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第1个回答 2020-03-19
（1） dy/dx = 1/(x+y), dx/dy - x = y,
x = e^(∫dy)[∫ye^(-∫dy)dy + C] = e^y[∫ye^(-y)dy + C]
= e^y[-∫yde^(-y) + C] = e^y[-ye^(-y) - e^(-y) + C]
通解 x = - y - 1 + Ce^y
（2） dy/dx - 2y/(x+1) = (x+1)^(5/2),
y = e^[∫2dx/(x+1)]{∫(x+1)^(5/2)e^[∫-2dx/(x+1)]dx + C}
= (x+1)^2[∫(x+1)^(1/2)dx + C]
= (x+1)^2[(2/3)(x+1)^(3/2) + C]本回答被网友采纳
x = e^(∫dy)[∫ye^(-∫dy)dy + C] = e^y[∫ye^(-y)dy + C]
= e^y[-∫yde^(-y) + C] = e^y[-ye^(-y) - e^(-y) + C]
通解 x = - y - 1 + Ce^y
（2） dy/dx - 2y/(x+1) = (x+1)^(5/2),
y = e^[∫2dx/(x+1)]{∫(x+1)^(5/2)e^[∫-2dx/(x+1)]dx + C}
= (x+1)^2[∫(x+1)^(1/2)dx + C]
= (x+1)^2[(2/3)(x+1)^(3/2) + C]本回答被网友采纳
第2个回答 2020-03-19
enn...太难了
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