Determine whether the function is a solution of the differential equation

y''' - 8y = 0

(a) y = cos(2x)

(b) y = e^(2x)

What exactly is one suppose to do here? Is this suppose to be the third degree because we only covered first degree DE?

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- Jan 31st 2010, 04:34 PMxterminal01Determine whether the function is a solution of the DE?
Determine whether the function is a solution of the differential equation

y''' - 8y = 0

(a) y = cos(2x)

(b) y = e^(2x)

What exactly is one suppose to do here? Is this suppose to be the third degree because we only covered first degree DE? - Jan 31st 2010, 04:36 PMpickslides
Find $\displaystyle y'''$ for each possibility and sub them back into the equation to check.

Sound good? - Jan 31st 2010, 04:37 PMlvleph
Well, you could differentiate the given functions and see if they satisfy the equation, i.e., you want to know if $\displaystyle y''' = 8y$.

- Jan 31st 2010, 04:52 PMxterminal01
- Jan 31st 2010, 04:53 PMlvleph
Your derivative is not correct. $\displaystyle y'= -2\cdot \sin(2x)$.

- Jan 31st 2010, 04:59 PMxterminal01
- Jan 31st 2010, 05:02 PMlvleph
$\displaystyle y'''$ means the third derivative. You are suppose to use the solutions in a) and b) and plug them in the original DEQ. If $\displaystyle y$ from a) or b) satisfies $\displaystyle y''' + 8y =0$ then it is a solution to the DEQ.

From a)

$\displaystyle y''' = 8\cos(2x)$

and so

$\displaystyle 8\cos(2x) - 8cos(2x) = 0$

So, a) is a solution to the DEQ. Now, check if b) works.