\begin{align*}
\mathcal{L}\{F'(t)\}&=\int_0^\infty e^{-st}F'(t)dt\\
&=e^{-st}F(t)\vert_0^\infty+s\int_0^\infty e^{-st}F(t)dt
\end{align*}
In order to ensure that $\mathcal{L}\{F'(t)\}$ exists, other than requiring that $F'(t)$ is piecewise continuous in every finite interval $[0,T]$ we also need to require that $F(t)$ be of order $e^{\alpha t}$, in terms of the big O notation we write $F(t)=O(e^{\alpha t})$ as $t\to\infty$, meaning $\frac{F(t)}{e^{\alpha t}}$ is bounded for large $t$ i.e. there exists $M>0$ such that $e^{-\alpha t}|F(t)|<M$ for large $t$. With this assumption,
$$|e^{-st}F(t)|<Me^{-(s-\alpha)t}\to 0$$
as $t\to\infty$ provided $s>\alpha$. Consequently, we have
\begin{equation}
\label{eq:laplace6}
\mathcal{L}\{F'(t)\}=s\mathcal{L}\{F(t)\}-F'(0)
\end{equation}
Using \eqref{eq:laplace6}
\begin{align*}
\mathcal{L}\{F^{\prime\prime}(t)\}&=s\mathcal{L}\{F'(t)\}-F'(0)\\
&=s(s\mathcal{L}\{F(t)\}-F'(0))-F'(0)\\
&=s^2\mathcal{L}\{F(t)\}-sF(0)-F'(0)
\end{align*}
Continuing we obtain
\begin{equation}
\label{eq:laplace7}
\mathcal{L}\{F^{(n)}(t)\}=s^n\mathcal{L}\{F(t)\}-s^{n-1}F(0)-s^{n-2}F'(0)-s^{n-3}F^{\prime\prime}(0)-\cdots -F^{(n-1)}(0)
\end{equation}
Some of the transforms we found here can be obtained using \eqref{eq:laplace7} as seen in the next couple of examples below.

Example. Let $F(t)=t$. Then $F'(t)=1$ so \eqref{eq:laplace6} results in $\mathcal{L}\{1\}=s\mathcal{L}\{s\}$ and thereby
$$\mathcal{L}\{t\}=\frac{1}{s^2}$$

Example. Let $F(t)=\sin kt$. Then $F'(t)=k\cos kt$ and $F^{\prime\prime}(t)=-k^2\sin kt$. So using \eqref{eq:laplace7} for $n=2$
$$\mathcal{L}\{F^{\prime\prime}(t)\}=s^2\mathcal{L}\{F(t)\}-sF(0)-F'(0)$$
we have
$$-k^2\mathcal{L}\{\sin kt\}=s^2\mathcal{L}\{\sin kt\}-k$$
that is,
$$\mathcal{L}\{\sin kt\}=\frac{k}{s^2+k^2}$$

Example. Using \eqref{eq:laplace7} we can also prove the formula
\begin{equation}
\label{eq:laplace8}
\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}
\end{equation}
where $n$ is a nonnegative integer. Recall that the formula \eqref{eq:laplace8} appeared here. Let $F(t)=t^n$. Then
\begin{align*}
F(0)&=F'(0)=\cdots=F^{(n-1)}(0)=0\\
F^{(n)}(t)&=n!\\
F^{(n+1)}(t)&=0
\end{align*}
So
$$\mathcal{L}\{F^{(n+1)}(t)\}=s^{n+1}\mathcal{L}\{F(t)\}-s^nF(0)-s^{n-1}F'(0)-\cdots – F^{(n)}(0)$$
reduces to
$$0=s^{n+1}\mathcal{L}\{t^n\}-n!$$
that is,
$$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$$

Example. For $k>-1$ real,
$$\mathcal{L}\{t^k\}=\int_0^\infty e^{-st}t^kdt$$
with $s>0$. Using the subsitution $x=st$ $\mathcal{L}\{t^k\}$ can be written as
\begin{equation}
\label{eq:laplace9}
\begin{aligned}
\mathcal{L}\{t^k\}&=\frac{1}{s^{k+1}}\int_0^\infty e^{-x}x^kdx\\
&=\frac{\Gamma(k+1)}{s^{k+1}}
\end{aligned}
\end{equation}
\begin{equation}
\label{eq:gamma}
\Gamma(k+1)=\int_0^\infty e^{-x}x^kdx
\end{equation}
is called the Gamma function or factorial function with the argument $k+1$. (There are different ways to define the Gamma function. This definition is due to Euler. For other definitions and for more details please see the reference [1] below.) $\Gamma (k+1)$ is also denoted by $k!$. In fact if $k=n$ is a positive integer,
$$k!=n!=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1$$
When $k=-\frac{1}{2}$ using the substitution $u=\sqrt{x}$ we see that $\Gamma\left(\frac{1}{2}\right)$ is just the Gaussian integral
$$\Gamma\left(\frac{1}{2}\right)=2\int_0^\infty e^{-u^2}du=\sqrt{\pi}$$

References:

[1] Mathematical Methods for Physicists, George Arfken, Third Edition, Academic Press, 1985

The Laplace transform is an integral transform that allows us to solve a linear differential equation by converting it to algebraic expressions such as rational functions.

The Laplace transform $\mathcal{L}\{F(t)\}$ of a function $F(t)$ is defined by
\begin{equation}
\label{eq:laplace}
\mathcal{L}\{F(t)\}=\int_0^\infty e^{-st}F(t)dt
\end{equation}
Since the improper integral in \eqref{eq:laplace} is a function of $s$, we also write $\mathcal{L}\{F(t)\}=f(s)$.

Example. Let $F(t)=1$, $t>0$.
$$\mathcal{L}\{F(t)\}=\int_0^\infty e^{-st}dt=\frac{1}{2}$$
provided $s>0$.

Example. Let $F(t)=e^{kt}$, $t>0$ and $k$ a constant.
\begin{align*}
\mathcal{L}\{F(t)\}&=\int_0^\infty e^{-st}e^{kt}dt\\
&=\int_0^\infty e^{-(s-k)}tdt\\
&=\frac{1}{s-k}
\end{align*}
provided $s>k$.

Example.
\begin{align*}
\mathcal{L}\{t\}&=\int_0^\infty te^{-st}dt\\
&=\frac{1}{s^2}
\end{align*}
for $s>0$.

Example. Again by the definition of the Laplace transform one obtains
\begin{align*}
\mathcal{L}\{t^2\}&=\frac{2!}{s^3}\\
\mathcal{L}\{t^3\}&=\frac{3!}{s^4}
\end{align*}
by simply calculating the required improper integrals. There appears to be a pattern and one may expect the formula
$$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$$
This is indeed true and we will prove this later.

Example. By the definition of the Laplace transform one obtains
\begin{align}
\label{eq:laplace2}
\mathcal{L}\{\sin kt\}&=\frac{k}{s^2+k^2}\\
\label{eq:laplace3}
\mathcal{L}\{\cos kt\}&=\frac{s}{s^2+k^2}
\end{align}

The Laplace transform is linear, namely
\begin{align*}
\mathcal{L}\{F(t)+G(t)\}&=\mathcal{L}\{F(t)\}+\mathcal{L}\{G(t)\}\\
\mathcal{L}\{cF(t)\}&=c\mathcal{L}\{F(t)\}
\end{align*}

Example. $\sinh kt=\frac{e^{kt}-e^{-kt}}{2}$ so by the linearity of the Laplace transform
\begin{align*}
\mathcal{L}\{\sinh kt\}&=\frac{1}{2}(\mathcal{L}\{e^{kt}\}-\mathcal{L}\{e^{-kt}\}\\
&=\frac{1}{2}\left(\frac{1}{s-k}-\frac{1}{s+k}\right)\\
&=\frac{k}{s^2-k^2}
\end{align*}
Hence we have
\begin{equation}
\label{eq:laplace4}
\mathcal{L}\{\sinh kt\}=\frac{k}{s^2-k^2}
\end{equation}
Similarly we also obtain
\begin{equation}
\label{eq:laplace5}
\mathcal{L}\{\cosh kt\}=\frac{s}{s^2+k^2}
\end{equation}

Depending on $g(t)$ often it is difficult to come up with a suitable trial solution of a given non-homogeneous equation in the method of undetermined coefficient discussed here. In this note, we discuss an alternative method called the method of variation of parameters. First assume that we know the general solution
$$x_{\mathrm{hom}}(t)=c_1x_1(t)+c_2x_2(t)$$
of the homogeneous equation $$\ddot{x}+p(t)\dot{x}+q(t)x=0$$ Let
\begin{equation}
\label{eq:vp}
x(t)=u_1(t)x_1(t)+u_2(t)x_2(t)
\end{equation}
be a solution of the non-homogeneous equation $$\ddot{x}+p(t)\dot{x}+q(t)x=g(t)$$ Differentiating \eqref{eq:vp}
$$\dot{x}(t)=\dot{u}_1x_1+u_1\dot{x}_1+\dot{u}_2x_2+u_2\dot{x}_2$$
Require that
\begin{equation}
\label{eq:vp2}
\dot{u}_1x_1+\dot{u}_2x_2=0
\end{equation}
so that we have
\begin{equation}
\label{eq:vp3}
\dot{x}(t)=u_1\dot{x}_1+u_2\dot{x}_2
\end{equation}
Differentiating \eqref{eq:vp3}
\begin{equation}
\label{eq:vp4}
\ddot{x}(t)=\dot{u}_1\dot{x}_1+u_1\ddot{x}_1+\dot{u}_2\dot{x}_2+u_2\ddot{x}_2
\end{equation}
Substituting $\ddot{x}$, $\dot{x}$, $x$ in the non-homogeneous equation with the corresponding expressions in \eqref{eq:vp}, \eqref{eq:vp3} and \eqref{eq:vp4}, respectively results in the equation
\begin{equation}
\label{eq:vp5}
\dot{u}_1\dot{x}_1+\dot{u}_2\dot{x}_2=g(t)
\end{equation}
Let
$$W(x_1,x_2)(t):=\begin{vmatrix}
x_1(t) & x_2(t)\\
\dot{x}_1(t) & \dot{x}_2(t)
\end{vmatrix}=x_1(t)\dot{x}_2(t)-x_2(t)\dot{x}_1(t)$$
$W(x_1,x_2)(t)$ is called the Wronskian of $x_1(t)$ and $x_2(t)$. If $x_1(t)$ and $x_2(t)$ are linearly dependent, so are the columns of $W(x_1,x_2)(t)$ hence $W(x_1,x_2)(t)=0$ for all $t$. This means that If $W(x_1,x_2)(t)\ne 0$ for some $t$, $x_1(t)$ and $x_2(t)$ are linearly independent. Also we have the following theorem holds.

Theorem. Let $x_1(t)$ and $x_2(t)$ be solutions of a homogeneous second-order linear differential equation. If $x_1(t)$ and $x_2(t)$ are linearly independent, then $W(x_1,x_2)(t)\ne 0$ for all $t$.

Since $x_1(t)$ and $x_2(t)$ are linearly independent, $W(x_1,x_2)\ne 0$ for all $t$ so by Cramer’s rule the solution of the system of linear equations \eqref{eq:vp2}, \eqref{eq:vp5} in $\dot{u}_1$ and $\dot{u}_2$ is given by
\begin{equation}
\begin{aligned}
\dot{u}_1(t)&=\frac{\begin{vmatrix}
0 & x_2\\
g(t) & \dot{x}_2
\end{vmatrix}
}{W(x_1,x_2)(t)}=-\frac{g(t)x_2(t)}{W(x_1,x_2)(t)}\\
\dot{u}_2(t)&=\frac{\begin{vmatrix}
x_1 & 0\\
\dot{x}_1 & g(t)
\end{vmatrix}
}{W(x_1,x_2)(t)}=\frac{g(t)x_1}{W(x_1,x_2)(t)}
\end{aligned}\label{eq:vp6}
\end{equation}
Integrating \eqref{eq:vp6}, $u_1(t)$ and $u_2(t)$ are determined to be
\begin{equation}
\label{eq:vp7}
\begin{aligned}
u_1(t)&=-\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt+c_1\\
u_2(t)&=\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt+c_2
\end{aligned}
\end{equation}
where $c_1$ and $c_2$ are constant.
Therefore
\begin{equation}
\label{eq:vp8}
\begin{aligned}
x(t)=c_1x_1(t)+&c_2x_2(t)+\left\{-x_1(t)\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt+\right.\\
&\left.x_2(t)\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt\right\}
\end{aligned}
\end{equation}
is indeed the general solution of the non-homegeneous equation for
\begin{equation}
\label{eq:vp9}
X(t)=-x_1(t)\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt+x_2(t)\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt
\end{equation}
being a particular solution of the non-homogeneous equation.

Example. Solve the non-homogeneous equation $\ddot{x}-3\dot{x}-4x=2\sin t$.

Solution. Recall that $x_1(t)=e^{-t}$, $x_2(t)=e^{4t}$. The Wronskian is
$$W(x_1,x_2)(t)=\begin{vmatrix}
e^{-t} & e^{4t}\\
-e^{-t} & 4e^{4t}
\end{vmatrix}=5e^{3t}$$
\begin{align*}
\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt&=\int\frac{e^{4t}(2\sin t)}{5e^{3t}}dt\\
&=-\frac{1}{5}e^t\cos t+\frac{1}{5}e^t\sin t
\end{align*}
and
\begin{align*}
\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt&=\int\frac{e^{-t}(2\sin t)}{5e^{3t}}dt\\
&=-\frac{2}{85}e^{-4t}\cos t-\frac{8}{85}e^{-4t}\sin t
\end{align*}
The general solution is
\begin{align*}
x(t)&=c_1e^{-t}+c_2e^{4t}\\
&-e^{-t}\left(-\frac{1}{5}e^t\cos t+\frac{1}{5}e^t\sin t\right)+e^{4t}\left(-\frac{2}{85}e^{-4t}\cos t-\frac{8}{85}e^{-4t}\sin t\right)\\
&=c_1e^{-t}+c_2e^{4t}+\frac{3}{17}\cos t-\frac{5}{17}\sin t.
\end{align*}

Example. Solve the non-homogeneous $\ddot{x}+4x=8\tan t$, $-\frac{\pi}{2}<t<\frac{\pi}{2}$

Solution. $x_1(t)=\cos(2t)$, $x_2(t)=\sin(2t)$. The Wronskian is
$$W(x_1,x_2)(t)=\begin{vmatrix}
\cos(2t) & \sin(2t)\\
-2\sin(2t) & 2\cos(2t)
\end{vmatrix}=2$$
\begin{align*}
\int\frac{x_2(t)g(t)}{W(x_1,x_2)(t)}dt&=\int\frac{\sin(2t)(8\tan t)}{2}dt\\
&=4t-2\sin(2t)
\end{align*}
and
\begin{align*}
\int\frac{x_1(t)g(t)}{W(x_1,x_2)(t)}dt&=\int\frac{\cos(2t)(8\tan t)}{2}dt\\
&=-2\cos(2t)+4\ln(\cos t)
\end{align*}
Thus the general solution is
$$x(t)=c_1\cos(2t)+c_2\sin(2t)-4t\cos(2t)+4\sin(2t)\ln(\cos t)$$

Let us consider the following second-order linear differential equation
\begin{equation}
\label{eq:nhde}
\ddot{x}+p(t)\dot{x}+q(t)x=g(t)
\end{equation}
Let $X_1(t)$ and $X_2(t)$ be two solution of \eqref{eq:nhde}. Then $X_1(t)-X_2(t)$ is a solution of the homogeneous equation
\begin{equation}
\label{eq:hde}
\ddot{x}+p(t)\dot{x}+q(t)x=0
\end{equation}
Thus $X_1(t)-X_2(t)=c_1x_!(t)+c_2x_2(t)$ where $x_1(t)$ and $x_2(t)$ are a fundamental set of solutions of \eqref{eq:hde}. This implies that the general solution of \eqref{eq:nhde} is given by
$$x(t)=c_1x_1(t)+c_2x_2(t)+X(t)$$
where $X(t)$ is a solution of \eqref{eq:nhde}. That is, solving the non-homogeneous equation \eqref{eq:nhde} boils down to finding a solution of \eqref{eq:nhde}. There are two methods of finding a solution of \eqref{eq:nhde}:

1.  Method of Undetermined Coefficients
2. Variation of Parameters

The Method of Undetermined Coefficients

This method finds a solution by guessing a particular solution with undetermined coefficients.

Example. Solve the non-homogeneous equation $\ddot{x}-3\dot{x}-4x=3e^{2t}$.

Solution. Let $X(t)=Ae^{2t}$ be a solution of the non-homogeneous equation. Then $\dot{X}=2Ae^{2t}$ and $\ddot{X}=4Ae^{2t}$. By substitution we obtain $\ddot{X}-3\dot{X}-4X=-6Ae^{2t}$ and so $A=-\frac{1}{2}$. Hence the general solution of the non-homogeneous equation is
$$x(t)=c_1e^{-t}+c_2e^{4t}-\frac{1}{2}e^{2t}$$

Example. Solve the non-homogeneous equation $\ddot{x}-3\dot{x}-4x=2\sin t$.

Solution. Let $X(t)=A\sin t$. Then $\dot{X}(t)=A\cos t$ and $\ddot{X}=-A\sin t$. By substitution we obtain $\ddot{X}-3\dot{X}-4X=-5A\sin t -3A\cos t$. This has to be the same as $2\sin t$ which leads to the equation $(5A+2)\sin t+3A\cos t$. Since $\sin t$ and $\cos t$ are linearly independent, $5A+2=0$ and $3A=0$, a contradiction! This time we assume that $X(t)=A\sin t+B\cos t$. Then $\dot{X}(t)=A\cos t-B\sin t$, $\ddot{X}(t)=-A\sin t-B\cos t$ and
$$\ddot{X}-3\dot{X}-4X=(-5A+3B)\sin t+(-3A-5B)\cos t$$
Comparing this with $2\sin t$ we get the system of linear equations
$$\left\{\begin{aligned}-5A+3B&=2\\
-3A-5B&=0
\end{aligned}\right.$$
of which solution is $A=-\frac{5}{17}$ and $B=\frac{3}{17}$. Therefore, the general solution is
$$x(t)=c_1e^{-t}+c_2e^{4t}-\frac{5}{17}\sin t+\frac{3}{17}\cos t$$

Example. Solve the non-homogeneous equation $\ddot{x}-3\dot{x}-4x=4t^2-1$.

Solution. Let $X(t)=At^2+Bt+C$ be a solution of the non-homogeneous equation. Then $A=-1$, $B=\frac{3}{2}$ and $C=-\frac{11}{8}$. Hence the general solution is
$$x(t)=c_1e^{-t}+c_2e^{4t}-t^2+\frac{3}{2}t-\frac{11}{8}$$

Example. Solve the non-homogeneous equation $\ddot{x}-3\dot{x}-4x=-8e^t\sin 2t$.

Solution.
\begin{align*}
\dot{X}(t)&=(-2A+B)e^t\sin 2t+(A+2B)e^t\cos 2t\\
\ddot{X}(t)&=(-3A+4B)e^t\cos 2t+(-4A-3B)e^t\sin 2t
\end{align*}
and
$$\ddot{X}-3\dot{X}-4X=(-10A-2B)e^t\cos 2t+(2A-10B)e^t\cos 2t$$
Comparing this with $-8e^t\cos 2t$ we get the system of linear equations
$$\left\{\begin{aligned}
-10A-2B&=-8\\
2A-10B&=0
\end{aligned}\right.$$
of which solution is $A=\frac{10}{13}$ and $B=\frac{1}{13}$.
Hence the general solution is
$$x(t)=c_1e^{-t}+c_2e^{4t}+\frac{10}{13}e^t\cos 2t+\frac{2}{13}e^t\sin t$$

When we studied second-order linear differential equations through undamped and damped harmonic motion, we made a hand waving argument that the general solution of a second-order linear differential equation is the linear combination of two distinct solutions that are linearly independent in the same sense as for vectors i.e. one solution is not a constant multiple of another solution. These solutions are called fundamental solutions. In this note, we discuss an important and intriguing relationship between second-order linear differential equations and linear algebra and explain why the general solution is given by the linear combination of two fundamental solutions. For the sake of simplicity, we limit our discussion to the case that characteristic equation has two distinct real solutions. In order to maintain this lecture note as much self-contained as possible, we include some of the basic concepts in linear algebra that we need for our discussion below.

Let us consider a second-order linear differential equation \begin{equation}\label{eq:2lde}\ddot{x}+p\dot{x}+rx=0\end{equation} \eqref{eq:2lde} can be written as a system of two first-order linear differential equations \begin{equation}\left\{\begin{aligned}\frac{dx}{dt}&=s\\\frac{ds}{dt}&=-ps-rx\end{aligned}\right.\label{eq:ldesys}\end{equation} Let $X=\begin{pmatrix}x\\s\end{pmatrix}$ and $A=\begin{pmatrix}0 & 1\\-r & -p\end{pmatrix}$. Then \eqref{eq:ldesys} can be written as the matrix differential equation $$\frac{dX}{dt}=AX$$

Definition. Let $A$ be a $2\times 2$ real matrix or equivalently a linear map $A: \mathbb{R}^2\longrightarrow\mathbb{R}^2$. A vector $v\in\mathbb{R}^2$ is called an eigenvector of $A$ if there exists a number $q\in \mathbb{R}$ such that $Av=qv$. The number $q$ is called an eigenvalue of $A$ belonging to the eigenvector $v$. We also say $v$ is an eigenvector associated with the eigenvalue $q$. Eigen is a German word and it means own or self. As you will see below, given an eigenvalue there are infinitely many eigenvectors that are associated with the eigenvalue but they are linearly dependent i.e one eigenvector is a scalar multiple of another. So the name makes sense.

How do we find eigenvalues of a matrix $A$? The equation $Av=qv$ is the same as $(A-qI)v=0$. In order for this equation to have a non-trivial solution ($v\ne 0$) it must be that \begin{equation}\label{eq:cheq}\det(A-qI)=0\end{equation} The equation \eqref{eq:cheq} is called the characteristic equation. You heard the name characteristic equation before when we discussed harmonic motion. While you must not see any resemblance, that characteristic equation and \eqref{eq:cheq} are the same thing, hence the name characteristic equation. For example, if $A=\begin{pmatrix}0 & 1\\-r & -p\end{pmatrix}$, $\det(A-qI)=q^2+pq+r$, so we see that $\eqref{eq:cheq}$ is the same as the characteristic equation $$q^2+pq+r=0$$ of the second-order linear differential equation \eqref{eq:2lde}. Again for the sake of simplicity, the rest of the discussion will be done using a simple example but the same idea applies to the general case. Let us now consider the second-order differential equation $$\frac{d^2}x{dt^2}+5\frac{dx}{dt}+6x=0$$ The matrix $A$ is $A=\begin{pmatrix}0 & 1\\-r & -p\end{pmatrix}=\begin{pmatrix}0 & 1\\-6 & -5\end{pmatrix}$ and $\det(A-qI)=q^2+5q+6=0$ has two distinct real solutions $q=-3, -2$. These are the eigenvalues of $A$. Now we find eigenvectors. For $q_1=-3$, $Av_1=q_1v_1$ with $v_1=\begin{pmatrix}a\\b\end{pmatrix}$ leads to the equation $b=-3a$. So we may choose $v_1=\begin{pmatrix}1\\-3\end{pmatrix}$. Similarly for $q_2=-2$, we find an eigenvector $v_2=\begin{pmatrix}1\\-2\end{pmatrix}$. These eigenvectors can be used to find solutions of $\frac{dX}{dt}=AX$. To see this let $Av=qv$ and $X(t)=f(t)v$. Suppose that $\frac{dX}{dt}=AX$. Then the function $f(t)$ is to be determined. \begin{align*}\frac{df(t)}{dt}v&=A(f(t)v)\\&=f(t)Av\\&=f(t)qv\end{align*} This implies that $$\frac{df(t)}{dt}=f(t)q$$ whose solution is $f(t)=Ae^{qt}$ where $A$ is a constant. So we see that $$X_1(t)=A_1e^{-3t}\begin{pmatrix}1\\-3\end{pmatrix}$$ and $$X_2(t)=A_2e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}$$ are solutions of $\frac{dX}{dt}=AX$. Since the equation is linear, their sum \begin{equation}\begin{aligned}X_1(t)+X_2(t)&=A_1e^{-3t}\begin{pmatrix}1\\-3\end{pmatrix}+A_2e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}\\&=\begin{pmatrix}A_1e^{-3t}+A_2e^{-2t}\\-3A_1e^{-3t}-2A_2e^{-2t}\end{pmatrix}\end{aligned}\label{eq:ldesyssol}\end{equation} is also a solution. It turns out that \eqref{eq:ldesyssol} is the most general solution of $\frac{dX}{dt}=AX$ meaning any solution would be in the form of \eqref{eq:ldesyssol}. To understand why this is the case let us first suppose that $A$ is a diagonal matrix $$A=\begin{pmatrix}q_1 & 0\\0 & q_2\end{pmatrix}$$ with $q_i\ne 0$, $i=1,2$. Let $X(t)=\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}$ be a solution of $\frac{dX}{dt}=AX$. Then $$\frac{dx_1(t)}{dt}=q_1x_1(t),\ \frac{dx_2(t)}{dt}=q_2x_2(t)$$ of which solutions are $$x_1(t)=A_1e^{q_1t},\ x_2(t)=A_2e^{q_2t}$$ Now $X(t)$ can be written as $$X(t)=\begin{pmatrix}A_1e^{q_1t}\\A_2e^{q_2t}\end{pmatrix}=A_1e^{q_1t}\begin{pmatrix}1\\0\end{pmatrix}+A_2e^{q_2t}\begin{pmatrix}0\\1\end{pmatrix}$$ Note that $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ are the eigenvectors of $A$ associated with the eigenvalues $q_1$ and $q_2$, respectively. Conversely, any matrix-valued function $X(t)$ of the form $X(t)=\begin{pmatrix}A_1e^{q_1t}\\A_2e^{q_2t}\end{pmatrix}$ satisfies the differential equation $\frac{dX}{dt}=AX$. Let $V$ be the set of all solutions of $\frac{dX}{dt}=AX$. Then $V$ is a vector space over $\mathbb{R}$. (The verification is straightforward and is left for readers.) The above argument shows that the linearly independent solutions $e^{q_1t}\begin{pmatrix}1\\0\end{pmatrix}$ and $e^{q_2t}\begin{pmatrix}0\\1\end{pmatrix}$ form a basis for $V$. So the dimension of $V$ is 2.

Remark. Let $V$ be the set of all infinitely differentiable functions. For $f,g\in V$ and $c\in\mathbb{R}$, define $f+g$ and $cf$ by \begin{align*}(f+g)(t)&=f(t)+g(t)\\(cf)(t)&=cf(t)\end{align*} Then $V$ forms a vector space over $\mathbb{R}$. So infinitely differentiale functions can be considered as vectors. The derivative $\frac{d}{dt}$ is a map $$\frac{d}{dt}: V\longrightarrow V;\ f\longmapsto \frac{df}{dt}$$ The well-known properties of the derivative: \begin{align*}\frac{d(f+g)}{dt}&=\frac{df}{dt}+\frac{dg}{dt}\\\frac{d(cf)}{dt}&=c\frac{df}{dt}\end{align*} ensure that $\frac{d}{dt}:V\longrightarrow V$ is indeed a linear map.  Let $\lambda\in\mathbb{R}$. Then $f(t)=e^{\lambda t}$ is an eigenvector of $\frac{d}{dt}$ associated with the eigenvalue $\lambda$ because $\frac{de^{\lambda t}}{dt}=\lambda e^{\lambda t}$.

In our case, $A=\begin{pmatrix}0 & 1\\-r & -p\end{pmatrix}$ is not a diagonal matrix so the previous argument does not apply straightforwardly. However if $A$ has two distinct eigenvalues. then it is diagonalizable namely there is an invertible matrix $M$ such that $MAM^{-1}$ is a diagonal matrix. Such a matrix $M$ is called change of basis matrix. Note that $MAM^{-1}$ has exactly the same eigenvalues as those of $A$: \begin{align*}\det(MAM^{-1}-qI)&=\det(MAM^{-1}-M(qI)M^{-1})\\&=\det[M(A-qI)M^{-1}]\\&=\det(M)\det(A-qI)\det(M^{-1})\\&=\det(A-qI)\end{align*} If $v$ is the eigenvector of $A$ associated with the eigenvalue $q$, then $Mv$ is the eigenvector of $MAM^{-1}$ associated with the same eigenvalue $q$. To see this let $Av=qv$. Then \begin{align*}(MAM^{-1})(Mv)&=(MA)v\\&=M(Av)\\&=M(qv)\\&=q(Mv)\end{align*} Let $V$ be the ssolution space of $\frac{dX}{dt}=AX$ and $W$ the solutions apce of $\frac{dY}{dt}=(MAM^{-1})Y$. Since $M$ is invertible, $M: V\longrightarrow W$ is an isomorphism. For example, let $A=\begin{pmatrix}0 & 1\\-6 & -5\end{pmatrix}$ and $M$ change of basis matrix such that $MAM^{-1}=\begin{pmatrix}-3 & 0\\0 & -2\end{pmatrix}$. Set \begin{equation}\begin{aligned}M\begin{pmatrix}1\\-3\end{pmatrix}&=\begin{pmatrix}1\\0\end{pmatrix}\\M\begin{pmatrix}1\\-2\end{pmatrix}&=\begin{pmatrix}0\\1\end{pmatrix}\end{aligned}\label{eq:isobasis}\end{equation} Let $M=\begin{pmatrix}a & b\\c & d\end{pmatrix}$. Then \eqref{eq:isobasis} results in the systems of linear equations $$\left\{\begin{aligned}a-3b&=1\\a-2b&=0\end{aligned}\right.$$ and $$\left\{\begin{aligned}c-3d&=0\\c-2d&=1\end{aligned}\right.$$ whose solutions are $a=-2$, $b=-1$, $c=1$, and $d=3$. That is, $M=\begin{pmatrix}-2 & -1\\3 & 1\end{pmatrix}$ and we have $$MAM^{-1}=\begin{pmatrix}-2 & -1\\3 & 1\end{pmatrix}\begin{pmatrix}0 & 1\\-6 & -5\end{pmatrix}\begin{pmatrix}1 & 1\\-3 & 2\end{pmatrix}=\begin{pmatrix}-3 & 0\\0 & -2\end{pmatrix}$$ as expected. Recall from linear algebra that an isomorphism $M: V\longrightarrow W$ maps a basis of $V$ to a basis of $W$. So $e^{-3t}\begin{pmatrix}1\\-3\end{pmatrix}$ and $e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}$ form basis for $V$. Therefore $X(t)=A_1e^{-3t}\begin{pmatrix}1\\-3\end{pmatrix}+A_2​e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}$ is the general solution of $\frac{dX}{dt}=\begin{pmatrix}0 & 1\\-6 & -5\end{pmatrix}X$ and consequently $x(t)=A_1e^{-3t}+A_2e^{-2t}$ is the general solution of the second-order linear differential equation $\ddot{x}+5\dot{x}+6=0$.