As the given equation has 3 different arbitrary constants so we can differentiate it thrice with respect to x
So, differentiating once with respect to x,
\(\frac{dy}{dx} = 2ax + b\)
Differentiating twice with respect to x,
\(\left(\frac{d^2y}{dx^2}\right)=2a\)
Now, differentiating thrice with respect to x we get,
\(\frac{d^3y}{dx^3}=0\)
Hence, \(\frac{d^3y}{dx^3}=0\) is the differential equation corresponding to y = ax2 + bx + c.