If dy/dt = ky and k is a nonzero constant, y could be

[lude1]

Homework Statement

If dy/dt = ky and k is a nonzero constant, than y could be

a. 2e^kty
b. 2e^kt
c. e^kt + 3
d. kty + 5
e. .5ky^2 + .5

Correct answer is b. 2e^kt

Homework Equations

The Attempt at a Solution

I don't really know what I should be doing. I have a feeling I have to work backwards, therefore do the integral of ky? But if that is the case, I have no idea how e^kt is created through that integral. 2e^kt reminded me of Ce^kt, but that doesn't have anything to do with this problem either.

 

[Squeezebox]

Look at the derivative of e^kt and see how that relates to e^kt.

 

[rock.freak667]

The 'correct' thing to do would be

dy/dt = ky

∫ dy/y = ∫ k dt


But since you have a multiple choice, you could quickly differentiate each one if you wanted to.

 

[lude1]

Hm.. this is what I got when I did ∫ dy/y = ∫ k dt

lny = kt
y = e^kt

Where does the 2 come from? I guessed it would have to be from a square, but when you do the integral of a number (since k is a nonzero constant) you multiply the constant with a variable with an exponent of 1, not 2.

 

[rock.freak667]

Hm.. this is what I got when I did ∫ dy/y = ∫ k dt

lny = kt
y = e^kt

Where does the 2 come from? I guessed it would have to be from a square, but when you do the integral of a number (since k is a nonzero constant) you multiply the constant with a variable with an exponent of 1, not 2.

you'd get

ln y = kt+A such that y=e^kt+A=Be^kt where B is a constant. So B can be any number, such as '2'.

 

[lude1]

Ah, I forgot about that. Thanks!

 

[HallsofIvy]

Homework Statement

If dy/dt = ky and k is a nonzero constant, than y could be

a. 2e^kty

If this means y=-2e^(kty) then its derivative is dy/dt= 2ke^(kty)(ky+ kt(dy/dt)).
dy/dt(1- 2kt e^(kty)= 2k^y e^(kty). No, that differential equation is not satisfied.

b. 2e^kt

If y= 2e^(kt), then dy/dt= 2(ke^(kt))= k(2e^(kt))= ky and satisfies the equation.
(I am assuming the parentheses combine these as you wanted- that you did not intend 2(e^k)t.)

c. e^kt + 3

If y= e^(kt)+ 3, then y'= ke^(kt) which is not ky= k(e^(kt)+ 3)

d. kty + 5

If y= kty+ 5, then dy/dt= ky+ kt dy/dt so dy/dt(1- kt)= ky and dy/dt= (k/(1-kt) y. That does not satisfy the given equation.

e. .5ky^2 + .5

If y= .5y^2+ .5, then dy/dt= 1.0 y dy/dt so dy/dt(1- y)= 0. Either dy/dt= 0 or y= 1. That does not satisfy the given equation.

Correct answer is b. 2e^kt

Homework Equations

The Attempt at a Solution

I don't really know what I should be doing. I have a feeling I have to work backwards, therefore do the integral of ky? But if that is the case, I have no idea how e^kt is created through that integral. 2e^kt reminded me of Ce^kt, but that doesn't have anything to do with this problem either.