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

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

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\text{'}+\mathrm{f}\text{'}-12$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{位}\right)={\mathrm{位}}^2+\mathrm{位}鈥�12$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} 位=\frac{鈥�b卤\sqrt{b^2鈥�4ac}}{2a} \\ 位=\frac{鈥�1卤\sqrt{{\left(1\right)}^2鈥�4\left(1\right)\left(鈥�12\right)}}{2\left(1\right)}\left\{鈭�a=1,,b=1,c=鈥�12\right\} \\ 位=\frac{鈥�1卤\sqrt{1+48}}{2} \\ 位=\frac{鈥�1卤\sqrt{49}}{2} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} 位=\frac{鈥�1卤7}{2} \\ 位=\frac{鈥�1卤7}{2},位=\frac{鈥�1鈥�7}{2} \\ 位=\frac{6}{2},位=\frac{鈥�8}{2} \\ 位=3,鈥�4 \end{gathered}$$

Therefore, the roots of the characteristic polynomials are 3 and $鈥�4$.

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

The exponential functions ${\mathrm{e}}^{3\mathrm{t}}$and${\mathrm{e}}^{-4\mathrm{t}}$ 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$\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{f}\text{'}\left(\mathrm{t}\right)鈥�12=0$ is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}$.