What would be a differential equation whose solutions include the parabolas in the picture?

General equation for vertical axis parabolas is (x−h)²=4a(y−k)

If a parabola goes through (−2,0) and (0,−2) then

(2+h)²=4a(−k) … (i)

h²=4a(−2−k) … (ii)

(i)−(ii) gives 4h+4=8a ⟹ h=2a−1 and from (i) k= −a−1−1/(4a)

Thus for any given a the family parabola is (x−2a+1)² = 4a(y+a+1+1/(4a))

Expand and simplify to 4a(y+x+2) = x²+2x … (ii)

Differentiate wrtx : 4a(y’+1) = 2x+2 … (iii)

Required ODE is found by eliminating a from (ii) and (iii)

(ii)÷(iii) gives (y+x+2)/(y’+1) = (x²+2x)/(2x+2) ⟹ x(x+2)y’−2(x+1)y=(x+2)²

y = (Ax - 1)(x + 2)