# Particular solution of mth degree polynomial

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!

• It's unclear what problem you're trying to solve. You say two equations, but you show five; presumably the three in the middle are your attempted solution, not part of the original problem? Are the first and last equation supposed to be related in any way, or are they for two completely independent functions? – David K Oct 20 '17 at 20:33
• @DavidK woops. The two non-bolded equations are what I'm trying to solve and are not related in any way (separate problems). The bolded ones are my work :) – Anthony Oct 20 '17 at 20:42

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.
• A term with $t^{m+4}$ is exactly what you want. If you write $y = A_{m+4}t^{m+4} + \cdots$ and differentiate four times like your first attempt, it should work. (You won't get that unwanted $t^4$ again.) – David K Oct 20 '17 at 22:51
• The idea of using an arbitrary degree is you would have started with something like $y=A_n t^n+\cdots$ and eventually you would have figured out that you want $n=m+4.$ But since you already figured out the $m+4,$ you can skip the "arbitrary degree" step. – David K Oct 20 '17 at 22:53