91Ó°ÊÓ

Consider the linear differential operator

$\mathrm{T}\left(\mathrm{f}\right)={\mathrm{f}}^{\left(\mathrm{n}\right)}+{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{f}}^{\left(\mathrm{n}-1\right)}+...+{\mathrm{a}}_1\mathrm{f}\text{'}+{\mathrm{a}}_0\mathrm{f}$.

The characteristic polynomial of is defined as

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{λ}\right)={\mathrm{λ}}^{\mathrm{n}}+{\mathrm{a}}_{\mathrm{n}-1}\mathrm{λ}+...+{\mathrm{a}}_1\mathrm{λ}+{\mathrm{a}}_0$.

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}"+3\mathrm{f}$'

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{λ}\right)={\mathrm{λ}}^2+3\mathrm{λ}$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} {\mathrm{λ}}^2+3\mathrm{λ}=0 \\ \mathrm{λ}\left(\mathrm{λ}+3\right)=0 \\ \mathrm{λ}=0,\left(\mathrm{λ}+3\right)=0 \\ \mathrm{λ}=0,\mathrm{λ}=-3 \end{gathered}$$

Therefore, the roots of the characteristic polynomials are and -3 .

Since, the roots of the characteristic equation is different real numbers.

The exponential functions ${\mathrm{e}}^{0\mathrm{t}}$and ${\mathrm{e}}^{-3\mathrm{t}}$ form a basis of the kernel of T.

Hence, they form a basis of the solution space of the homogenous differential equation is T (f) = 0 .

Thus, the general solution of the differential equation $\mathrm{f}"\left(\mathrm{t}\right)+\mathrm{f}\left(\mathrm{t}\right)=0$ is$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1+{\mathrm{C}}_2{\mathrm{e}}^{-3\mathrm{t}}$ .