INV problem for $ \frac{dy}{dt} = 2y+3\cos4t $ Consider the differential equation
$$ \frac{dy}{dt} = 2y+3\cos4t $$
For what initial values $y(0)$ = $y_0$ are the solutions bounded for all t?
My approach is to calculate a general solution,which is $y = ke^{2t} - \frac{3}{5}cos4t + \frac{3}{5}sin4t$.
But I don't know how to make it bounded.
Please help me to solve this question. Thank you!
 A: As you're only asking about the solutions being bounded, I'll assume you can find the solutions.  To find it, I just used WolframAlpha.  I can help with finding the general solution by hand if you want though.
The general solution for this is $y(t)=c_1 e^{2t}+\frac{3}{5}\sin 4t-\frac{3}{10}\cos 4t$.  Let's see what $y_0$ looks like here by calculating $y(0)$:
$$y(0)=c_1e^{0}+\frac{3}{5}\sin(0)-\frac{3}{10}\cos 0=c_1-\frac{3}{10}=y_0\implies c_1=y_0+\frac{3}{10}$$ 
So, if we want to write our solution in terms of $y_0$, we get $y(t)=\left(y_0+\frac{3}{10}\right)e^{2t}+\frac{3}{5}\sin 4t-\frac{3}{10}\cos 4t$.
So, what values of $y_0$ lead to this being bounded?  Looking at each term, we see that $\sin 4t$ is bounded, that $\cos 4t$ is bounded, and that $e^{2t}$ is decidedly unbounded.  How are the sums of these impacted?  Per the triangle inequality, if we have that $|f(t)|<M$ and $|g(t)|<N$, then $|f(t)+g(t)|<|f(t)|+|g(t)|<M+N$, so the sum of bounded functions is bounded.  This is NOT true for unbounded functions.  So, we want $y_0$ such that the coefficient of $e^{2t}$ is zero, so our solution is the sum of bounded functions.  This will be true when $y_0=-\frac{3}{10}$.
