Second order linear homogeneous ODE with constant coefficients In homework I was asked to find all solutions to the following ODE:
$$x''+ax'+bx = 0$$
After reading, I know the following.  
(a)
If $t^2+at+b = (t-\lambda_1)(t-\lambda_2)$ with $\lambda_1\ne\lambda_2$, then
$$x(t) = c_1e^{\lambda_1 t}+c_2e^{\lambda_2 t} $$ 
is the general solution.
(b)
If $t^2+at+b = (t-\lambda)^2$ then $x(t) = (c_1+c_2t)e^{\lambda t}$
is the general solution.

EDIT
I know why and when these are solutions, but this is not my question. My question is how to show if $x(t)$ satisfies this equation, then it must be in one of the two forms. Not that these two forms are solutions.
EDIT
Set $y = (x,x')$, then $y' = F(y) = (y(2),-a\cdot y(2)-b\cdot y(1))$
If I know $F(y)$ is locally Lipschitz then by Picard–Lindelöf theorem solution is unique.
 A: One could quote a more general uniqueness theorem, and even prove it. But we will stick to the particular type of equation. And, in order to use only first-year calculus material, we will assume that $r^2+ra+b=0$ does not have non-real solutions. 
Given any initial conditions $x(0)=p$, $x'(0)=q$, there is a solution of the type you mention. 
Call that solution $x_0$. We want to show that that there is no other solution. Suppose that $x_1$ is a solution of the same initial value problem. Let $y=x_1-x_0$. Then $y''+ay'+by=0$ and $y(0)=y'(0)=0$. 
We want to show that this forces $y=0$. Consider the function $z=ye^{kt}$. Substituting, after some calculation we get 
$$z'' +(2k+a)z' + (k^2+ak+b)z=0.$$
Choose $k$ so that $k^2+ak+b=0$. 
Then we have arrived at an equation of the form 
$$z'' -cz'=0.$$
Put $w=z'$.  We are looking at the equation $w'=cw$, with the initial condition $w(0)=0$. 
Any solution of $w'=cw$ has shape $Ae^{ct}$. The usual way to prove this is to consider the function $f(t)=\frac{w}{e^{ct}}$. Differentiate. We get that $f'(t)$ is identically $0$. So by the Mean Value Theorem, $f(t)$ is a constant. Thus $w=Ae^{ct}$ for some $A$. But since $w(0)=0$, we have $A=0$. 
So $w$ is identically $0$. Thus $z'$ is identically $0$. It follows that $z$ is a constant. Since $z(0)=0$, this constant is $0$. 
A: Inspired by user1709828. Transform into first order ODE.
$$y' = F(y) = (y(2), -a\cdot y(2)-b\cdot y(1))$$
Claim $F$ is Lipschitz.  
Proof:
$|F(x)-F(y)| = \sqrt{|x_2-y_2|^2+(a(x_2-y_2)+b(x_1-y_1))^2} = \sqrt{(a^2+1)|x_2-y_2|^2+b^2|x_1-y_1|^2+2ab\cdot |x_2-y_2|\cdot|x_1-y_1|} \\
\le \sqrt{\max(a^2+1,b^2)\cdot (|x_1-y_1|^2+|x_2-y_2|^2)+|ab|\cdot (|x_1-y_1|^2+|x_2-y_2|^2)}\\
< L \sqrt{|x_1-y_1|^2+|x_2+y_2|^2} = L |x-y|$  
where $L> \sqrt{\max(a^2+1,b^2)+|ab|}$
By Picard-Lindel$\ddot{o}$f theorem $F(y)$ is Lipschitz, given any initial condition $y(t_0) = (x_0,x'_0)$ solution exists and is unique.
For case (1) $\forall (x_0,x'_0)\in \mathbb{C}^2.\;Ae^{\lambda_1 t_0}+ Be^{\lambda_2 t_0} = x_0, \lambda_1Ae^{\lambda_1 t_0}+ \lambda_2Be^{\lambda_2 t_0} = x'_0$ has unique pair $(Ae^{\lambda_1 t_0},Be^{\lambda_2 t_0})$ that satisfy the equation since $(1,1)$ and $(\lambda_1,\lambda_2)$ are linearly independent vectors. Therefore unique pair $(A,B)$ that both satisfy the initial condition and the ODE by above theorem.  
Therefore for any given initial condition $(x_0,x'_0)$ solutions are in the form $x(t) = Ae^{\lambda_1 t}+Be^{\lambda_2 t}$. Similar for case (2).
A: you need  to consider
1.when   determinant  of  characteristic equation (D)  is positive  ,in this case  you have two  real distinct root and colution is
  $c_1*e^{k_1*x}+c_2*e^{k_2*x}$
where     $k_1,k_2$  are roots of quadratic characteristic equation.
2.when  $D=0$
$y(t)=(c_1+t*c_2)*e^{k*x}$
again  $k$  root of  characteristic equation(actual you have  two solution  with $k_1=k_2$)
(algebraic multiplicity)
$D<0$
complex solution  $k_1=p+q*i$
$k_2=p-q*i$
$y(t)=e^{-p*t}*(c_1*cos q*t+c_2*sin q*t)$
