Why does a linear homogeneous ODE have only a solution of summed exponentials?

Question

There is a linear homogeneous ODE (let's pick a second order one, but it can be in any order):

\begin{align*} af'' + bf' + cf = 0 \end{align*}

We know, that

\begin{align*} f(t)=e^{\lambda t} \end{align*}

is a solution, and we need to find two $\lambda$'s, so the general solution is (if $\lambda$'s are real and distinct):

\begin{align*} f(t)=c_1e^{\lambda _1 t} + c_2e^{\lambda _2 t} \end{align*}

My question is, why do the only solution is in the form of summed exponentials? What is the proof, that there is no other solution in some other form, a non-exponential one?

(I understand, that if $f_1$ and $f_2$ are solutions, then $c_1f_1+c_2f_2$ is a solution too, but I don't understand, why $f_*$ have to be exponential)

Answer 1

There is a linear homogeneous ODE (let's pick a second order one, but it can be in any order):

Since you only took the second order case as an example, I'll elaborate on the more general case.

• You can show that the solutions to an $n$-th order linear ODE form a vector space with dimension at most $n$; such a space is spanned by (at most) $n$ linearly independent functions.
• Plugging $e^{\lambda t}$ into an $n$-th order, linear, homogeneous ODE with constant coefficients will result in an $n$-th degree polynomial which has exactly $n$ (possibly complex) solutions, if we take the multiplicities into account.
• If the $r$ distinct roots are $\lambda_1 , \ldots , \lambda_r$ with respective multiplicities $m_1,\ldots,m_r$ (and thus we have $m_1+\ldots+m_r=n$), then you can show that for all $i$ with $1 \le i \le r$, the functions $e^{\lambda_i},te^{\lambda_i},\ldots,t^{m_i}e^{\lambda_i}$ are solutions to the ODE; there are in total $n$ such functions.
• The $n$ functions from above are linearly independent and thus span an $n$-dimensional vector space so this contains all the solutions to the ODE; in other words: any solution will be a linear combination of these exponential functions above.

This argument is a bit indirect in the sense that it doesn't provide a direct intuition as to why the solutions have to be exponential, but it does show that there cannot be any other: all the solutions are in this vector space which is spanned by the "exponentials" (including those of the form $t^ke^{\lambda t}$ which technically aren't exponentials).

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See for example here (Theorem 8.3) or here (Theorem 4.1).

Answer 2

I don't if this answers your question. Let me impose the following IVP $$ af''+bf'+cf=0,f(0)=f_0,f'(0)=f_1. \tag{1}$$ Let $y(t)=c_1e^{\lambda _1 t} + c_2e^{\lambda _2 t}$ and then there are $c_1,c_2$ such that $y(0)=f_0,y'(0)=f_1$. Clearly $y(t)$ satisfies the following equation $$ ay''+by'+cy=0,y(0)=f_0,y'(0)=f_1 \tag{2}. $$ Let $X(t)=f(t)-y(t)$. Then $X(t)$ is the solution of the following IVP $$ aX''+bX'+cX=0,X(0)=0,X'(0)=0 \tag{3}. $$ By the existence and uniqueness theorem, (3) has the solution $X(t)\equiv0$. Namely $$ f(t)=y(t)=c_1e^{\lambda _1 t} + c_2e^{\lambda _2 t}. $$