Answer to Question #315649 in Differential Equations for Sheshank

Question

2. Solve by variation of parameters:

(D ^ 2 - 2D + 1) * y = x ^ (3/2) * e ^ x

Answer 1

$Homogeneous\,\,equation:\\\left( D^2-2D+1 \right) y=0\\\lambda ^2-2\lambda +1=0\Rightarrow \lambda _{1,2}=1\Rightarrow \\\Rightarrow y=C_1e^x+C_2xe^x\\u_1\left( x \right) =e^x,u_2\left( x \right) =xe^x\\y\left( x \right) =A\left( x \right) u_1\left( x \right) +B\left( x \right) u_2\left( x \right) \\\left[ \begin{matrix} u_1\left( x \right)& u_2\left( x \right)\\ u_1'\left( x \right)& u_2'\left( x \right)\\\end{matrix} \right] \left[ \begin{array}{c} A'\left( x \right)\\ B'\left( x \right)\\\end{array} \right] =\left[ \begin{array}{c} 0\\ f\left( x \right)\\\end{array} \right] \\\left[ \begin{matrix} e^x& xe^x\\ e^x& e^x+xe^x\\\end{matrix} \right] \left[ \begin{array}{c} A'\left( x \right)\\ B'\left( x \right)\\\end{array} \right] =\left[ \begin{array}{c} 0\\ x^{3/2}e^x\\\end{array} \right] \Rightarrow \\\Rightarrow \left[ \begin{array}{c} A'\left( x \right)\\ B'\left( x \right)\\\end{array} \right] =\left[ \begin{matrix} e^x& xe^x\\ e^x& e^x+xe^x\\\end{matrix} \right] ^{-1}\left[ \begin{array}{c} 0\\ x^{3/2}e^x\\\end{array} \right] =\\=e^{-2x}\left[ \begin{matrix} e^x\left( 1+x \right)& -xe^x\\ -e^x& e^x\\\end{matrix} \right] \left[ \begin{array}{c} 0\\ x^{3/2}e^x\\\end{array} \right] =\left[ \begin{array}{c} -x^{5/2}\\ x^{3/2}\\\end{array} \right] \\A\left( x \right) =-\frac{2}{7}x^{7/2},B=\frac{2}{5}x^{5/2}\\y\left( x \right) =-\frac{2}{7}x^{7/2}e^x+\frac{2}{5}x^{7/2}e^x=-\frac{4}{35}x^{7/2}e^x-particular\,\,solution\\y\left( x \right) =-\frac{4}{35}x^{7/2}e^x+C_1e^x+C_2xe^x-general\,\,solution$