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求微分方程dx/y+dy/x=0满足初始条件y(4)=2特解的为?
人气:384 ℃ 时间:2019-08-19 20:11:16
解答
(1/y)dx + (1/x)dy = 0(1/y)dx = - (1/x)dy等号两边各乘以 xyxdx = -ydy积分(1/2)x^2 + (C1) = -(1/2)y^2 +(C2)化简x^2 +y^2 = C代入初试条件4^2+2^2 = CC = 20微分方程为x^2 + y^2 = 20
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