y'+2y-y^2=0 помогите с решением

y''+ y'+2=0; y(0) = 0; y'(0) = -2;. 2) 4y'+ (y'')2 = 4xy''; ...1) yy''-(y')2 = y2 lny;. 2) y(1 -lny)y''+ (1 + lny)(y')2 = 0: ...2) 3y'y'' = 2y; y(0) = y'(0) = 1;. 3) 2y'' = 3y2; y(-2)