Let y=y(x) be the solution of the differential equation (x+y+2)^2 d x=d y, y(0)=-2 . Let the maximum and minimum values of the function y=y(x) in [0, 3 ] be and , respectively. If (3 + )^2+ ^2= + 3 , , Z , then + equals ______

Step-by-step solution

d y d x =(x+y+2)^2 ...(1) y(0)=-2 Let x + y +2= v 1+ dy dx = dv dx from (1) d v d x =1+v^2 aligned & d v 1+v^2 = d x & ⁻¹(v)=x+C & ⁻¹(x+y+2)=x+C & at x=0 y=-2 C=0 aligned aligned & ⁻¹( x + y +2)= x & y = x - x -2 & f ( x )= x - x -2, x [0, 3 ] aligned aligned & f^ (x)= ^2 x-1>0 f(x) & f_ =f(0)=-2= & f_ =f ( 3 )= 3 - 3 -2= aligned now (3 + )^2+ ^2= + 3 aligned & (3 + )^2+ ^2=(3 3 -6)^2+4 & + 3 =67-36 3 & =67 and =-36 + =31 aligned