Question

y"+3y'+2y=e^(-t) , y(0)=0, y'(0)=0

Answer 1

$y''+3y+2y=e^{-t},y\left( 0 \right) =0,y'\left( 0 \right) =0\\\lambda ^2+3\lambda +2=0\Rightarrow \lambda \in \left\{ -1,-2 \right\} \\y=C_1e^{-t}+C_2e^{-2t}-the\,\,solution\,\,of\,\,homogeneous\,\,equation\\The\,\,partial\,\,solution\,\,of\,\,the\,\,form\,\,Ate^{-t},\sin ce\,\,-1 is\,\,a\,\,characteristic\,\,number\,\,of\,\,multiplicity\,\,1\\y'=\left( -At+A \right) e^{-t}\\y''=\left( At-2A \right) e^{-t}\\\left( At-2A \right) e^{-t}+3\left( -At+A \right) e^{-t}+2Ate^{-t}=e^{-t}\\Ae^{-t}=e^{-t}\Rightarrow A=1\\Thus\,\,\\y=te^{-t}+C_1e^{-t}+C_2e^{-2t}\\y'=e^{-t}-te^{-t}-C_1e^{-t}-2C_2e^{-2t}\\\left\{ \begin{array}{c} y\left( 0 \right) =0\\ y'\left( 0 \right) =0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} C_1+C_2=0\\ 1-C_1-2C_2=0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} C_1=-1\\ C_2=1\\\end{array} \right. \\y=te^{-t}-e^{-t}+e^{-2t}$