y′′+4y′=sin5x+x−1y=y1+y21)y′′+4y′=0λ2+4λ=0λ1=0,λ2=−4y1=c1+c2e−4x2)y2=acos5x+bsin5x+x(kx+p)y2′=−5asin5x+5bcos5x+2kx+py2′′=−25acos5x−25bsin5x+2k−25acos5x−25bsin5x+2k−20asin5x++20bcos5x+8kx+4p=sin5x+x−1cos5x:−25a+20b=0sin5x:−25b−20a=1x:8k=1x0:2k+4p=−1a=−2054b=−411k=81p=−165y2=−2054cos5x−411sin5x+x(81x−165)y=c1+c2e−4x−−2054cos5x−411sin5x+x(81x−165)
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