Answer to Question #297354 in Differential Equations for Nicolas

Question

{F} Find the solution of

(D^3+3D^2-4)y=0

Answer 1

The auxiliary equation is $m^3+3m^2-4=0$

Performing synthetic division,

$\begin{array}{c|rrr} 1&1&3&0&-4\\ & & 1 & 4&4\\ \hline &1&4&4&\mid 0 \end{array}$

We get $m-1$ as a factor. Therefore,

$m^3+3m^2-4= (m-1)(m^2+4m+4)=0\\ \implies m=1~\&~ m^2+4m+4=0\\ i.e., m=1~\&~ m=-2(twice)\\$

The solution is $y = c_1 e^x + (c_2+c_3x)e^{-2x}$