Question

Find the solution of the following differential equation that passes through (-1,-1). You may use separation of variables.

dy/dt = [ 2t(y+1) ] / [y]

This is what I have after separation of variables and integration of both sides:

y - ln(y+1) = t^2 +c

Plugging in (-1,-1) would not work as ln(0) does not exist. Is there another way to solve this problem?

Thanks in advance!

Answer 1

this is weird my solution also has ln(y+1)

Answer 2

it seems like they would give points for which the function is defined

Answer 3

my solution doesn't.

Answer 4

what did you get?

Answer 5

it has to be that it is undefined. i feel confident in my answer

Answer 6

let me ask loiskin though one sec

Answer 7

oh he left nvm

Answer 8

hint: it is NON-LINEAR ordinary diffeq

Answer 9

okay we have y dy/(y+1)=2t dt. Ingratiating both sides gives us y+1-ln(y+1)=t^2/2+C

Answer 10

for the y part, I used a substitution y+1=u so dy=du and we we can find y=u-1 so we have int([u-1]/u, u)=u-lnu= y+1-ln(y+1)

Answer 11

therefore, the function does not exist at (-1,-1)

Answer 12

oops the t side is both to have t^2+C, that was a type-0

Answer 13

suppose to not both to lol stupid type-0s

Answer 14

nope. you didn't finish the problem.

Answer 15

ln(y+1) does not exist for y=-1

Answer 16

(-1,-1) is not in the domain or range

Answer 17

keep going.

Answer 18

just tell me what you are thinking. I'm not the one who needs this problem done. lol

Answer 19

that guy left along time ago it looks like

Answer 20

did you want me to say there isn't a constant that exist with the initial condition y(-1)=-1. Is this what you wanted me to say?

Answer 21

I got that answer, but I'm looking at something.

Answer 22

Hmm, I have to say, given that this is a first order differential equation, and you've found a solution, and solutions are unique, if the solution satisfies the d.e., then that's it! Couldn't it be the case you have a typo. for the BC's?

Answer 23

he already left. I just wanted to make sure I said what I needed to about the solution since quantish felt I was wrong about the solution or something. I'm not really sure what quantish thinks. Thanks for looking lokisan.

Answer 24

Do you have maple, mathmatica, or matlab?

Answer 25

i have maple

Answer 26

I've run a few different things in Wolfram Alpha, and I'm pretty sure no solutions exist.

Answer 27

so wolframalpha said no solution exist?

Answer 28

Wolfram Alpha just confirmed what I worked out above. I tried to get it to repeat the process and solve given the initial condition and it reports "No solutions exist".

http://www.wolframalpha.com/input/?i=-1+%3D+-W%28-e^%28-x-1-1%29%29-1+

Answer 29

There's something wrong with your question, myininaya.

Answer 30

it isn't mine lol

Answer 31

i was just worried about it because it makes me worry when people disagree with me

Answer 32

Oh, well, I think you have the weight of evidence on your side ;)

Answer 33

ok thanks for supporting my solution.

Answer 34

No problems.

Answer 35

http://www.wolframalpha.com/input/?i=DSolve%5B%7By%27%5Bx%5D+%3D%3D+%28%282x%28y%2B1%29%29%2F%28y%29%29%2C+y%5B-1%5D+%3D%3D+-1%7D%2C+y%5Bx%5D%2C+x%5D

Answer 36

http://www.wolframalpha.com/input/?fp=1&i=-1%3d-productlog%5bz%5d(-e%5e(-c-x%5e2-1))-1&recompute=splat&_=1302422465085&incTime=true

Answer 37

http://www.wolframalpha.com/input/?i=-1%3D%28-productlog%5Bz%5D%29%28-e%5E%28-C%5B1%5D-x%5E2-1%29%29-1%2C+x%3D-1

Answer 38

I played around with that too and kept getting similar answers. I'll have to ask my professor. :/