Question

大一高数求微分方程通解,yy''-(y')^2+y'=0

Answer 1

令p=y'

则y"=dp/dx=dp/dy*dy/dx=pdp/dy

代入原方程：ypdp/dy-p^2+p=0

得：p=0或ydp/dy-p+1=0

p=0得：dy/dx=0,即：y=c

ydp/dy-p+1=0,得：dp/(p-1)=dy/y,得：ln(p-1)=lny+c1,得：p-1=cy

得：dy/dx=cy+1,

得：dy/(cy+1)=cx,

得：ln(cy+1)=cx^2/2+c2

cy+1=e^(cx^2/2+c2)

y=[e^(cx^2/2+c2)-1]/c