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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.

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  • $\begingroup$ actually @mezhang as i know,solution of such kind of differental equation is intuitely,i was told like this by my professors,that why you may think why exponential solution is because of this.let wait from others what they will say $\endgroup$ – dato datuashvili Mar 29 '13 at 20:56
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    $\begingroup$ @dato I'm thoroughly convinced with the intuition and they are solutions. But I'm not convinced that no other solutions can exist. I am told they are the only solutions, but that does not solve my problem. I want a proof that no other solutions can exist. Thank you for your help though. $\endgroup$ – mez Mar 29 '13 at 21:02
  • $\begingroup$ because such kind of equation is depend on characteristic equation,then it could not be happen that this has equation has more then 3 case,but if we imagine except discriminant method,there is another one,then it will be discovering new method for such equation $\endgroup$ – dato datuashvili Mar 29 '13 at 21:04
  • $\begingroup$ With suitable substitution, you should be able to convert the $2^{nd}$ order ODE to first. Picard lindeloff theorem guarantees the uniqueness, conditionally. $\endgroup$ – user45099 Mar 29 '13 at 21:22
  • $\begingroup$ @user1709828 yes this is the answer that I was waiting for. Thank you. $\endgroup$ – mez Mar 29 '13 at 21:29
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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 $0$.

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  • $\begingroup$ Please check if what I wrote below is good. $\endgroup$ – mez Mar 29 '13 at 22:07
  • $\begingroup$ umm, if we don't have initial values, but asking about general form of solutions, how do I transform my argument such that I show "uniqueness" of the general form? $\endgroup$ – mez Mar 29 '13 at 22:14
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    $\begingroup$ @mezhang: The argument looks correct. Whether it is viewed as good for a homework solution depends on what background theorems you have by this time. As in my solution, we can always specify generic initial values. $\endgroup$ – André Nicolas Mar 29 '13 at 22:15
  • $\begingroup$ this is for analysis and we have proved Picard–Lindelöf theorem using Banach fixed point theorem. So I think it is adequate. $\endgroup$ – mez Mar 29 '13 at 22:16
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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).

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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 $k_1=p+q*i$

$k_2=p-q*i$

$y(t)=e^{-p*t}*(c_1*cos q*t+c_2*sin q*t)$

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    $\begingroup$ you tell me what I know. I want to prove that all solutions are in these form, not that these forms are solutions. $\endgroup$ – mez Mar 29 '13 at 20:52
  • $\begingroup$ see my comment above please $\endgroup$ – dato datuashvili Mar 29 '13 at 20:59

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