Differential equation problem

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  • #1
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Homework Statement



Find a one-parameter family of solutions to the DE:

[itex]\frac{dy}{dt} = -y^3(t+1)[/itex]


Then find solutions to the DE that satisfies each of the following initial conditions y(0) = 2 and y(0) = -1, and give the range of t for which each solution exists.


The Attempt at a Solution



My main trouble is with the last part of the question which is very difficult. Here's my attempt so far:

I used the method of separation to get the general solution:

[itex]\frac{dy}{-y^3} = (t+1)dt[/itex]

[itex]\int \frac{1}{-y^3} dy = \int t+1 dt[/itex]

[itex]\frac{1}{2y^2} = \frac{t^2}{2} + t +c[/itex]

[itex]\frac{y^{-2}}{2} = \frac{t^2}{2} + t +c[/itex]

[itex]y(t) = \frac{1}{\sqrt{t^2 + 2t + 2c}}[/itex]

We write the constant more compactly:

[itex]y = \frac{1}{\sqrt{t^2 + 2t + k}}[/itex]

Is this general solution correct?

Now I found the solutions to those initial conditions:

[itex]y(0) =2 \implies \ 2=\frac{1}{\sqrt{k}} \implies k=\frac{1}{4}[/itex]

[itex]y(t) = \frac{1}{\sqrt{t^2 + 2t + 1/4}}[/itex]

[itex]y(0) =-1 \implies \ -1=\frac{1}{\sqrt{k}} \implies k=1[/itex]

[itex]y(t) = \frac{1}{\sqrt{t^2 + 2t + 1}}[/itex]

How do I find the range for which each solution exists? I think for the first initial condition we must have that t2+2t+1>0, and that's only 0 when t = -1, so the range is [-1,∞)?

For the second one, we must have t2+2t+1/4 > 0, and it's only 0 when t=-0.133, -1.86. So what can we say about that?

I appreciate any guidance.
 

Answers and Replies

  • #2
HallsofIvy
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Since you are given the value at x= 0, which is between those two points at which the solution does not exist, your solution exists between those values of x.

By the way, the it is better to give the exact values, [itex](2- \sqrt{3})/2[/itex] and [itex](2+ \sqrt{3})/2[/itex] rather than "-0.133" and "1.86".
 
  • #3
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For your solution when k=1, why do you have [-1,\infty)?

If t = -1, you are dividing by zero.

What is wrong with the (-\infty, -1)?

(-2)^2-4+1>0 Is this true for the entire set?
 
  • #4
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Thank you very much guys, I got the right answer. :)
 

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