Question

The particular solution of the differential equation dy/ dt = y/ 4 for which y(0) = 20 is

y = 20e^−0.25t
y = 19 + e^0.25t
y = 20 e^0.25t
y = 20e^4t

Answer 1

@iGreen @Luigi0210 @hartnn @bibby

Answer 2

you can do this one by separation

Answer 3

what would i get as an answer? :)

Answer 4

\[\frac{dy}{dt}=\frac{1}{4}y \Longrightarrow \frac{dy}{y}=\frac{1}{4}x\\
\int \frac{dy}{y}=1/4\int dx \Longrightarrow \ln y=\frac{x}{4}+C \]

Answer 5

take it from there...

Answer 6

solve for y
using exp properties

Answer 7

i think its C, y = 20 e^0.25t

Answer 8

idk you have to solve to figure out

Answer 9

y = e^(c+x/4)

Answer 10

that's what i got : )

Answer 11

go further
use the powers rules
to get y=Ce^{x/4}

Answer 12

oh by the way that should t not x

Answer 13

just a typo in the variable we have

Answer 14

y = approximately 2.7182 ^ C +0.2500x

Answer 15

2.7183*

Answer 16

what are you doing?

Answer 17

\[y=e^{t/4 +C}=e^{t/4}e^C=Ce^{t/4}\] for y(t=0)=20
we plug in t=0
\[20=Ce^0=C\Longrightarrow C=20\]

Answer 18

1/4=0.25 see what equation is good

Answer 19

y = 20 e^0.25t

Answer 20

yes!

Answer 21

thank u :)

Answer 22

YAY! I GOT A 100% thank u!!!!!!!!!!

Answer 23

YW