Using the method of variation of parameters find a particula

Using the method of variation of parameters find a particular solution of y\" + y\' - 2y = te^t. Next, write the general solution of this equation. Finally, given the vertical values y(0) = 0 and y\'(0) = 0, write the solution of the given equation.

Solution

The\\ ODE\\ is:\\\\
y\'\'+y\'-2y=te^t , \\ \\ \\ \\ y(0)=0, y\'(0)=0\\\\\\\\
This\\ is\\ a\\ 2nd\\ order\\ non-homogeneous\\ DE\\ whose\\ general\\ solution\\\\
is\\ given\\ by\\\\
y=y_h+y_p,\\ \\ \\ where\\ y_h\\ is\\ the\\ homogeneous\\ solution\\ and\\ y_p\\ is\\\\
the\\ particular\\ solution\\\\\\\\
Step\\ 1:\\ To\\ find\\ the\\ homogeneous\\ solution\\ y_h:\\\\
The\\ characteristic\\ equaiton\\ of\\ the\\ ODE\\ is\\ r^2+r-2=0\\\\
The\\ roots\\ of\\ the\\ above\\ equation \\ are\\ r_1=1\\ and\\ r_2=-2\\\\
Hence\\ the\\ homogeneous\\ solution\\ is\\\\
y_h=c_1e^t+c_2e^{-2t}\\\\\\\\
Step2:\\ To\\ find\\ the\\ particular\\ solution \\ using\\ variation\\ of\\ parameters\\\\
Based\\ on\\ the\\ homogeneous\\ solution\\ we\\ assume\\ the\\ particular\\ solution\\\\
y_p\\ in\\ the\\ form:\\\\
y_p(t)=u_1(t)e^t+u_2(t)e^{-2t}\\\\
The \\ variation \\ of \\ parameters \\ formula \\ gives \\ the \\ system\\\\
u_1\'(t)e^t+u_2\'(t)e^{-2t}=0\\\\
=>u_1\'y_1+u_2\'y_2=0\\\\
u_1\'(t)e^t-2u_2\'(t)e^{-2t}=te^t\\\\
=>=>u_1\'y\'_1+u_2\'y\'_2=te^t\\\\
\\\\
Then \\the\\ solution\\ is\\ given\\ by\\\\
u_1(t)=-\\int \\frac{y_2.g(t)}{W(y1,y2)}\\\\
where\\ g(t)= wronskian \\ of\\ y1\\ and\\ y2\\\\\\\\

W(y1,y2)=| \\begin{array}{cc} e^{t} & e^{2 t} \\\\\\\\ e^{t} & 2 e^{2 t} \\end{array} |=-3e^{3 t}\\\\

so \\ u_1(t)=\\frac{1}{3}(-\\frac{1}{4}e^{-4t}t-\\frac{e^{-4t}}{16})
and\\ u_2(t)=\\int \\frac{y_1.g(t)}{W(y1,y2)}\\\\
=\\int \\frac{e^{t}.te^t}{-3e^{3 t}}}=