Question

Solve the initial value problem in$\mathbf{f}\mathbf{\text{'}\text{'}}\left(t\right)\mathbf{‐}\mathbf{9}\mathbf{f}\left(t\right)\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{f}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{f}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

Short answer

The solution is $f\left(t\right)=\cos \left(3\mathrm{t}\right)$.

Step-by-step solution

Definition of characteristic polynomial

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

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

Determination of the solution

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.

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

Solve the characteristic polynomial and find the roots as follows.

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

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

Explanation of the solution

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 $\mathrm{f}\text{'}\left(\mathrm{t}\right)-9\mathrm{f}\text{'}\left(\mathrm{t}\right)=0\mathrm{is}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{t}}$.

Explanation of the solution with initial value problem

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

Simplify further.

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

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