Question

Solve the differential equation$\mathit{f}\mathbf{"}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$and find all the real solutions of the differential equation.

Short answer

The solution is$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1+{\mathrm{C}}_2{\mathrm{e}}^{-3\mathrm{t}}$ .

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.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}"+3\mathrm{f}$'

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^2+3\mathrm{\lambda }$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} {\mathrm{\lambda }}^2+3\mathrm{\lambda }=0 \\ \mathrm{\lambda }\left(\mathrm{\lambda }+3\right)=0 \\ \mathrm{\lambda }=0,\left(\mathrm{\lambda }+3\right)=0 \\ \mathrm{\lambda }=0,\mathrm{\lambda }=-3 \end{gathered}$$

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

Conclusion of the solution

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

The exponential functions ${\mathrm{e}}^{0\mathrm{t}}$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 T (f) = 0 .

Thus, the general solution of the differential equation $\mathrm{f}"\left(\mathrm{t}\right)+\mathrm{f}\left(\mathrm{t}\right)=0$ is$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1+{\mathrm{C}}_2{\mathrm{e}}^{-3\mathrm{t}}$ .