Question

Which of the following functions is a solution to the differential equation y" + 2y'

Answer 1

take their first and second derivatives and sub them in for y'', and y'

Answer 2

What do you mean? What am I supposed to take the derivative of?

Answer 3

they should have gave you functions

Answer 4

i.e. there is not enough information to solve this as you asked it..

Answer 5

a) y=e^t
b)-e^t
c) y=e^-t
d) e^-2t

Answer 6

Okay so first do I take the derivatives of each of these?

Answer 7

so for a)
\(y'=e^t, y''=e^t\)

now plug them into the thing you have and see if makes since

\(y''+2y'+y=0\implies e^t+2e^t+e^t=0 \) for all \(t\), this is not true, so its not a solution.

Answer 8

The derivative of -e^t is 0

Answer 9

no
\((-e^t)'=-e^t\)

Answer 10

-e^t+2(-e^t)+-e^t

Answer 11

and so is the second derivative, so
\(y''+2y'+y=-e^t+2(-e^t)+(-e^t)=-4e^t\ne 0\)

Answer 12

try the next one:)

Answer 13

anything that is e raised to the power has the same derivative right?

Answer 14

no, only \(e^x\) is its own derivative

Answer 15

e^-t +2(e^-t) +(e^-t)=

Answer 16

right?

Answer 17

=4e^-t

Answer 18

\(y=e^{-t}\\y'=-e^{-t}\\y''=-(-e^{-t})=e^{-t}\)

plugging this in

\(y''+2y'+y=e^{-t}+2(-e^{-t})+e^{-t}=2e^{-t}-2e^{-t}=0\)

Answer 19

One minute I am trying to write all of this down so that I can try and do it by myself again. Haha. okay so for some reason I though that e^anything was always the same

Answer 20

But I guess it is only e^x. Right?

Answer 21

no, there is the chain rule

\(y=e^{ax}\implies\frac{dy}{dx}=e^{ax}(ax)'=e^{ax}a=ae^{ax}\)

Answer 22

so correct the only two things that will ever look the same after you take the derivative is


\(\large y= e^x \text{ and } y=-e^x\)

Answer 23

but not \(\large e^{-x}\)

Answer 24

oh okay that makes sense