Y2+x2dy/dx=xy dy/dx 解齐次方程

eallysky2022-10-04 11:39:541条回答

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枫叶孤城 共回答了17个问题 | 采纳率94.1%
设t=y/x,则dy/dx=t+xdt/dx
∵y²+x²dy/dx=xydy/dx
==>(y/x)²=(1-y/x)dydx
==>t²=(1-t)(t+xdt/dx)
==>t(2t-1)=x(1-t)dt/dx
==>(1-t)dt/[t(2t-1)]=dx/x
==>ln|x|=∫[1/(2t-1)-1/t]dt
==>ln|x|=(1/2)ln|2t-1|-ln|t|+ln|C1| (C1是积分常数)
==>ln|x|=ln|C1√(2t-1)/t|
==>x=C1√(2t-1)/t
==>x=C1√(2y/x-1)/(y/x)
==>y=C1√(2y/x-1)
==>y√x=C1√(2y-x)
==>xy²=C(2y-x) (令C=C1²,∴C也是积分常数)
∴原微分方程的通解是:xy²=C(2y-x) (C是积分常数)
1年前

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