Question

Solve the differential equation$\mathit{f}\mathbf{"}\left(t\right)\mathbf{+}\mathbf{2}\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$and find all the real solutions of the differential equation.

Short answer

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

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.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}"-2\mathrm{f}\text{'}+\mathrm{f}$

Then the characteristic polynomial is as follows.

${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^2-2\mathrm{\lambda }+1$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} {\mathrm{\lambda }}^2+2\mathrm{\lambda }+1=0 \\ {\mathrm{\lambda }}^2+\mathrm{\lambda }+\mathrm{\lambda }+1=0 \\ \mathrm{\lambda }\left(\mathrm{\lambda }+1\right)+1\left(\mathrm{\lambda }+1\right)=0 \\ \left(\mathrm{\lambda }+1\right)\left(\mathrm{\lambda }+1\right)=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{\lambda }+1=0 \\ \mathrm{\lambda }=-1 \\ \mathrm{\lambda }+1=0 \\ \mathrm{\lambda }=-1 \end{gathered}$$

Therefore, the roots of the characteristic polynomials are -1 and -1 .

Conclusion of the solution

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

The exponential functions$e^{-t}$ and${\mathrm{te}}^{-\mathrm{t}}$ 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$f"\left(t\right)+2f\text{'}\left(t\right)=0$ is$\mathrm{f}\left(\mathrm{t}\right)+{\mathrm{e}}^{-\mathrm{t}}\left({\mathrm{C}}_1+{\mathrm{C}}_2\mathrm{t}\right)$ .