General solution to linear second-order homogeneous ODE question

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.

Answer 1

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 $c_1y_1+c_2y_2$.

Answer 2

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 $$ A = \begin{pmatrix} 0 & 1\\ -b(t) & -a(t) \end{pmatrix}\,. $$

Answer 3

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.