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求微分方程(x^2-1)y'+2xy-cosx=0的通解
人气:465 ℃ 时间:2020-04-09 12:37:29
解答
(x^2-1)y'+2xy-cosx=0
(x^2-1)dy+2xydx-cosxdx=0
d(yx^2-y-sinx)=0
yx^2-y-sinx=C
y=(sinc+C)/(x^2-1)
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