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

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).