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anonymous
 5 years ago
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
 5 years ago
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!

This Question is Closed

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0this is weird my solution also has ln(y+1)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0it seems like they would give points for which the function is defined

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0it has to be that it is undefined. i feel confident in my answer

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0let me ask loiskin though one sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hint: it is NONLINEAR ordinary diffeq

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0okay we have y dy/(y+1)=2t dt. Ingratiating both sides gives us y+1ln(y+1)=t^2/2+C

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0for the y part, I used a substitution y+1=u so dy=du and we we can find y=u1 so we have int([u1]/u, u)=ulnu= y+1ln(y+1)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0therefore, the function does not exist at (1,1)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0oops the t side is both to have t^2+C, that was a type0

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0suppose to not both to lol stupid type0s

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope. you didn't finish the problem.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ln(y+1) does not exist for y=1

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0(1,1) is not in the domain or range

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0just tell me what you are thinking. I'm not the one who needs this problem done. lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0that guy left along time ago it looks like

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0did 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
 5 years ago
Best ResponseYou've already chosen the best response.0I got that answer, but I'm looking at something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm, 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
 5 years ago
Best ResponseYou've already chosen the best response.0he 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
 5 years ago
Best ResponseYou've already chosen the best response.0Do you have maple, mathmatica, or matlab?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've run a few different things in Wolfram Alpha, and I'm pretty sure no solutions exist.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so wolframalpha said no solution exist?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wolfram 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%28e^%28x11%29%291+

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's something wrong with your question, myininaya.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i was just worried about it because it makes me worry when people disagree with me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, well, I think you have the weight of evidence on your side ;)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks for supporting my solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I played around with that too and kept getting similar answers. I'll have to ask my professor. :/
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