Answer to Question #154062 in Differential Equations for Anthony Masirag

Question

( D^3 + D^2 - 4D - 4) y = 3e^-4x -6

Answer 1

$( D^3 + D^2 - 4D - 4) y = 3e^{-4x -6}$

Auxiliary equation is: $m^3+m^2-4m-4=0\\$

$\Rightarrow (m+2)(m-2)(m+1)=0\\ \Rightarrow m=2,-2,-1\\\;\\ \therefore\text{C.F.}=c_1e^{2x}+c_2e^{-2x}+c_3e^{-x}\\\;\\ \text{P.I.}=\dfrac{1}{f(D)}f(x)=\dfrac{1}{f(D)}e^{-4x}(3e^{-6})\\\;\\ =e^{-4x}\times\dfrac{1}{f(D-4)}\times(3e^{-6})\\\;\\ =\dfrac{3e^{-4x-6}}{(D-2)(D-6)(D-3)}\\\;\\ =3e^{-4x-6}\left[\dfrac{1}{12(D-6)}-\dfrac{1}{3(D-3)}+\dfrac{1}{4(D-2)}\right]\\\;\\ =3e^{-4x-6}\left[\dfrac{e^{6x}}{12}\int e^{-6x}dx-\dfrac{e^{3x}}{3}\int e^{-3x}dx+\dfrac{e^{2x}}{4}\int e^{-2x}dx\right]\\\;\\ =3e^{-4x-6}\left[-\dfrac{1}{72}+\dfrac{1}{9}-\dfrac{1}{8}\right]\\\;\\ =-\dfrac{e^{-4x-6}}{12}$

$\therefore \text{The complete solution is} :-\\ y=\text{C.F.+P.I}.=c_1e^{2x}+c_2e^{-2x}+c_3e^{-x}-\dfrac{e^{-4x-6}}{12}$

So,

$$\boxed{\pmb{y=c_1e^{2x}+c_2e^{-2x}+c_3e^{-x}-\dfrac{1}{12}e^{-4x-6}}}$$