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!

- anonymous

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- myininaya

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

- myininaya

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

- anonymous

my solution doesn't.

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## More answers

- myininaya

what did you get?

- myininaya

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

- myininaya

let me ask loiskin though one sec

- myininaya

oh he left nvm

- anonymous

hint: it is NON-LINEAR ordinary diffeq

- myininaya

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

- myininaya

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)

- myininaya

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

- myininaya

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

- myininaya

suppose to not both to lol stupid type-0s

- anonymous

nope. you didn't finish the problem.

- myininaya

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

- myininaya

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

- anonymous

keep going.

- myininaya

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

- myininaya

that guy left along time ago it looks like

- myininaya

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?

- anonymous

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

- anonymous

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?

- myininaya

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.

- anonymous

Do you have maple, mathmatica, or matlab?

- myininaya

i have maple

- anonymous

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

- anonymous

so wolframalpha said no solution exist?

- anonymous

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+

- anonymous

There's something wrong with your question, myininaya.

- myininaya

it isn't mine lol

- myininaya

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

- anonymous

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

- myininaya

ok thanks for supporting my solution.

- anonymous

No problems.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

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