Q.The number of arbitrary constants in the particular solution of a differential equation of second order is (are)
(A) $0$
(B) $1$
(C) $2$
(D) $3$

The Intuition: Why "Particular" and Why "Constants"?

Imagine you're solving a puzzle where the answer isn't a single number, but a whole family of numbers. The equation $y' = 2x$ has infinitely many solutions: $y = x^2 + C$, where $C$ can be any real number. Each different $C$ gives a different curve — all parallel parabolas shifted up or down.

Now suppose the puzzle gives you an extra condition: "the curve must pass through $(1, 3)$". That pins down exactly one member of the family: $3 = 1^2 + C \implies C = 2$, so the answer is $y = x^2 + 2$.

That $C = 2$ is a particular solution constant — the specific value of the arbitrary constant that makes the solution fit a given condition (an initial condition or boundary condition).

The Precise Statement

When solving a differential equation, you get a general solution of the form:

where $C_1, C_2, \dots, C_n$ are arbitrary constants. The number of constants equals the order of the differential equation (first-order → one constant, second-order → two constants, etc.).

A particular solution is obtained by assigning specific numerical values to these constants. Those specific numbers are the particular solution constants, determined by imposing initial conditions or boundary conditions.

A Concrete Example (First-Order)

Solve: $\frac{dy}{dx} = 3x^2$, given $y(0) = 5$.

Step 1 — General solution: $y = \int 3x^2 \, dx = x^3 + C$.

Step 2 — Apply the condition: $y(0) = 5$ means when $x = 0$, $y = 5$: $5 = 0^3 + C \implies C = 5$.

Step 3 — Particular solution: $y = x^3 + 5$. The particular solution constant is $C = 5$.

A Second-Order Example (Two Constants)

Solve: $\frac{d^2y}{dx^2} = 6x$, given $y(0) = 1$ and $y'(0) = 2$.

Step 1 — General solution: Integrate twice.

First integration: $y' = \int 6x \, dx = 3x^2 + C_1$

Second integration: $y = \int (3x^2 + C_1) \, dx = x^3 + C_1 x + C_2$

Step 2 — Apply conditions:

From $y'(0) = 2$: $2 = 3(0)^2 + C_1 \implies C_1 = 2$

From $y(0) = 1$: $1 = 0^3 + 2(0) + C_2 \implies C_2 = 1$ …