# Determine whether the function is a solution of the DE?

• Jan 31st 2010, 05:34 PM
xterminal01
Determine 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, 05:36 PM
pickslides
Find $y'''$ for each possibility and sub them back into the equation to check.

Sound good?
• Jan 31st 2010, 05:37 PM
lvleph
Well, you could differentiate the given functions and see if they satisfy the equation, i.e., you want to know if $y''' = 8y$.
• Jan 31st 2010, 05:52 PM
xterminal01
$cos(2x)-8y=0$
I end up with
y = 1/8 cos(2 x) + C
What should i do with this solution?

Quote:

Originally Posted by pickslides
Find $y'''$ for each possibility and sub them back into the equation to check.

Sound good?

• Jan 31st 2010, 05:53 PM
lvleph
Your derivative is not correct. $y'= -2\cdot \sin(2x)$.
• Jan 31st 2010, 05:59 PM
xterminal01
So i am suppose to replace a&b with Y''' or with 8(y)
and what does the 3 ''' mean?

Quote:

Originally Posted by lvleph
Your derivative is not correct. $y'= -2\cdot \sin(2x)$.

• Jan 31st 2010, 06:02 PM
lvleph
$y'''$ means the third derivative. You are suppose to use the solutions in a) and b) and plug them in the original DEQ. If $y$ from a) or b) satisfies $y''' + 8y =0$ then it is a solution to the DEQ.
From a)
$y''' = 8\cos(2x)$
and so
$8\cos(2x) - 8cos(2x) = 0$
So, a) is a solution to the DEQ. Now, check if b) works.