Question

Solve the differential equation$\mathbf{f}\mathbf{\text{'}\text{'}\text{'}}\left(t\right)\mathbf{-}\mathbf{f}\mathbf{\text{'}\text{'}}\left(t\right)\mathbf{-}\mathbf{4}\mathbf{f}\mathbf{\text{'}}\left(t\right)\mathbf{+}\mathbf{4}\mathbf{f}\left(t\right)\mathbf{=}\mathbf{0}$and find all the real solution of the differential equation.

Short answer

The solution is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{C}}_3{\mathrm{e}}^{-2\mathrm{t}}$.

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 T is defined as

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

Determination of the solution

The characteristic polynomial of the operator as follows.

$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\text{'}\text{'}\text{'}‐\mathrm{f}\text{'}\text{'}‐4\mathrm{f}\text{'}+4\mathrm{f}$

Then the characteristic polynomial is as follows.

role="math" localid="1660798972108" ${\mathrm{P}}_{\mathrm{T}}\left(\mathrm{\lambda }\right)={\mathrm{\lambda }}^3‐{\mathrm{\lambda }}^2‐4\mathrm{\lambda }+4$

Solve the characteristic polynomial and find the roots as follows.

$$\begin{gathered} {\lambda }^3‐{\lambda }^2‐4\lambda +4=0 \\ {\lambda }^2\left(\lambda ‐1\right)‐4\left(\lambda ‐1\right)=0 \\ \left({\lambda }^2‐4\right)\left(\lambda ‐1\right)=0 \\ \left(\lambda +2\right)\left(\lambda ‐2\right)\left(\lambda ‐1\right)=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \lambda ‐1=0 \\ \lambda =1 \\ \lambda +2=0 \\ \lambda =‐2 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \lambda ‐2=0 \\ \lambda =2 \end{gathered}$$

Therefore, the roots of the characteristic polynomials are 1,2 and $‐2$.

Conclusion of the solution

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

The exponential functions ${\mathrm{e}}^{\mathrm{t}},{\mathrm{e}}^{2\mathrm{t}}$and${\mathrm{e}}^{-2\mathrm{t}}$ 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 $\mathrm{f}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)‐\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)‐4\mathrm{f}\text{'}\left(\mathrm{t}\right)+4\mathrm{f}\text{'}\left(\mathrm{t}\right)=0$is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{C}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{C}}_2{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{C}}_3{\mathrm{e}}^{-2\mathrm{t}}$.