Solving the Integral Equation Set Problem

We need to find the number of elements in the set $S = \left\{ x : x \in [0, 100] \text{ and } \int_{0}^{x} t^2 \sin(x-t)dt = x^2 \right\}$.

Let the integral be $F(x) = \int_{0}^{x} t^2 \sin(x-t)dt$. The equation becomes $F(x) = x^2$. We can evaluate $F(x)$ using differentiation under the integral sign (Leibniz rule).

Deriving the Differential Equation

Using the Leibniz integral rule $\frac{d}{dx} \int_{a(x)}^{b(x)} f(x, t) dt = f(x, b(x)) \frac{db}{dx} - f(x, a(x)) \frac{da}{dx} + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x, t) dt$:

$F'(x) = \frac{d}{dx} \int_{0}^{x} t^2 \sin(x-t)dt$

$F'(x) = t^2 \sin(x-t) \big|_{t=x} \cdot (1) - t^2 \sin(x-t) \big|_{t=0} \cdot (0) + \int_{0}^{x} \frac{\partial}{\partial x} [t^2 \sin(x-t)] dt$

$F'(x) = x^2 \sin(0) + \int_{0}^{x} t^2 \cos(x-t) dt = \int_{0}^{x} t^2 \cos(x-t) dt$

Differentiating again:

$F''(x) = \frac{d}{dx} \int_{0}^{x} t^2 \cos(x-t) dt$

$F''(x) = t^2 \cos(x-t) \big|_{t=x} \cdot (1) + \int_{0}^{x} \frac{\partial}{\partial x} [t^2 \cos(x-t)] dt$

$F''(x) = x^2 \cos(0) + \int_{0}^{x} t^2 (-\sin(x-t)) dt$

$F''(x) = x^2 - \int_{0}^{x} t^2 \sin(x-t) dt$

$F''(x) = x^2 - F(x)$

This gives the second-order linear differential equation: $F''(x) + F(x) = x^2$.

Finding the Solution F(x)

We need initial conditions.

• $F(0) = \int_{0}^{0} t^2 \sin(0-t)dt = 0$.
• $F'(0) = \int_{0}^{0} t^2 \cos(0-t) dt = 0$.

The differential equation is $F''(x) + F(x) = x^2$. The characteristic equation for the homogeneous part ($F'' + F = 0$) is $r^2 + 1 = 0$, giving $r = \pm i$. The homogeneous solution is $F_h(x) = C_1 \cos(x) + C_2 \sin(x)$.

For the particular solution $F_p(x)$, assume $F_p(x) = Ax^2 + Bx + C$. Then $F_p'(x) = 2Ax + B$ and $F_p''(x) = 2A$. Substituting into the DE:

$2A + (Ax^2 + Bx + C) = x^2$

$Ax^2 + Bx + (2A + C) = x^2$

Comparing coefficients: $A=1$, $B=0$, $2A+C=0 \implies 2(1)+C=0 \implies C=-2$.

So, $F_p(x) = x^2 - 2$. The general solution is $F(x) = F_h(x) + F_p(x) = C_1 \cos(x) + C_2 \sin(x) + x^2 - 2$.

Using initial conditions:

• $F(0) = 0 \implies C_1 \cos(0) + C_2 \sin(0) + 0^2 - 2 = 0 \implies C_1 - 2 = 0 \implies C_1 = 2$.
• $F'(x) = -C_1 \sin(x) + C_2 \cos(x) + 2x$.
• $F'(0) = 0 \implies -C_1 \sin(0) + C_2 \cos(0) + 2(0) = 0 \implies C_2 = 0$.

The solution is $F(x) = 2\cos(x) + x^2 - 2$.

Finding Solutions x in [0, 100]

The original equation is $F(x) = x^2$. Substitute the solution for $F(x)$:

$2\cos(x) + x^2 - 2 = x^2$

$2\cos(x) - 2 = 0$

$2\cos(x) = 2$

$\cos(x) = 1$

The general solution for $\cos(x) = 1$ is $x = 2k\pi$, where $k$ is an integer ($k \in \mathbb{Z}$).

We need solutions in the interval $[0, 100]$. So, we must have:

$0 \le 2k\pi \le 100$

$0 \le k \le \frac{100}{2\pi} = \frac{50}{\pi}$

Using $\pi \approx 3.14159$, we get $\frac{50}{\pi} \approx 15.915$.

Since $k$ must be an integer, the possible values for $k$ are $0, 1, 2, \ldots, 15$.

The number of possible integer values for $k$ is $15 - 0 + 1 = 16$.

Therefore, there are 16 values of $x$ in the interval $[0, 100]$ that satisfy the given integral equation.

The number of elements in the set $S$ is 16.