求微分方程通解 y''+y'=x^2+cosx
人气:229 ℃ 时间:2020-05-10 22:00:52
解答
齐次特征方程r^2+r=0r=0,r=-1所以齐次通解是y=C1+C2e^(-x)非齐次分两部分y''+y'=x^2和y''+y'=cosx设第一部分特解是y1=ax^4+bx^3+cx^2+dx+ey'=4ax^3+3bx^2+2cx+dy''=12x^2+6bx+2c代入得12x^2+6bx+2c+4ax^3+3bx^2+2cx+...
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