I'm trying to solve the following initial value problem.
$$ \begin{cases} y'=y-y^3,\ t>0 \\ y(0)= \frac{1}{2} \end{cases} $$
Consider whether their solutions remain bounded for every $t ≥ 0$. If not, you can determine the 'burst time'.
Solution
We Solve the Bernoulli equation $\dfrac{dy(t)}{dt}-y(t)= - y(t)^3,\ t>0$ (1), such that $y(0) = \frac{1}{2}$.
We set $$v=y^{1-3}$$ Then we solve for y, which gives $$y=v^{-\frac{1}{2}}\,(2).$$ We differentiate the (2), so we have that $$\frac{dy}{dt}=-\frac{1}{2}v^{-\frac{3}{2}}\frac{dv}{dt}.$$
We plug all those into equation (1) and we have $$-\frac{1}{2}v^{-\frac{3}{2}}\frac{dv}{dt}-v^{-\frac{1}{2}}=-(v^{-\frac{1}{2}})^3$$
We multiply both sides by $-2v^{\frac{3}{2}}$ and we received $$\frac{dv}{dt}+2v=2.$$
Let $$p(t)=e^{\int 2 dt}=e^{2t} $$
Multiply both sides with p(t) so we have: $$\frac{dv}{dt}e^{2t}+e^{2t}2v=2e^{2t}$$ We integrate $$\int (e^{2t}v)' \, dt= \int 2 e^{2t} dt $$ and we have $$e^{2t}v=\frac{2e^{2t}v}{2}+c.$$ Then we solve for v $$v=1+c e^{-2t}.$$ Finally we solve for y and our solution is $$y=\frac{1}{\sqrt{1+c e^{-2t}}}$$
Then we plug in the initial condition $y(0) = \frac{1}{2}$ and we find that $$y(t) = \frac{1}{\sqrt{1+3e^{-2 t } }}$$
$$ 1+3e^{-2 t } \geq 0\Leftrightarrow e^{2t}\leq -3$$
My question Is this solution bounded or what is the burst time? I have tried to find the domain of the solution but it doesn't give me an interval