Question

求微分 2yy''-(y’)^2+1=0 的通解！ 给出过程

Answer 1

y''=dy'/dx=(dy'/dy)(dy/dx)=y'(dy'/dy)
带入得2yy'(dy'/dy)=y'^2-1
2y'dy'/(y'^2-1)=dy/y
所以ln|y'^2-1|=ln|y|+c
即y'^2-1=cy
dy/dx=(cy+1)^(1/2)
dy/((cy+1)^(-1/2)=dx
所以2(cy+1)^(1/2)=cx+c'

Answer 2

2yy''-(y')^2+1=0
(y'^2)'-(y')^2+1=0
y'^2=p
p'-p+1=0
dp/dx=p-1
dp/(p-1)=dx
ln(p-1)=x+C
p=C'e^x+1
(y')^2=C'e^x+1
1
y'=√(C'e^x+1)
dy/dx=√(C'e^x+1)
y=∫√(C'e^x+1)dx
C'e^x+1=t^2
x=ln[(t^2-1)/C']
dx=2tdt/[C'(t^2-1)]
y=∫2t^2dt/[C'(t^2-1)]=(1/C')2t+(1/C')∫2dt/[(t-1)(t+1)]=(1/C')2t+(1/C')ln[(t-1)/(t+1)]+C0
y=(2/C')√(C'e^x+1)+(1/C')ln[(√(C'e^x+1)-1)/(√(C'e^x+1)+1)]+C0
2
y'=-√(C'e^x+1)
y=(-2/C')√(C'e^x+1)+(-1/C')ln[(√(C'e^x+1)-1)/(√(C'e^x+1)+1)]+C1

Answer 3

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