# General solution to linear second-order homogeneous ODE question

I understand that the solution to a linear second-order homogeneous ODE is in the form $y=c_1y_1+c_2y_2$ where $y_1$ and $y_2$ are solutions to the ODE. I also understand why a linear combination of two unique solutions is also a solution. I do not understand why $y=c_1y_1+c_2y_2$ will give the general solution which will take into account all particular solutions. Could anyone please clear up my misunderstanding? Thank you.

• A hand-wavy explanation: It might help to think of $y''(x)=0$. The general solution only depends on two arbitrary constants, $c_1$ and $c_2$, since you can only integrate the equation twice. If you have two linearly independent solutions $u$ and $v$, then $y= c_1 u + c_2 v$ is a solution for any $c_1$ and $c_2$, so we have used up our two degrees of freedom, and this $y$ must be the general solution. Jun 1, 2017 at 7:21

A linear second-order homogeneous ODE is has the form

$y''+a(x)y'+b(x)y=0$,

where $a$ and $b$ are continuous functions on an intervall $I$.

Let $L$ be the set of all functions in $C^2(I)$, which are solutions of this ODE.

Then $L$ is a vector space and $\dim L=2$.

If $\{y_1,y_2\}$ is a basis of $L$ , then each solution of the ODE has the form $c_1y_1+c_2y_2$.

• So basically the span of {y1,y2} is the set $L$ which has elements that satisfy the second order differential equation? And any $y_1$ and $y_2$ will be an acceptable basis as long as they are unique because their span will be describe the same set of all solutions? Then the initial conditions will lead to one particular element in this set of possible solutions? Jun 1, 2017 at 16:43

This is part of the general theory on first-order linear homogeneous systems of ODEs.

Namely, if you have a system in $\mathbb{R}^n$ of the form $x' = A(t) x$, with $A\colon I \to M_n$ continuous on some interval $I$ (here $M_n$ denotes the vector space of $n\times n$ matrices with real elements), then the set of all maximal solutions is a vector subspace of $C^1(I, \mathbb{R}^n)$ of dimension $n$.

Given this result, you have only to recast your second-order homogeneous linear equation to a first-order linear system in $\mathbb{R}^2$. Indeed, if your equation is of the form $$y'' + a(t) y' + b(t) y = 0$$
you can define $x_1 = y$, $x_2 = y'$, obtaining the system $$\begin{cases} x_1' = x_2, \\ x_2' = - a(t) x_2 - b(t) x_1, \end{cases}$$ i.e. $x' = A x$ with $$A = \begin{pmatrix} 0 & 1\\ -b(t) & -a(t) \end{pmatrix}\,.$$

Let $y_0$ and $y_1$ be solutions with $$(y_0(0),y'_0(0))=(1,0)\text{ and }(y_1(0),y'_1(0))=(0,1).$$ These exist by the existence theorem.

For any other solution $y$ form the function $$u(x)=y(0)y_0(x)+y'(0)y_1(x).$$ Then this function is also a solution as a linear combination of solutions and it has the same initial values as $y$ in $x=0$. By the uniqueness theorem $u=y$ everywhere.

This proves that the dimension of the solution space is $2$, as $(y_0,y_1)$ is a basis.