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!

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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!

Mathematics
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this is weird my solution also has ln(y+1)
it seems like they would give points for which the function is defined
my solution doesn't.

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what did you get?
it has to be that it is undefined. i feel confident in my answer
let me ask loiskin though one sec
oh he left nvm
hint: it is NON-LINEAR ordinary diffeq
okay we have y dy/(y+1)=2t dt. Ingratiating both sides gives us y+1-ln(y+1)=t^2/2+C
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)
therefore, the function does not exist at (-1,-1)
oops the t side is both to have t^2+C, that was a type-0
suppose to not both to lol stupid type-0s
nope. you didn't finish the problem.
ln(y+1) does not exist for y=-1
(-1,-1) is not in the domain or range
keep going.
just tell me what you are thinking. I'm not the one who needs this problem done. lol
that guy left along time ago it looks like
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?
I got that answer, but I'm looking at something.
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?
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.
Do you have maple, mathmatica, or matlab?
i have maple
I've run a few different things in Wolfram Alpha, and I'm pretty sure no solutions exist.
so wolframalpha said no solution exist?
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+
There's something wrong with your question, myininaya.
it isn't mine lol
i was just worried about it because it makes me worry when people disagree with me
Oh, well, I think you have the weight of evidence on your side ;)
ok thanks for supporting my solution.
No problems.
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
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
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
I played around with that too and kept getting similar answers. I'll have to ask my professor. :/

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