Question

Let g: R → R be a continuous function. Which one of the following is the solution of the differential equation
$\frac{d^{2} y}{d x^{2}} + y = g (x) for x \in R,$
satisfying the conditions y(0) = 0, y’(0) = 1 ?

Answer: $y (x) = sin x + \int_{0}^{x} sin (x - t) g (t) d t$

Option (1)
we have $y (x) = sin x - \int_{0}^{x} sin (x - t) g (t) d t \dots \dots (1)$
Differentiate eq(1) w. r. to $x$ , we get

\begin{aligned} \frac{d (y (x))}{d x} & = & \frac{d (sin x)}{d x} - \frac{d}{d x} (\int_{0}^{x} sin (x - t) g (t) d t) \\ y^{'} (x) & = & cos x - [\int_{0}^{x} \frac{\partial {sin (x - t) g (t)}}{\partial x} d t + sin (x - x) g (x) \frac{d x}{d x} - \\ sin (x - 0) g (0) \frac{d (0)}{d x}] (using leibniz rule of two variable) \\ y^{'} (x) & = & cos x - [\int_{0}^{x} cos (x - t) g (t) d t + sin 0 \cdot g (x) \cdot 1 - sin x \cdot g (0) \cdot 0] \\ y^{'} (x) & = & cos x - [\int_{0}^{x} cos (x - t) g (t) d t + 0 - 0] \\ y^{'} (x) & = & cos x - \int_{0}^{x} cos (x - t) g (t) d t \dots \dots (2) \end{aligned}

Now, again Differentiate eq(2) w. r. to $x$ , we get

\begin{aligned} \frac{d (y^{'} (x))}{d x} & = & \frac{d (cos x)}{d x} - \frac{d}{d x} (\int_{0}^{x} cos (x - t) g (t) d t) \\ y^{′′} (x) & = & - sin x - [\int_{0}^{x} \frac{\partial {cos (x - t) g (t)}}{\partial x} d t + cos (x - x) g (x) \frac{d x}{d x} - \\ cos (x - 0) g (0) \frac{d (0)}{d x}] \\ y^{′′} (x) & = & - sin x - [\int_{0}^{x} (- sin (x - t)) g (t) d t + cos 0 \cdot g (x) \cdot 1 - 0] \\ y^{′′} (x) & = & - sin x + \int_{0}^{x} sin (x - t) g (t) d t - 1 \cdot g (x) \\ y^{′′} (x) & = & - (sin x - \int_{0}^{x} sin (x - t) g (t) d t) - g (x) \\ y^{′′} (x) & = & - y - g (x) \\ ∴ y^{′′} (x) + y & = & - g (x) \dots \dots (3) \end{aligned}

Now, we can see that differential equation of eq(3) of option (1) is not same as question’s differential equation.
Hence, option (1) is wrong.

Now, Option (2)

we have $y (x) = sin x + \int_{0}^{x} sin (x - t) g (t) d t \dots \dots (1)$
Differentiate eq(1) w. r. to $x$ , we get

\begin{aligned} \frac{d (y (x))}{d x} & = & \frac{d (sin x)}{d x} + \frac{d}{d x} (\int_{0}^{x} sin (x - t) g (t) d t) \\ y^{'} (x) & = & cos x + [\int_{0}^{x} \frac{\partial {sin (x - t) g (t)}}{\partial x} d t + sin (x - x) g (x) \frac{d x}{d x} - \\ sin (x - 0) g (0) \frac{d (0)}{d x}] (using leibniz rule of two variable) \\ y^{'} (x) & = & cos x + [\int_{0}^{x} cos (x - t) g (t) d t + sin 0 \cdot g (x) \cdot 1 - sin x \cdot g (0) \cdot 0] \\ y^{'} (x) & = & cos x + [\int_{0}^{x} cos (x - t) g (t) d t + 0 - 0] \\ y^{'} (x) & = & cos x + \int_{0}^{x} cos (x - t) g (t) d t \dots \dots (2) \end{aligned}

Now, again Differentiate eq(2) w. r. to $x$ , we get

\begin{aligned} \frac{d (y^{'} (x))}{d x} & = & \frac{d (cos x)}{d x} + \frac{d}{d x} (\int_{0}^{x} cos (x - t) g (t) d t) \\ y^{′′} (x) & = & - sin x + [\int_{0}^{x} \frac{\partial {cos (x - t) g (t)}}{\partial x} d t + cos (x - x) g (x) \frac{d x}{d x} - \\ cos (x - 0) g (0) \frac{d (0)}{d x}] \\ y^{′′} (x) & = & - sin x + [\int_{0}^{x} (- sin (x - t)) g (t) d t + cos 0 \cdot g (x) \cdot 1 - 0] \\ y^{′′} (x) & = & - sin x - \int_{0}^{x} sin (x - t) g (t) d t + 1 \cdot g (x) \\ y^{′′} (x) & = & - (sin x + \int_{0}^{x} sin (x - t) g (t) d t) + g (x) \\ y^{′′} (x) & = & - y + g (x) \\ ∴ y^{′′} (x) + y & = & g (x) \dots \dots (3) \end{aligned}

Now, we can see that differential equation of eq(3) of option (2) is same as question’s differential equation.
Hence, option (2) is correct.

Now, Option (3)

we have $y (x) = sin x - \int_{0}^{x} cos (x - t) g (t) d t \dots \dots (1)$
Differentiate eq(1) w. r. to $x$ , we get

\begin{aligned} \frac{d (y (x))}{d x} & = & \frac{d (sin x)}{d x} - \frac{d}{d x} (\int_{0}^{x} cos (x - t) g (t) d t) \\ y^{'} (x) & = & cos x - [\int_{0}^{x} \frac{\partial {cos (x - t) g (t)}}{\partial x} d t + cos (x - x) g (x) \frac{d x}{d x} - \\ cos (x - 0) g (0) \frac{d (0)}{d x}] (using leibniz rule of two variable) \\ y^{'} (x) & = & cos x - [\int_{0}^{x} (- sin (x - t)) g (t) d t + cos 0 \cdot g (x) \cdot 1 - cos x \cdot g (0) \cdot 0] \\ y^{'} (x) & = & cos x - [\int_{0}^{x} (- sin (x - t)) g (t) d t + g (x) - 0] \\ y^{'} (x) & = & cos x + \int_{0}^{x} sin (x - t) g (t) d t - g (x) \dots \dots (2) \end{aligned}

Now, again Differentiate eq(2) w. r. to $x$ , we get

\begin{aligned} \frac{d (y^{'} (x))}{d x} & = & \frac{d (cos x)}{d x} + \frac{d}{d x} (\int_{0}^{x} sin (x - t) g (t) d t) - \frac{d (g (x))}{d x} \\ y^{′′} (x) & = & - sin x + [\int_{0}^{x} \frac{\partial {sin (x - t) g (t)}}{\partial x} d t + sin (x - x) g (x) \frac{d x}{d x} - \\ sin (x - 0) g (0) \frac{d (0)}{d x}] - g^{'} (x) \\ y^{′′} (x) & = & - sin x + [\int_{0}^{x} cos (x - t) g (t) d t + sin 0 \cdot g (x) \cdot 1 - 0] - g^{'} (x) \\ y^{′′} (x) & = & - sin x + \int_{0}^{x} cos (x - t) g (t) d t + 0 \cdot g (x) - g^{'} (x) \\ y^{′′} (x) & = & - (sin x - \int_{0}^{x} cos (x - t) g (t) d t) - g^{'} (x) \\ y^{′′} (x) & = & - y - g^{'} (x) \\ ∴ y^{′′} (x) + y & = & - g^{'} (x) \dots \dots (3) \end{aligned}

Now, we can see that differential equation of eq(3) of option (3) is not same as question’s differential equation.
Hence, option (3) is also wrong.

Now, Option (4)

we have $y (x) = sin x + \int_{0}^{x} cos (x - t) g (t) d t \dots \dots (1)$
Differentiate eq(1) w. r. to $x$ , we get

\begin{aligned} \frac{d (y (x))}{d x} & = & \frac{d (sin x)}{d x} + \frac{d}{d x} (\int_{0}^{x} cos (x - t) g (t) d t) \\ y^{'} (x) & = & cos x + [\int_{0}^{x} \frac{\partial {cos (x - t) g (t)}}{\partial x} d t + cos (x - x) g (x) \frac{d x}{d x} - \\ cos (x - 0) g (0) \frac{d (0)}{d x}] (using leibniz rule of two variable) \\ y^{'} (x) & = & cos x + [\int_{0}^{x} (- sin (x - t)) g (t) d t + cos 0 \cdot g (x) \cdot 1 - cos x \cdot g (0) \cdot 0] \\ y^{'} (x) & = & cos x + [\int_{0}^{x} (- sin (x - t)) g (t) d t + g (x) - 0] \\ y^{'} (x) & = & cos x - \int_{0}^{x} sin (x - t) g (t) d t + g (x) \dots \dots (2) \end{aligned}

Now, again Differentiate eq(2) w. r. to $x$ , we get

\begin{aligned} \frac{d (y^{'} (x))}{d x} & = & \frac{d (cos x)}{d x} - \frac{d}{d x} (\int_{0}^{x} sin (x - t) g (t) d t) + \frac{d (g (x))}{d x} \\ y^{′′} (x) & = & - sin x - [\int_{0}^{x} \frac{\partial {sin (x - t) g (t)}}{\partial x} d t + sin (x - x) g (x) \frac{d x}{d x} - \\ sin (x - 0) g (0) \frac{d (0)}{d x}] + g^{'} (x) \\ y^{′′} (x) & = & - sin x - [\int_{0}^{x} cos (x - t) g (t) d t + sin 0 \cdot g (x) \cdot 1 - 0] + g^{'} (x) \\ y^{′′} (x) & = & - sin x - \int_{0}^{x} cos (x - t) g (t) d t - 0 \cdot g (x) + g^{'} (x) \\ y^{′′} (x) & = & - (sin x + \int_{0}^{x} cos (x - t) g (t) d t) + g^{'} (x) \\ y^{′′} (x) & = & - y + g^{'} (x) \\ ∴ y^{′′} (x) + y & = & g^{'} (x) \dots \dots (3) \end{aligned}

Now, we can see that differential equation of eq(3) of option (4) is not same as question’s differential equation.
Hence, option (4) is also wrong.

So, correct answer is option-2, $\to y (x) = sin x + \int_{0}^{x} sin (x - t) g (t) d t$