91影视

The given differential equation is:

\begin{array}{l} {{\bf{f}}^{{\bf{'''}}}}\left( {\bf{t}} \right){\bf{ - 3}}{{\bf{f}}^{{\bf{''}}}}\left( {\bf{t}} \right){\bf{ + 3}}{{\bf{f}}^{\bf{'}}}\left( {\bf{t}} \right){\bf{ = 0}}\end{array}

The characteristic equation of the given differential equation is

\begin{array}{l}{{\bf{\lambda }}^{\bf{3}}}{\bf{ - 3}}{{\bf{\lambda }}^{\bf{2}}}{\bf{ + 2\lambda = 0}}\end{array}

Now simplify the above differential equation as follows:

\[\begin{array}{l}\lambda \left( {{\lambda ^2} - 3\lambda + 2} \right) = 0\\\lambda \left( {{\lambda ^2} - \lambda - 2\lambda + 2} \right) = 0\\\lambda \left( {\lambda \left( {\lambda - 1} \right) - 2\left( {\lambda - 1} \right)} \right) = 0\end{array}\]

Further, simplify as follows:

\begin{array}{l}\lambda \left( {\left( {\lambda - 1} \right)\left( {\lambda - 2} \right)} \right) = 0\\\lambda \left( {\lambda - 1} \right)\left( {\lambda - 2} \right) = 0\\\lambda = 0{\rm{ or }}\lambda - 1 = 0{\rm{ or }}\lambda - 2 = 0\\\lambda = 0{\rm{ or }}\lambda = 1{\rm{ or }}\lambda = 2\end{array}

Thus, the characteristic values are 0,1,2

The solution of the given differential equation using the characteristic values is

Thus, the solution of the differential equation

\begin{array}{l} {{\bf{f}}^{{\bf{'''}}}}\left( {\bf{t}} \right){\bf{ - 3}}{{\bf{f}}^{{\bf{''}}}}\left( {\bf{t}} \right){\bf{ + 3}}{{\bf{f}}^{\bf{'}}}\left( {\bf{t}} \right){\bf{ = 0}}\end{array}

is

\[\begin{array} f\left( t \right) = {c_1} + {c_2}{e^t} + {c_3}{e^{2t}}\end{array}