Intuition behind general solution of linear differential equation? If $y_1$, $y_2$ and so on till $y_n$ are solutions of a linear differential equation of order $n$, it is a theorem that $c_1y_1+ c_2y_2+\ldots +c_ny_n$ is the general solution for it. I have seen this in many places but I don't get how or why this is true. Would someone please give me some intuition in this result?
The obvious thing is that a linear combination of solutions is a solution. This is pretty much what it means for the equation to be linear and homogeneous: by assumption $L \left ( \sum_i c_i y_i \right ) = \sum_i c_i L(y_i)$ and the equation is $Ly=0$ so if each $L(y_i)=0$ then $L(y)=0$. Note that a linear inhomogeneous equation doesn't have this property, but it can be addressed in a similar way by summing up the general solution to the homogeneous problem with any particular solution to the inhomogeneous problem.

The less obvious thing is that the solution space has dimension $n$, so that if your $y_k$ are linearly independent then you have the general solution. To me the intuition for this is to parametrize the solutions by initial conditions at some arbitrary point; if you assume uniqueness of solutions to IVPs then it is apparent that the space of initial conditions has dimension $n$ so the same is true of the space of solutions. The problem then reduces to showing that a linear ODE of order $n$ with sufficiently nice coefficients and forcing has unique solutions to its IVPs.
A homogeneous linear differential equation of order $n$ looks like:
$$
\sum_{k = 0}^n c_ky^{(k)} = 0
$$
where $y^{(k)}$ is the $k$-th derivative of $y$ and $y^{(0)} = y$ by convention.
Observe that we may recursively define $u_0 = y$ and $u_k = u_{k - 1}'$ so that $u_k = y^{(k)}$. With this definition, the original equation becomes
$$
\sum_{k = 0}^n c_k u_k = 0
$$
Rearranging, we have
$$
u_{n} = -\sum_{k = 0}^{n-1}  \frac{c_k}{c_n} u_k
$$
Note that $c_n \neq 0$ by assumption; otherwise, the original equation would have had at most order $n - 1$. Defining $d_k = -\frac{c_k}{c_n}$ we can further rewrite the above as
$$
u_{n - 1}' = \sum_{k = 0}^{n-1} d_k u_k
$$
Meanwhile, we have the $n - 1$ other equations
$$
u_{k - 1}' = u_k
$$
with $1 \leq k \leq n - 1$. This system of equations can be transformed to the matrix equation
$$
\begin{bmatrix}
u_0' \\
\vdots\\
u_{k - 1}'\\
\vdots\\
u_{n - 1}'
\end{bmatrix}
=
\begin{bmatrix}
0 &amp; 1 &amp; 0 &amp; \dots &amp; 0 &amp; 0 &amp; 0 &amp; \dots&amp;0\\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\
0 &amp; 0 &amp; 0 &amp; \dots &amp; 0 &amp; 1 &amp; 0 &amp; \dots &amp; 0\\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\
d_0 &amp; d_1 &amp; d_2 &amp; \dots &amp; d_{k - 1} &amp; d_k &amp; d_{k + 1} &amp; \dots &amp; d_{n - 1}
\end{bmatrix}
\begin{bmatrix}
u_0 \\
\vdots \\
u_k \\
\vdots \\
u_{n - 1}
\end{bmatrix}
$$
Define $\bf{u} = \begin{bmatrix} u_0 &amp; \dots &amp; u_{n - 1} \end{bmatrix}^T$ and $\bf{u}' = \begin{bmatrix}u_0' &amp; \ldots &amp; u_{n - 1}' \end{bmatrix}^T$. Let $A$ be the $n \times n$ matrix above, so that
$$
\bf{u}' = A\bf{u}
$$
Now $\bf{u}$ is a solution of the above equation with $u_0 = y$ if and only if $y$ is a solution of our original differential equation.
We will let $t$ be the variable with respect to which we have been differentiating. Thus, more explicitly, we have
$$
\bf{u}'(t) = A\bf{u}(t)
$$
Define
$$
\bf{v}(t) = e^{tA}\bf{v}(0)
$$
(See https://en.wikipedia.org/wiki/Matrix_exponential if the matrix exponential is unfamiliar). One can check that (the proof is identical to the proof that $\frac{d}{dt} e^{tx} = xe^{tx}$):
$$
\bf{v}'(t) = Ae^{tA} \bf{v}(0) = A\bf{v}(t)
$$
So $\bf{v}(t)$ as we have defined it is a solution to the matrix equation with initial condition $\bf{v}(0)$. Suppose $\bf{u}(t)$ is another solution to the matrix equation with $\bf{u}(0) = \bf{v}(0)$. Put $\bf{a}(t) = e^{-tA} \bf{u}(t)$. Then
\begin{align}
\bf{a}'(t) &amp;= -Ae^{-tA} \bf{u}(t) + e^{-tA}\bf{u}'(t)
\\&amp;= -Ae^{-tA} \bf{u}(t) + e^{-tA}A\bf{u}(t)
\\&amp;= e^{-tA}(-A + A) \bf{u}(t)
\\&amp;= 0
\end{align}
So it must be the case that $\bf{a}(t)$ is constant. But $\bf{a}(0) = e^{-0A}\bf{u}(0) = \bf{u}(0) = \bf{v}(0)$. Thus, $e^{-tA} \bf{u}(t) = \bf{a}(t) = \bf{v}(0)$, which shows $\bf{u}(t) = e^{tA} \bf{v}(0) = \bf{v}(t)$. Thus, $\bf{v}(t)$ is the unique solution to the matrix equation with initial condition $\bf{v}(0)$.
Now, if $y$ solves the original differential equation, then $\bf{u}$ with $u_0 = y$ solves the matrix equation. That means, by the above remarks, that $\bf{u}(t) = e^{tA} \bf{u}(0)$. Let $y_1(t) , \ldots , y_n(t)$ be the entries across the first row of $e^{tA}$. Then, reading off from $\bf{u}(t) = e^{tA}\bf{u}(0)$, we find
$$
y(t) = u_0(t) = \sum_{k = 0}^{n - 1} u_k(0) y_{k + 1}(t) = \sum_{k = 0}^{n - 1} y^{(k)}(0) y_{k + 1}(t)
$$
From this, we see that every solution to the differential equation is in the span of $y_1, \ldots, y_n$, with the coefficients determined by the initial conditions of $y$.
But it is worth noting that the $y_1, \ldots, y_n$ were not arbitrary -- they were just entries of $e^{tA}$. If we are handed an arbitrary set of solutions $f_1, \ldots, f_n$ how can we be sure that they span? Since the differential equation is linear (by definition) its solutions form a vector space. Because we showed that every solution is in the span of $n$ particular solutions, the dimension of the vector space can be no greater than $n$. Thus, any set of $n$ linearly independent (in the sense that no linear combination of the functions over $\mathbb{R}$ is identically $0$) must also span the solution space. This is just a restatement of the fact that in an $n$ dimensional vector space, a set of $n$ linearly independent vectors must span the space.
Concluding Remarks
This has been a long post, so let me summarize. We began with a homogeneous linear differential equation of order $n$. We transformed it to a matrix equation. We showed that the matrix equation has unique solutions for every choice of initial conditions. Using this solution of the matrix equation, we showed that any solution to the original differential equation had to live in the span of $n$ (not necessarily distinct!) solutions.
It all comes down to the fact that we have existence and uniqueness theorems for solutions of homogeneous linear differential equations and that the initial condition of a solution is specified by $n$ constants.