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 T 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.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}"-4\mathrm{f}\text{'}+13\mathrm{f}$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{λ}\right)={\mathrm{λ}}^2-4\mathrm{λ}+13$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} \mathrm{λ}=\frac{-\mathrm{b}Â\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ \mathrm{λ}=\frac{4Â\pm \sqrt{{\left(4\right)}^2-4\left(1\right)\left(13\right)}}{2\left(1\right)}\left\{âˆ\mu \mathrm{a}=1,,\mathrm{b}=-4,\mathrm{c}=13\right\} \\ \mathrm{λ}=\frac{4Â\pm \sqrt{16-52}}{2} \\ \mathrm{λ}=\frac{4Â\pm \sqrt{-36}}{2} \end{gathered}$$

Simplify further as follows.

$\begin{array}{l}\mathrm{λ}=\frac{4Â\pm 6\mathrm{i}}{2}\\ \mathrm{λ}=\frac{4+6\mathrm{i}}{2},\mathrm{λ}=\frac{4-6\mathrm{i}}{2}\\ \mathrm{λ}=2+3\mathrm{i},2-3\mathrm{i}\end{array}$

Therefore, the roots of the characteristic polynomials are 2+3i and 2-3i .

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

The exponential functions${\mathrm{e}}^{\left(2+3\mathrm{i}\right)\mathrm{t}}$and${\mathrm{e}}^{\left(2-3\mathrm{i}\right)\mathrm{t}}$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 .

The exponential function${\mathrm{e}}^{\left(2+3\mathrm{i}\right)\mathrm{t}}$and${\mathrm{e}}^{\left(2-3\mathrm{i}\right)\mathrm{t}}$can be written as follows.

$$\begin{gathered} {\mathrm{e}}^{\left(2+3\mathrm{i}\right)\mathrm{t}}={\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{c}}_1\cos 3\mathrm{t}+{\mathrm{c}}_2\sin 3\mathrm{t}\right) \\ {\mathrm{e}}^{\left(2-3\mathrm{i}\right)\mathrm{t}}={\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{c}}_3\cos \left(-3\mathrm{t}\right)+{\mathrm{c}}_4\sin \left(-3\mathrm{t}\right)\right) \\ {\mathrm{e}}^{\left(2-3\mathrm{i}\right)\mathrm{t}}={\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{c}}_3\cos 3\mathrm{t}-{\mathrm{c}}_4\sin 3\mathrm{t}\right) \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\left(2+3\mathrm{i}\right)\mathrm{t}}+{\mathrm{e}}^{\left(2-3\mathrm{i}\right)\mathrm{t}} \\ ={\mathrm{c}}_1{\mathrm{e}}^{2\mathrm{t}}\cos 3\mathrm{t}+{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{t}}\sin 3\mathrm{t}+{\mathrm{c}}_3{\mathrm{e}}^{2\mathrm{t}}\cos 3\mathrm{t}-{\mathrm{c}}_4{\mathrm{e}}^{2\mathrm{t}}\sin 3\mathrm{t} \\ =\left({\mathrm{c}}_1+{\mathrm{c}}_3\right){\mathrm{e}}^{2\mathrm{t}}\cos 3\mathrm{t}-\left({\mathrm{c}}_{2-}{\mathrm{c}}_4\right){\mathrm{e}}^{2\mathrm{t}}\sin 3\mathrm{t} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{C}}_1\cos 3\mathrm{t}+{\mathrm{C}}_2\sin 3\mathrm{t}\right)\left\{âˆ\mu {\mathrm{C}}_1={\mathrm{c}}_1+{\mathrm{c}}_3,{\mathrm{C}}_2={\mathrm{c}}_2-{\mathrm{c}}_4\right\} \end{gathered}$$

Thus, the general solution of the differential equation$\mathrm{f}"\left(\mathrm{t}\right)-4\mathrm{f}\text{'}\left(\mathrm{t}\right)+13\mathrm{f}=0$is $\mathrm{f}\left(\mathrm{t}\right)-{\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{C}}_1\cos 3\mathrm{t}+{\mathrm{C}}_2\sin 3\mathrm{t}\right)$.