Question

Find the particular solution for the following equation y"-y'-12y=te^4t

Answer 1

$y''-y'-12y=te^{4t}\\\lambda ^2-\lambda -12=0\Rightarrow \lambda \in \left\{ 4,-3 \right\} \\Thus\,\,the\,\,form\,\,of\,\,particular\,\,solution\,\,is\,\,tP\left( t \right) e^{4t},P\,\,is\,\,a\,\,polynomial\,\,of\,\,power\,\,1\\y=t\left( A+Bt \right) e^{4t}=\left( At+Bt^2 \right) e^{4t}\\y'=\left( A+\left( 2B+4A \right) t+4Bt^2 \right) e^{4t}\\y''=\left( 8A+2B+\left( 16A+16B \right) t+16Bt^2 \right) e^{4t}\\\left( 8A+2B+\left( 16A+16B \right) t+16Bt^2-A-\left( 2B+4A \right) t-4Bt^2-12At-12Bt^2 \right) e^{4t}=te^{4t}\\\left\{ \begin{array}{c} 7A+2B=0\\ 14B=1\\ 0B=0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} A=-\frac{1}{49}\\ B=\frac{1}{14}\\\end{array} \right. \\y=\left( -\frac{1}{49}t+\frac{1}{14}t^2 \right) e^{4t}\\$