91影视

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 T is defined as

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

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\text{'}\text{'}鈥�\mathrm{f}\text{'}\text{'}鈥�4\mathrm{f}\text{'}+4\mathrm{f}$

Then the characteristic polynomial is as follows.

role="math" localid="1660798972108" ${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{位}\right)={\mathrm{位}}^3鈥�{\mathrm{位}}^2鈥�4\mathrm{位}+4$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} {位}^3鈥�{位}^2鈥�4位+4=0 \\ {位}^2\left(位鈥�1\right)鈥�4\left(位鈥�1\right)=0 \\ \left({位}^2鈥�4\right)\left(位鈥�1\right)=0 \\ \left(位+2\right)\left(位鈥�2\right)\left(位鈥�1\right)=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} 位鈥�1=0 \\ 位=1 \\ 位+2=0 \\ 位=鈥�2 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} 位鈥�2=0 \\ 位=2 \end{gathered}$$

Therefore, the roots of the characteristic polynomials are 1,2 and $鈥�2$.

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

The exponential functions ${\mathrm{e}}^{\mathrm{t}},{\mathrm{e}}^{2\mathrm{t}}$and${\mathrm{e}}^{-2\mathrm{t}}$ form a basis of the kernel of T .

Hence, they form a basis of the solution space of the homogenous differential equation is $\mathrm{T}\left(\mathrm{f}\right)=0$.

Thus, the general solution of the differential equation $\mathrm{f}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)鈥�\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)鈥�4\mathrm{f}\text{'}\left(\mathrm{t}\right)+4\mathrm{f}\text{'}\left(\mathrm{t}\right)=0$is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{C}}_3{\mathrm{e}}^{-2\mathrm{t}}$.