Question

微分方程求解

一共三道题，求详细步骤

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Answer 1

(1)[1+e^(-x/y)]dx=(x/y -1)dy
dx/dy =(x/y -1) / [1+e^(-x/y)]
令u=x/y，x=yu
则dx/dy=u+ydu/dy

代入原方程得
u+ydu/dy =(u-1)/[1+e^(-u)]
ydu/dy=-(e^u +u)/(e^u +1)
(e^u +1)du/(e^u +u) =-dy/y
d(e^u +u)/(e^u +u)=-dy/y
ln|e^u +u| =-ln|y|+ln|C|
e^u +u =C/y
即ye^(-x/y)+x=C
(2)
2dx/(x²-4) +2dy/y =0
d(x²-4)/(x²-4) +2dy/y =0
ln|x²-4|+2ln|y|=ln|C|
(x²-4)y²=C
将x=1，y=2代入得C=-12
故所求特解为(x²-4)y²+12=0
(3)令u=x+y-3，则y=u-x+3
y'=u'-1
代入原方程得u'-1=(x-u+x-3+1)/u
整得u'=2(x-1)/u
2udu=4(x-1)dx
u²=2x²-4x+C1
(x+y-3)²=2x²-4x+C1
化简得x²-y²-2xy+2x+6y=C

Answer 2

Answer 3

Answer 4