anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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myininaya
  • myininaya
this is weird my solution also has ln(y+1)
myininaya
  • myininaya
it seems like they would give points for which the function is defined
anonymous
  • anonymous
my solution doesn't.

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

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