Use variation of parameters to solve the initial value problem, given the solutions y1,y2\displaystyle {y}_{{1}},{y}_{{2}}y1,y2 of the complementary equation
(x−1)2y′′−2(x−1)y′+2y=(x−1)2,y(0)=3,y′(0)=−6\displaystyle {\left({x}-{1}\right)}^{{2}}{y}{''}-{2}{\left({x}-{1}\right)}{y}'+{2}{y}={\left({x}-{1}\right)}^{{2}},{y}{\left({0}\right)}={3},\quad{y}'{\left({0}\right)}=-{6}(x−1)2y′′−2(x−1)y′+2y=(x−1)2,y(0)=3,y′(0)=−6
y1=x−1,y2=x2−1\displaystyle {y}_{{1}}={x}-{1},\quad{y}_{{2}}={x}^{{2}}-{1}y1=x−1,y2=x2−1
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