是否存在这样一个初等函数：它的三阶导数是其本身，而一、二阶导数不是其本身？ 第1页

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If there is a function that satisfies , it must be a solution of the differential equation , which is a third-order autonomous ordinary differential equation.

Its characteristic equation has a real root and two complex conjugate roots, they are , and .

The real root suggests that is a linearly independent solution; the complex conjugate roots suggest that and are two linearly independent solutions of the differential equation.

Therefore, the general solution of the differential equation is .

Using Euler's identity, we can rewrite the last two terms as and .

Regrouping and simplification of the constants gives the final result,

.

Due to , there must be .

Some of my other answers on differential equations: