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: 8


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.
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