Prove that the polynomial $q$ exists. Suppose $p \in P(R)$. Prove that there exists a polynomial $q \in P(R)$ such that $5q''+3q' = p.$
I have already proved this using algebra, does anyone know how to do this using linear algebra. Would I have to show that p is in the span of $q''$ and $q'$?
 A: Suppose that $p$ has degree $n$, and consider the linear operator $T(q) = 5q'' + 3q'$ from the vector space of degree $n+1$ polynomials to the vector space of degree $n$ polynomials.
By the rank-nullity theorem, it is enough to show that the kernel of $T$ has dimension $1$.  But, if we look at the highest degree monomial of $q$, it is straightforward to show that $T(q)=0$ if and only if $q$ is a constant.
Note that, on the much larger space of twice-differentiable functions $\mathbb{R}\to\mathbb{R}$, the kernel of $T$ also contains $e^{-\frac{3}{5}X}$.
A: Given any polynomial $r$, it is easy to find a polynomial $q$ such that $q' = r$. This means the problem reduces to: 

Given a polynomial $p$, how to find a polynomial $r$ such that $5r'+3r = p$?

Let $m = \deg p$. 
Let $X = \big\{\; f \in \mathbb{R}[x] : \deg f \le m \;\big\}$ be the sub-ring of polynomials with degree at most $m$. Define a map on $D : X \to X$ by
sending 
$$f = f_m x^m + f_{m-1} x^{m-1} + \cdots + f_0 \quad\text{to}\quad D(f) = m f_m x^{m-1} + (m-1) f_{m-1} x^{m-2} + \cdots + f_1$$
In terms of $D$, the problem becomes how to find a $r \in X$ such that $(3 + 5D)r = p$.
Notice $D$ is nilpotent. More precisely, $D^{m+1} = 0$. The formal power series for the inverse of $3+5D$ terminates at finite step
$$(3 + 5D)^{-1} 
= \frac13 \sum_{k=0}^\infty\left(-\frac53\right)^k D^k
= \frac13 \sum_{k=0}^n\left(-\frac53\right)^k D^k
$$
As a result, $(3 + 5D)^{-1}$ is well defined and $3+5D$ is an invertible linear map.
We can apply $(3 + 5D)^{-1}$ to $p$ to obtain the desired polynomial $r$:
$$r = \frac13 \sum_{k=0}^n\left(-\frac53\right)^k D^k(p)$$
A: You have an ordinary linear differential equation, so you can use linear algebra to help understand. If you can solve
$$ 5 f'' + 3 f' = x^k $$
for all $k$, then you can take the appropriate linear combinations to obtain solutions to your equation of interest.
If you want to be really sneaky, you can select a different basis for $P(\mathbb{R})$ that makes the equation really easy to solve, such as
$$ 5f'' + 3f' = 5(k+2)(k+1) x^{k} + 3(k+2) x^{k+1} $$
(which has solution $f = x^{k+2}$), along with a solution to
$$ 5f'' + 3f' = 1$$
which is also easy to solve.
It may be better to think of this as turning the question around: rather than asking whether, for any polynomial $p$, you can find a polynomial $q$ such that $5q'' + 3q' = p$, you should instead ask the questions "which polynomials $p$ can be the value of $5q'' + 3q'$?" That is, find the image of the linear transformation $q \mapsto 5q'' + 3q'$.
A: If you can show that the map defined by $q \rightarrow 5q'' + 3q'$ is a surjective linear operator on $P(R)$, then you are done.
If you have a surjective linear operator $T : V \rightarrow V$, then it can be shown that $\alpha T$ for any $\alpha \not= 0$ is also surjective. Moreover, You can show that given another surjective linear operator $S: V \rightarrow V$, that $ST$ and $S + T$ are also surjective.
With these observations, if you can show that the derivative operator $q \rightarrow q'$ is surjective, then you can show that $q \rightarrow 3q'$ is surjective, as well as $q \rightarrow 5q''$ (the latter is $q \rightarrow q'$ composed with itself, and then multiplied by 5). You can then show that $q \rightarrow 5q'' + 3q'$ is surjective. In other words, given any $p \in P(R)$, there exists $q \in P(R)$ such that $5q'' + 3q' = p$. 
