Use variation of parameters to find a particular solution, given the solutions y1,y2\displaystyle {y}_{{1}},{y}_{{2}}y1,y2 of the complementary equation
x2y′′−4xy′+(x2+6)y=x4\displaystyle {x}^{{2}}{y}{''}-{4}{x}{y}'+{\left({x}^{{2}}+{6}\right)}{y}={x}^{{4}}x2y′′−4xy′+(x2+6)y=x4
y1=x2cos(x),y2=x2sin(x)\displaystyle {y}_{{1}}={x}^{{2}}{\cos{{\left({x}\right)}}},{y}_{{2}}={x}^{{2}}{\sin{{\left({x}\right)}}}y1=x2cos(x),y2=x2sin(x)
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