Question

求微分方程的解 yy''-(y')的平方+y'=0

Answer 1

此为不显含x的方程，令y'=p,y''=p(dp/dy),化为p对于自变量y的方程
ypp'-p²+p=0
p(yp'-p+1)=0 => p=0或 yp'-p+1=0 => y=C或者yp'-p+1=0
1/(p-1)dp=1/ydy
ln|p-1|=ln|y|+C
p-1=C1y
y'=C1y+1
[ln|C1y+1|]/C1=x+C
y=(C2e^(C1x)-1)/C1 或 y=C

Answer 2

yy''-y'^2+y'=0
x'y'=1 y'=1/x' y''=-x''/(x')^2
y*(-x''/(x')^2-(1/x')^2+1/x'=0
x''y+1-x'=0
x''y-x'=-1
x''y-x'=-1
x''/y-x'/y^2=-1/y^2
(x'/y)'=-1/y^2
x'/y=1/y+C0
x'=1+C0y
dx/dy=1+C0y
x=y+C1y^2+C2