Question

If dy/dt=-2y and if y=-1 when t=0, what is the value of t for which y=-1/2?

Answer 1

Take the antiderivative of both sides.

Answer 2

int (dy/y) = int(-2dt)
so would that be lny=-2t+c?

Answer 3

When you take the antiderivative of both sides, it should look like\[\int\limits_{}^{}\frac{ dy }{ dt }=\int\limits_{}^{}-2ydt\]

Answer 4

I don't think it's the way to solve this problem

Answer 5

Correction: the left side should be \[\int\limits_{}^{}\frac{ dy }{ dt }dt\]

Answer 6

If you simplify that, it becomes y, and the right side simplifies to -2yt+C.
So, the equation is y=-2yt+C

Answer 7

dy/dt = -2y it means y' + 2y =0
t =0 y = -1 mean y(0) =-1

Answer 8

solve for this you have y (t) = \(e^{-2t} +C\)
replace y(0) = -1 you have C = -2
therefore, y (t) = \(e^{-2t} -2\)
now calculate y(-1/2) = \(e^ {-1}-2 = \dfrac{1}{e} -2\)

Answer 9

how did you get y(t)=e^(-2t)+C

Answer 10

I guesss... hihihi... (joke) I believe that @amoodarya will give you the perfect explanation for that

Answer 11

\[\frac{ dy }{ dt }=-2y \\ \frac{ dy }{ y }=-2dt\\ \int\limits \frac{ dy }{ y }=\int\limits -2tdt\\ln y=-2t +c\\y=e^{-2t+c} \\y=e^{-2t} e^c=e^{-2t} a\]

Answer 12

where did the t come from in the third part?

Answer 13

i type with mistake
there is not t in 3rd part

Answer 14

oh. so the integral of -2dt is -2t?

Answer 15

yes

Answer 16

like
\[\int\limits -2dx=-2x+const\]

Answer 17

oh ok. and e^c is always a constant?

Answer 18

yes

Answer 19

so now how would I find the value of t for which y=-1/2?

Answer 20

like
e^2 or e^3 ...
or e^(ln5) =5

Answer 21

ok. i understand. how would I find the value of t for which y=-1/2?
the answer choices are
a) -ln2/2
b) -1/4
c) ln2/2
d)sqrt(2)/2
e) ln2

Answer 22

put t=0 and y=-1
y=a(e^(0))
-1=a *1
a=-1

so y(t)=-e^(-2t)
y=-1/2
when
-1/2=-1 e^(-2t)
so
e^(-2t)=1/2
ln e^(-2t)=ln 1/2
-2t=- ln 2
t=1/2 ln 2