Question

f(u)'f(u)''+f(u)^2=0,f(0)=1,f(0)'=1/2,求f(u). f(u)'f(u)''+f(u)^2=0,f(0)=1,f(0)'=1/2,求f(u).

Answer 1

令y=f(u),p=y'=dy/du,则y''=dp/du=(dp/dy)(dy/du)=p(dp/dy).代入到方程中得p^2(dp/dy)=-y^2,分离变量得p^2dp=-y^2dy,两边积分得p=-y+c,代入初始条件,x=0时1/2=-1+c,c=3/2,因此dy/du=3/2-y,再次分离变量,dy/(3/2-y)=du,积分得u=-lnC(3/2-y),代入u=0,y=1得C=2,故f(u)=y=[3-e^(-u)]/2