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解微分方程dx/dy=-x-y^2
人气:390 ℃ 时间:2020-02-24 20:47:54
解答
x'+x=-y²(e^y)(x'+x)=-y²e^y[xe^y]'=-y²e^yxe^y=∫-y²e^ydy=-y²e^y+∫2ye^y=-y²e^y+2ye^y-∫2e^ydy=-y²e^y+2ye^y-2e^y+Cx=-y²+2y-2+Ce^(-y)
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