Question

DIff Eq. initial value problem.
dy/dt = -y + 5 y(o) = yo

Answer 1

So dt/dy = 1/(-y+5)

Answer 2

Why flip it like that?

Answer 3

So you can now integrate lol

Answer 4

I dont see hw that helps..sorry I dont understand.

Answer 5

Integrate 1/(-y+5) lol

Answer 6

dy of course

Answer 7

so - ln(-y + 5)

Answer 8

Yep

Answer 9

So t=-ln(-y+5)

Answer 10

Is that because we were actually evaluating dy/(-y + 5) = dt. then integrated each side so t = -ln(-y+5)

Answer 11

Yep you can say that

Answer 12

okay so t = -ln(-y+5)....

Answer 13

+C

Answer 14

okay...

Answer 15

So I need to get this in terms of y?

Answer 16

Yep

Answer 17

But you find the constant first I think?

Answer 18

how does e^(-ln(-y+5) simplify?

Answer 19

\[\Large e^{-\ln(-y+5)}=\frac1{-y+5}\]

Answer 20

okay I shouldn't have skipped so many steps

Answer 21

\[\Large e^{-\ln(-y+5)}=\left[e^{\ln(-y+5)}\right]^{-1}=(-y+5)^{-1}=\frac1{-y+5}\]

Answer 22

ahhh I always wondered how that worked out so nicely like that.

Answer 23

c wont be part of e's power?

Answer 24

It would be better to find the value of C before simplifying all that

Answer 25

How would I do that?

Answer 26

That's where the initial value is useful

Answer 27

Substitute the initial value

Answer 28

for c?

Answer 29

Well I actually don't know what's y(o)=yo

Answer 30

Y(0) = Y naught or however it's spelled (;

Answer 31

What's Y0?

Answer 32

Means you plug zero into the function y

Answer 33

but what's y0?

Answer 34

What you just said was y(0)

Answer 35

it is y naught aka the inital y value.

Answer 36

Uh...

Answer 37

What's the initial value?

Answer 38

Okay I get it

Answer 39

So plug that in now

Answer 40

as in instead of having a integer as the initial value we have a variable name for that integer.

Answer 41

t=0 and y=y0