Showing a solution to 3rd order differential equations forms a subspace Let $S$ denote the set of all solutions of the following differential equation defined on $C^3[0,\infty)$;
$$
\begin{align}
  \frac{d^3x}{dt^3} + b \frac{d^2x}{dt^2} + c \frac{dx}{dt} + dx = 0
\end{align}
$$
Show that $S$ is a linear subspace of $C^3[0, \infty)$

I know that subspaces are closed under addition and scalar multiplication and also contan the zero vector, but I'm not sure how to show that all solutions form a linear subspace. (I'm not sure how to solve this equation either)
Final Answer
Note that:
$$
\begin{align}
   0 &= \frac{d^3x}{dt^3} + b \frac{d^2x}{dt^2} + c \frac{dx}{dt} + dx \\
     &= \alpha \left(\frac{d^3x}{dt^3} + b \frac{d^2x}{dt^2} + c \frac{dx}{dt} + dx\right)
\end{align}
$$
Where $\alpha$ is a constant.
Consider arbitrary functions $f$ and $g$ and scalar $\alpha$. By linearity of operators we have:
$$
\begin{align}
\frac{d^3}{dt^3}(\alpha f + g ) &= \alpha \frac{d^3f}{dt^3} + \frac{d^3g}{dt^3} \\
\frac{d^2}{dt^2}(\alpha f + g ) &= \alpha \frac{d^2f}{dt^2} + \frac{d^2g}{dt^2} \\
\frac{d}{dt}(\alpha f + g ) &= \alpha \frac{df}{dt} + \frac{dg}{dt} \\
d(\alpha f + g) &= \alpha df + dg
\end{align}
$$
Now assign $f$ and $g$ to arbitrary individual solutions to our differential equation, we have:
$$
\begin{align}
  0 &= \frac{d^3}{dt^3}(\alpha f + g) + b \frac{d^2}{dt^2}(\alpha f + g) + c \frac{d}{dt}(\alpha f + g) + d(\alpha f + g) \\
  &= \alpha(\frac{d^3}{dt^3}f + b \frac{d^2}{dt^2}f + c \frac{d}{dt}f + df ) + (\frac{d^3}{dt^3}g + b \frac{d^2}{dt^2}g + c \frac{d}{dt}g + dg) \\
  &= \alpha \pmb f + \pmb g 
\end{align}
$$
This shows that $\alpha \pmb f + \pmb g$ solves the ODE which proves that $\alpha \pmb f + \pmb g \in \mathcal S$
The addition holds because the sum that describes $\pmb f$ sums to $0$. Likewise for $\pmb g$.
Therefore, all arbitrary combinations of $\pmb f$ and $\pmb g$ are in $\mathcal S$ (This can be seen by also allowing $\pmb f$ be $\pmb g$ and $\pmb g$ be $\pmb f$).
Finally, $\pmb 0$ is also in the set of solutions.
Thus, the set of all solutions forms a subspace.
 A: You don't need to solve the ODE. Just imagine a solution $f$. Plug $\alpha f$ in and see it verifies the ODE, and the same with $f+g$ for the sum of solutions. Verify $0$ solves the ODE if you please (this is key in the case that the diffeq might have no solutions: see discussion below), and you know the solution space forms a subspace of the vector space, say, of all $C^3$ functions on some appropriate domain.
A: It is easy to see that $S\subset \mathcal{C}^3[0,\infty)$.
All you need do is show that if $f$ and $g$ are in $S$ and $\alpha$ is a scalar then $\alpha f+g$ is a solution of the ODE. In other words, $\alpha f+g \in S$.
Clearly $\frac{d^3}{dt^3}(\alpha f+g) = \alpha \frac{d^3f}{dt^3} + \frac{d^3g}{dt^3}$, $\frac{d^2}{dt^2}(\alpha f+g) = \alpha \frac{d^2f}{dt^2} + \frac{d^2g}{dt^2}$, and
$\frac{d}{dt}(\alpha f+g) = \alpha \frac{df}{dt} + \frac{dg}{dt}$.
So,
$\frac{d^3}{dt^3}(\alpha f+g) + \frac{d^2}{dt^2}(\alpha f+g) + \frac{d}{dt}(\alpha f+g) +d(\alpha f+g)$
$ = \alpha \bigg(\frac{d^3f}{dt^3} + \frac{d^2f}{dt^2}+\frac{df}{dt} + df\bigg) +\frac{d^3g}{dt^3}+ \frac{d^2g}{dt^2}+ \frac{dg}{dt} + dg = 0$
