Answer to Question #297354 in Differential Equations for Nicolas

Question #297354

{F} Find the solution of




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

1
Expert's answer
2022-02-20T08:06:14-0500

The auxiliary equation is m3+3m24=0m^3+3m^2-4=0

Performing synthetic division,


113041441440\begin{array}{c|rrr} 1&1&3&0&-4\\ & & 1 & 4&4\\ \hline &1&4&4&\mid 0 \end{array}


We get m1m-1 as a factor. Therefore,

m3+3m24=(m1)(m2+4m+4)=0    m=1 & m2+4m+4=0i.e.,m=1 & m=2(twice)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=c1ex+(c2+c3x)e2xy = c_1 e^x + (c_2+c_3x)e^{-2x}



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