Are they Bernoulli Equations? I'm trying to brush up on my knowledge of partial differential equations.
So I'm trying to solve the following initial value problems.
$$ \begin{cases}
y'=y-y^3,\ t>0 \\
y(0)= \frac{1}{2}
\end{cases} $$
and
$$ \begin{cases}
y'=y+y^3,\ t>0 \\
y(0)= \frac{1}{2}
\end{cases} $$
Consider whether their solutions remain bounded for every $t ≥ 0$. If not, you can
determine the 'burst time'.
Could someone suggest what theory I should read to understand how to solve this problem?
 A: To consider the first equation in the form, with constants $a$ and $b$,
$$y^{'} = a \, y - b \, y^3 \quad \text{with} \, y(0) = c $$
then one can follow the guide of Paul's Online Notes, or many other reference sites and/or books, to cast the equation into the form
$$ y^{-3} \, y^{'} - a \, y^{-2} = -b.$$
Make the substitution $v = y^{-2}$ to obtain
$$ - \frac{1}{2} \, v^{'} - a \, v = - b \quad \text{or} \\
v^{'} + 2 \, a \, v = 2 \, b. $$
Now make the change to $v = f + \frac{b}{a}$ to obtain
$$ f^{'} + 2 \, a \, f = 0. $$
This equation has the solution $f = c_{0} \, e^{- 2 a \, t}$ and returning to $y(t)$ one obtains
$$ y^{-2} = \frac{b}{a} + c_{0} \, e^{-2 a \, t} \quad \text{or} \\
y(t) = \left(\frac{b}{a} + c_{0} \, e^{-2 a \, t}\right)^{-1/2}. $$
Applying the boundary condition leads to
$$ y(t) = \left( \frac{b}{a} + \left(\frac{1}{c^2} - \frac{b}{a}\right) \, e^{-2 a \, t}\right)^{-1/2}. $$
A similar pattern leads to the other equation's solution. This is left as an exercise to the reader.
It should be noted that, in this case,
$$ \lim_{t \to \infty} \, y(t) = \sqrt{\frac{a}{b}}, $$
where $a$ and $b$ are constants. This says that the solution for large $t$ approaches a constant.
A: I Solved the separable equation $\dfrac{dy(t)}{dt} = y(t)^3 + y(t)$, such that $y(0) = \frac{1}{2}$:
Divided both sides by $y^{3}(t) + y(t)$:
$$\frac{\frac{dy(t)}{dt}}{y(t)^3 + y(t)} = 1$$
Integrated both sides with respect to $t$:
$$ \int \frac{\frac{dy(t)}{dt}}{y(t)^3 + y(t)} \, dt = \int 1 \, dt$$
Evaluated the integrals:
$$-\frac{1}{2} \, \log(y(t)^2 + 1) + \log(y(t)) = t + c_1,$$ where $c_1$ is an arbitrary constant.
Solve for $y(t)$:
$$y(t) = -\frac{i e^{t + c_1}}{\sqrt{e^{2 (t + c_1)} - 1}}$$ or
$$y(t) =\frac{i e^{t + c_1}}{ \sqrt{e^{2 (t + c_1)} - 1}}$$
And I have to solve these two cases for the initial condition $y(0) = \frac{1}{2}$
