Question

微分方程y''+2yy'^2=0满足条件y'(0)=-1,y(0)=1的解是（ ）?

Answer 1

yy"+y'^2 =0 (yy')'=0 yy'=C1 y|(x=0)=1，y'|(x=0)=1/2 1·(1/2)=C1 得C1=1/2 yy'=1/2 (1/2)(y2 )'=1/2 y2=x+C2 y|(x=0)=1 12=0+C2 C2=1 y2=x+1 y|(x=0)=1 y=√(x+1) 所以 yy"+y'^2 =0满足初始条件y|(x=0)=1，y'|(x=0)=1/2的特解是 y=√(x+1)