91Ó°ÊÓ

Consider the linear differential operator

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

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.

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

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{λ}\right)={\mathrm{λ}}^2+\mathrm{λ}-12$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} \mathrm{λ}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{λ}=\frac{-1Â\pm \sqrt{{\left(1\right)}^2-4\left(1\right)(-12)}}{2\left(1\right)}\left\{âˆ\mu a=1,,b=1,c=-12\right\} \\ \mathrm{λ}=\frac{-1Â\pm \sqrt{1-(-48)}}{2} \\ \mathrm{λ}=\frac{-1Â\pm \sqrt{1+48}}{2} \end{gathered}$$

Simplify further,

$$\begin{gathered} \mathrm{λ}=\frac{-1Â\pm \sqrt{49}}{2} \\ \mathrm{λ}=\frac{-1Â\pm 7}{2} \\ \mathrm{λ}=\frac{6}{2},\frac{-8}{2} \\ \mathrm{λ}=3,-4 \end{gathered}$$

Therefore, the roots of the characteristic polynomials are 3 and -4.

Since, the roots of the characteristic equation are 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 role="math" localid="1660799899986" $\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{f}\text{'}\left(\mathrm{t}\right)-12\mathrm{f}\left(\mathrm{t}\right)=0$is. $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}$.

Consider the general solution of the $\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{f}\text{'}\left(\mathrm{t}\right)-12\mathrm{f}\left(\mathrm{t}\right)=0$with the initial value problem$f\left(0\right)=f\text{'}\left(0\right)=0$ is as follows

$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}$

Hence the solution is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}$.