Question

Annihilator method, question

Answer 1

\(\color{#0000aa}{\displaystyle y''+2y'-3=xe^x }\)\({\tiny \\[0.5em]}\)
\(\color{#0000aa}{\displaystyle (D^2+2D-3)[y]=xe^x }\)
\({\tiny \\[0.8em]}\)
\(\color{#000000}{\displaystyle (D-1)^2=D^2-2D+1=(2e^x+xe^x)-2(e^x+xe^x)+xe^x=0 }\)
\({\tiny \\[0.8em]}\)
\(\color{#0000aa}{\displaystyle (D-1)(D-1)(D+1)(D-3)[y]=0 }\) \({\tiny \\[0.5em]}\)
\(\color{#0000aa}{\displaystyle y=c_1e^x+c_2xe^x+c_3e^{-x}+c_4e^{3x} }\)

Answer 2

(I am just checking my work)

Answer 3

Then, I plug in the solutions that resulted from annihilating the right side of the equation, that is, I plug in the particular-solution-part: \(\color{#0000aa}{\displaystyle y_p=c_1e^x+c_2xe^x }\) into the original equation.

Answer 4

\(\color{red}{\bf ATTENTION}\)
Sorry for this mistake, my equation, is \(\color{red}{y''+2y'-3y=xe^x}\).