Trying to solve the above two non-bolded, unrelated equations (functions of t) for only their particular solutions. I recognize that the right-hand side of the first equation is a polynomial of degree m, so the particular solution must have that form too. So I took 4 derivatives in order to plug it into the differential equation. But then that just gives me a very general form of a solution, with no ability to solve for the constants without setting all of them except one to be arbitrarily zero. Then my A_m would require a t-term in it, which doesn't make sense to me (isn't the solution to it supposed to be a constant?)
I don't believe my work is right and know I shouldn't continue to the next problem (as it seems to be very similar. Any help would be great!
For $y^{(4)} = t^m,$ observe that the derivative of a polynomial always has a lower degree than the original polynomial. You can't possibly differentiate an $m$th degree polynomial four times and still have an $m$th degree polynomial. You will have to start with a higher-degree polynomial. Try one of an arbitrary degree, differentiate it four times, and try to set the fourth derivative equal to $t^m.$ You've done most of the work, except you wrote $m$ in several places where you should have written the arbitrary degree of your unknown polynomial. Change those $m$s to something else (maybe $n$) and you may get somewhere.
For the second problem, notice that if $v = t^n e^t$ (notice that's an arbitrary exponent $n,$ not necessarily the same exponent $m$ as in the problem), then $$v' = n t^{n-1}e^t + t^n e^t = n t^{n-1}e^t + v.$$ We can rewrite this as $v' - v = n t^{n-1}e^t.$ If we then consider the operator $\frac{d}{dt} - 1$ (that is, "differentiate the function and subtract the original function from the result"), we have $$ \left(\frac{d}{dt} - 1\right) \left(t^n e^t\right) = v^{(1)} - v = n t^{n-1}e^t. $$
Try applying the operator $\frac{d}{dt} - 1$ to $v$ twice and see what happens. It won't be like the equation you need, but perhaps you can see a pattern developing. (As another hint, think about the binomial formula for expanding $(x - 1)^k$.) What happens if you repeat the application of $\frac{d}{dt} - 1$ multiple times?
Again, for the solution you'll have to start with something where $t$ has a higher exponent than $m,$ because every time you apply $\frac{d}{dt} - 1$ the exponent of $t$ is reduced by one.