解答题 10.求解y"=e2y+ey,且y(0)=0,y'(0)=2.
【正确答案】
p2=e2y+2ey+C,
即 y'2=e2y+2ey+C.
又y(0)=0,y'(0)=2,有C=1,所以
y'2=e2y+2ey+1=(ey+1)2
y'=ey+1(y'(0)=2>0),
【答案解析】