}{5}e^x -\frac{8}{5}e^{-4x}) = \frac{2}{5}e^x + \frac{32}{5}e^{-4x} + \frac{8}{5}e^x - \frac{32}{5}e^{-4x} = \frac{10}{5}e^x=2e^x.$$ The Right side is just $$2e^x.$$ We see that the left and the right sides are equal and this verifies that 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.