Question

The general solution of \(\frac{d^2 y}{dx^2} = 0\) is

• A. General solution of \(\frac{d^3y}{dx^3} = 0\)
• B. General solution of \(\frac{dy}{dx} = a + b\)
• C. General solution for both \(\frac{d^2y}{dx^2}=0\) and \(\frac{d^3y}{dx^3}=0\)
• D. General solution for \(\frac{d^2y}{dx^2} = 0\)

Hint:

For a second-order differential equation of the form \(y'' = 0\), integrate twice to find general solution containing two arbitrary constants.

Answer 1

The Correct Option is D

The order of a differential equation tells you exactly how many independent arbitrary constants its general solution must carry. The equation \(\frac{d^2y}{dx^2}=0\) is second order, so its general solution needs precisely two independent constants, no more and no fewer.

Integrating once gives \(\frac{dy}{dx}=C_1\), and integrating again gives \(y=C_1x+C_2\). This is a complete, self-contained solution built entirely from the equation \(\frac{d^2y}{dx^2}=0\) itself, with two constants matching its order.

This rules out linking it to the third-order equation \(\frac{d^3y}{dx^3}=0\), whose general solution would need three constants (an added quadratic term), so the same two-constant family cannot double as the general solution of that higher-order equation. It also is not properly described as merely \(\frac{dy}{dx}=a+b\), since that phrase does not by itself pin down \(y\) up to the two constants the second-order equation demands.

So the correct choice is option (4), the plain and direct statement that \(y=C_1x+C_2\) is the general solution for \(\frac{d^2y}{dx^2}=0\).