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 is defined as

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

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}-9$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{位}\right)={\mathrm{位}}^2-9$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} {\mathrm{位}}^2-9=0 \\ {\mathrm{位}}^2=9 \\ \mathrm{位}=\sqrt{9} \\ \mathrm{位}=卤3 \end{gathered}$$

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

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

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

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

Thus, the general solution of the differential equation $\mathrm{f}\text{'}\left(\mathrm{t}\right)-9\mathrm{f}\text{'}\left(\mathrm{t}\right)=0\mathrm{is}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{t}}$.

Consider the general solution of the $\mathrm{f}\text{'}"\left(\mathrm{t}\right)-9\mathrm{f}"\left(\mathrm{t}\right)=0$with the initial value problem $f\left(0\right)=0,f\text{'}\left(0\right)=1$is as follows

role="math" localid="1660809514458" $$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{t}} \\ \mathrm{f}\left(\mathrm{t}\right)=0\left({\mathrm{e}}^{3\mathrm{t}}\right)+1\left({\mathrm{e}}^{-3\mathrm{t}}\right) \\ \mathrm{f}\left(\mathrm{t}\right)=0+{\mathrm{e}}^{-3\mathrm{t}} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-3\mathrm{t}} \end{gathered}$$

$=\cos \left(3t\right)$

Hence the solution is $f\left(t\right)=\cos \left(3t\right)$.