Answer to Question #319376 in Differential Equations for Sharlene

Question

Obtain the solution of the differential equation

y"+3y'+2y=e^(2t) using the method of variation of parameters

Answer 1

$y''+3y'+2y=e^{2t}\\Homogeneous\,\,equation:\\y''+3y'+2y=0\\\lambda ^2+3\lambda +2=0\\\lambda _1=-2,\lambda _2=-1\\y=C_1e^{-2t}+C_2e^{-t}\\The\,\,fundamental\,\,system\,\,of\,\,solutions:\\y_1=e^{-t},y_2=e^{-2t}\\y\left( t \right) =C_1\left( t \right) y_1+C_2\left( t \right) y_2\\By\,\,the\,\,method:\\\left\{ \begin{array}{c} y_1\left( t \right) C_1'\left( t \right) +y_2\left( t \right) C_2'\left( t \right) =0\\ y_1'\left( t \right) C_1'\left( t \right) +y_2'\left( t \right) C_2'\left( t \right) =e^{2t}\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} e^{-t}C_1'\left( t \right) +e^{-2t}C_2'\left( t \right) =0\\ -e^{-t}C_1'\left( t \right) -2e^{-2t}C_2'\left( t \right) =e^{2t}\\\end{array} \right. \Rightarrow \\\Rightarrow \left\{ \begin{array}{c} C_2'\left( t \right) =-e^tC_1'\left( t \right)\\ e^{-t}C_1'\left( t \right) =e^{2t}\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} C_2'\left( t \right) =-e^tC_1'\left( t \right)\\ C_1'\left( t \right) =e^{3t}\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} C_2'\left( t \right) =-e^{4t}\\ C_1'\left( t \right) =e^{3t}\\\end{array} \right. \\\Rightarrow \left\{ \begin{array}{c} C_1\left( t \right) =\frac{1}{3}e^{3t}+C_1\\ C_2\left( t \right) =-\frac{1}{4}e^{4t}+C_2\\\end{array} \right. \Rightarrow y\left( t \right) =\left( \frac{1}{3}e^{3t}+C_1 \right) e^{-t}+\left( -\frac{1}{4}e^{4t}+C_2 \right) e^{-2t}=\\=\frac{1}{12}e^{2t}+C_1e^{-t}+C_2e^{-2t}$