Question

求微分方程y"+2/（1-y）*（y'）^2=0的通解

Answer 1

y"+2/（1-y）*（y'）^2=0

y''/y'+2y'(1-y)=0

y''/y'=2y'/(y-1)

(lny')'=2(ln(y-1))'

lny'=2ln(y-1) +C

=ln(y-1)^2+C

=ln(y-1)^2+lnC1

=lnC1(y-1)^2

y'=C1(y-1)^2

y=C1(y-1)^3/3 +C2

=C3(y-1)^3+C2