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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 t

^{2}+2t+1>0, and that's only 0 when t = -1, so the range is [-1,∞)?

For the second one, we must have t

^{2}+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.