Figuring out the Behavior of this Differential Equation I've been working on some very basic differential equations, but I came to a
problem where I need to figure out the behavior of $y(t)$ as $t \rightarrow
\infty$ Given that
$$\frac{dy}{dt} = \frac{3t}{1+2e^{y}}.$$
In this case, it was very apparent to me that I would not be able to solve for
a simple solution of $y(t)$ due to the equation $1+2e^y$ in the denominator.
However, solving for this was rather straightforward:
$$\frac{dy}{dt}(1+2e^y) = 3t$$
$$\implies \frac{dy}{dt}+2e^y\frac{dy}{dt} = 3t$$
$$\implies \int \frac{dy}{dt}+2e^y\frac{dy}{dt} dt= \int 3t dt$$
$$\implies y(t) + 2e^{y(t)} = \frac{3}{2}t^2 + C.$$
However, it now has become quite difficult for me to figure out how to figure
out the behavior of $y(t)$ as $t \rightarrow \infty$. Someone suggested that
I should look for a particular in equality, but I am not sure how I could
manipulate the right-hand side to provide me with the desired information.
Any suggestions on this?
 A: As $t \rightarrow \infty$, the right-hand side of your (implicit) solution goes to infinity, so the left-hand side must also.
Note that $\dfrac{\mathrm{d}}{\mathrm{d}u} u + 2 \mathrm{e}^u = 1 + 2 \mathrm{e}^u$, which is positive for all values of $u$.  Consequently, if $y(t)$ increases, the left-hand side increases and if $y$ decreases, the left-hand side decreases.  Therefore, since the left hand-side must go to infinity as $t \rightarrow \infty$ and the left-hand side is finite for every positive value of $y(t)$, we must have $y(t) \rightarrow \infty$ as well.
You can see some of this from the original differential equation: $$
\dfrac{\mathrm{d}y}{\mathrm{d}t} = \frac{3t}{1+2 \mathrm{e}^y}  \text{.}
$$  Since the denominator always positive (greater than $1$, even), as $t \rightarrow \infty$, the numerator is also positive, so the derivative of $y$ is always positive, so $y$ strictly monotonically increases (maybe not to $\infty$).  (Since by the previous paragraph, we know $y \rightarrow \infty$, it must be that $y$ does not grow too fast, otherwise the slope of $y$ would get too flat and would fail to escape to $\infty$.)
A: since $y'>0$ when $t>0.$   $y(t)$ is constantly increasing for all $t>0$
If $y(t)$ were bounded $y'(t)$ would either have to oscillate around $0,$ or the limit as $t$ goes to infinity of $y(t)$ would equal $0.$
A: I will give you a precise proof describing the behaviour of $y(t)$ when $t$ goes to $+\infty$. Observe that no matter what your initial condition at $t=0$ is, you will have $y'(t)>0$ for $t>0$. You get this information by checking the sign of the right hand side of $$\frac{dy}{dt} = \frac{3t}{1+2e^{y}}$$
The limit $\lim_{t\to +\infty}{y(t)}$ exists because $y$ is a monotone function and, because $y$ is increasing, it can be finite or $+\infty$. Let's prove that the finite case can't happen. Assume on the contrary that $\lim_{t\to +\infty}{y(t)}$ is finite. This implies that $1+2e^{y(t)}<M$ for every positive $t$ and certain positive constant $M$. So $$\frac{dy}{dt} = \frac{3t}{1+2e^{y}} > \frac{3t}{M}$$
for every positive $t$. Integrating both sides from $0$ to $t$ the inequality is preserved and you get $$y(0) + \frac{1}{2M} t^2 \leq y(t)$$ and that means that $\lim_{t\to +\infty}{y(t)} = +\infty$, a contradiction. So the limit $\lim_{t\to +\infty}{y(t)}$ is not finite and thus $\lim_{t\to +\infty}{y(t)} = +\infty$.
====== EDIT ======
Once you have proven $\lim_{t\to +\infty}{y(t)} = +\infty$, you might want to quantify that claim. Now is when your prior work comes in handy. Consider $$ y(t) + 2e^{y(t)} = \frac{3}{2}t^2 + C$$ where $C$ depends on the initial condition. Because $\lim_{t\to +\infty}{y(t)} = +\infty$, you know $y(t)$ is positive for big enough $t$ so $$ 2e^{y(t)} < y(t) + 2e^{y(t)} = \frac{3}{2}t^2 + C$$ and then $$ e^{y(t)} < \frac{3}{4}t^2 + \frac{C}{2}.$$ From this results $$ y(t) < \log{\big( \frac{3}{4}t^2 + \frac{C}{2} \big)}$$ for big enough $t$.
Similarly you have $$  \frac{3}{2}t^2 + C = y(t) + 2e^{y(t)} < 3e^{y(t)}$$ for big enough $t$. You can conclude $$ \log{\big( \frac{1}{2}t^2 + \frac{C}{3} \big)} < y(t)$$ for big enough $t$.
Combine those bounds to get
$$\log(\frac{1}{2}) + 2\log{\big(t \big)} + \log{\big( 1 + \frac{2C}{3t^2} \big)} < y(t) < \log(\frac{3}{4}) + 2\log{\big(t \big)} + \log{\big( 1 + \frac{2C}{3t^2} \big)}$$ for big enough $t$. This is a much more precise claim, giving you the order of $y$ at $+\infty$. We can say that $$\lim_{t\to +\infty}{\frac{y(t)}{2\log{ \big( t \big) }}} = 1$$ to simplify.
