Initial Value Problem / Differential Equation

  • Thread starter enkrypt0r
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  • #1
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Thanks for clicking!

So, I've got a problem here that I'm stuck on. I need to find the general solution to

y' = (y3 + 6y2 + 9y)/9

I found this to be

ln|y| + (3/(y+3)) - ln|y+3| = x + c

but I would appreciate it if you would check my work. Anywho, once I have the general solution I need to solve it for y(2) = -3. Obviously, plugging -3 in for y creates a problem in the second and third terms above.

Is there a way to rewrite the general solution or solve it in such a way that it won't contain any undefined terms when evaluated like it does now?

Thanks guys.

EDIT: Shoot, sorry for the non-descriptive title. I accidentally clicked post before finishing it. I meant it to say "[Calc II] Initial Value Problem / Differential Equation." My bad.
 
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Answers and Replies

  • #2
I got the same general solution, what do you mean by undefined terms?
 
  • #3
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I mean that when I plug -3 in for y, I get a zero in the denominator of the second term and ln(0) for the third term, both of which are undefined. I was wondering if there is any way to rewrite this or to solve this differently so -3 will not cause this discontinuity.
 
  • #4
tiny-tim
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hi enkrypt0r! :smile:
y' = (y3 + 6y2 + 9y)/9

erm :redface:look at the equation! :rolleyes:

if y = -3, what is y' ? :wink:
 
  • #5
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hi enkrypt0r! :smile:


erm :redface:look at the equation! :rolleyes:

if y = -3, what is y' ? :wink:

Hi, Tim!

I never thought to look at the original equation! This evaluates to 0 = 0. I'm afraid I don't know what this says about the differential equation. Does this mean that it's undefined, as my attempts to evaluate it have shown?
 
  • #6
tiny-tim
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oooh :cry:think! :smile:

if y' = 0, then … ? :wink:
 
  • #7
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Ummm... I'm feeling stupid here, haha. Sorry. I'm brainstorming but coming up with nothing beyond the obvious. We could integrate both sides and see that y is a constant, but we already know that. I'm sorry, I'm at a loss.
 
  • #8
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If y = 0 or y = -3, then y' = 0, which says that y is a constant.

In your work, the tacit assumption is that y != 0 and y != -3.
 
  • #9
Yeah, y=0 and y=-3 are "equilibria" solutions to the de. This sort of de is known as autonomous, because there are no t's on the right side of the equation. So, no matter what t is, the slope at y=0 and y=-3 is always zero.

So, why can't we plug in to solve the initial value problem?
 
  • #10
If you look at the direction field, I think you might find that more than one solution curve would satisfy the initial condition. Thus, there is no particular value of c that would net you a unique solution curve. Someone correct me if this is wrong, because I have never seen this case before! That's my analysis though.
 
  • #11
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If you look at the direction field, I think you might find that more than one solution curve would satisfy the initial condition. Thus, there is no particular value of c that would net you a unique solution curve. Someone correct me if this is wrong, because I have never seen this case before! That's my analysis though.

For the initial value problem, y' = (1/9)(y3 + 6y2 + 9y), y(2) = -3, we see that y' = (1/9)y(y + 3)2.

If y = -3, then y' = 0, so y(x) [itex]\equiv[/itex] C. Since y(2) = -3, C = -3, so the solution to the IVP is y(x) [itex]\equiv[/itex] -3. Are you saying that you think there are other solutions to this initial value problem?
 
  • #12
tiny-tim
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hi enkrypt0r! :smile:

(just got up :zzz:)

the solution comes in three parts …

y = -3 (for all t) is a solution

y = 0 (for all t) is a solution

and the other solutions are all the solutions of

dy/(y3 + 6y2 + 9y) = dt/9

(you multiplied by 0/0 when you got that equation, if y = -3 or 0, so you illegitmately lost those solutions :wink:)

if you draw a graph of some specimen solutions, you'll probably find that they "asymptote up to" the vertical lines at -3 and 0,

and (i'm guessing here :redface:) that one of y = 0 and v = -3 is stable and the other is unstable :smile:
 

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