91Ó°ÊÓ

Consider the linear differential equation $\frac{{\mathrm{d}}^3\mathrm{x}}{{\mathrm{dt}}^3}+\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}-\frac{\mathrm{dx}}{\mathrm{dt}}-x=0$.

Assume $D^n=\frac{d^{n}x}{d{t}^n}$, substitute the value $D^3$for $\frac{d^{3}x}{d{t}^3}$, $D^2$for $\frac{d^{2}x}{d{t}^2}$, $D$for $\frac{dx}{dt}$and 1 for x in the equation $\frac{{\mathrm{d}}^3\mathrm{x}}{{\mathrm{dt}}^3}+\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}-\frac{\mathrm{dx}}{\mathrm{dt}}-x=0$as follows.

$$\begin{gathered} \frac{{\mathrm{d}}^3\mathrm{x}}{{\mathrm{dt}}^3}+\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}-\frac{\mathrm{dx}}{\mathrm{dt}}-x=0 \\ {D}^3+D^2-D-1=0 \\ {D}^3-D^2+2D^2-2D+D-1=0 \\ {D}^2\left(D-1\right)+2D\left(D-1\right)+\left(D-1\right)=0 \\ \left(D-1\right)\left(D^2+2D+1\right)=0 \end{gathered}$$

Further, simplify the equation as follows.

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

The solution of the differential equation ${\left(DÂ\pm a\right)}^n=0$is $f\left(t\right)=e^{\sout{T}at}\left\{\underset{i=0}{\overset{n-1}{∑}}{c}_{i+1}{t}^i\right\}$.

If a is the solution of $\left(D+m\right)=0$, b is the solution of $\left(D+n\right)=0$, c is the solution of $\left(D+n\right)=0$then $a+b+c$is solution of$\left(D+m\right)\left(D+n\right)\left(D+o\right)=0$

Using the definition, the solution of the linear differential equation $\left(D-1\right){\left(D+1\right)}^2=0$is defined as follows.

$$\begin{gathered} x\left(t\right)=e^t\left\{\underset{i=0}{\overset{1-1}{∑}}{c}_{i+1}{t}^i\right\}+e^{-t}\left\{\underset{i=0}{\overset{2-1}{∑}}{b}_{i+1}{t}^i\right\} \\ x\left(t\right)=e^t{c}_1+e^{-t}\left\{b_1{t}^0+b_2{t}^1\right\} \\ x\left(t\right)=e^t{c}_1+e^{-t}\left\{b_1+b_{2}t\right\} \end{gathered}$$

As $b_i$and $c_i$are arbitrary constant, assume $b_1=c_2$and $b_2=c_3$.

Substitute the values for and for in the equation as follows.

$$\begin{gathered} x\left(t\right)=e^t{c}_1+e^{-t}\left\{b_1+b_{2}t\right\} \\ x\left(t\right)=e^t{c}_1+e^{-t}\left\{c_2+c_{3}t\right\} \end{gathered}$$

Hence, the solution of the linear differential equation is $\frac{{\mathrm{d}}^3\mathrm{x}}{{\mathrm{dt}}^3}+\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}-\frac{\mathrm{dx}}{\mathrm{dt}}-x=0$is

$x\left(t\right)=e^t{c}_1+e^{-t}\left\{c_1+c_{3}t\right\}$