The solution of the differential equation d 2 y d x 2 + y = 0 is

Given, $\frac{d^{2}y}{d{x}^2}+y=0$

We will check if $y=3\sin x+4\cos x$ is solution of the differential equation.

Differentiating this w.r.t $x$

$\frac{dy}{dx}=3\cos x-4\sin x$

Again differentiating w.r.t. $x$

$\frac{d^{2}y}{d{x}^2}=-3\sin x-4\cos x$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\left(3\sin x+4\cos x\right)=-y$

$\Rightarrow \frac{d^{2}y}{d{x}^2}+y=0$

Hence, $y=3\sin x+4\cos x$ is solution of the differential equation $\frac{d^{2}y}{d{x}^2}+y=0$.

Alternative Method:

Given is a second-order linear ordinary differential equation.

Its general solution is of the form $y=c{e}^{sx}$, where $c$ is constant.

Differentiating this twice w.r.t. $x$, we get

$\frac{d^{2}y}{d{x}^2}=s^{2}c{e}^{sx}+c{e}^{sx}=\left(s^2+1\right)c{e}^{sx}$

$\Rightarrow \left(s^2+1\right)c{e}^{sx}=0$, which is true if $s^2=-1\Rightarrow s=\pm i$

Therefore, we get $y=c{e}^{\pm ix}$.

Using Euler's formula, this is equal to $y=c\left(\cos x\pm i\sin x\right)$

From this, we see that the general solution to the original differential equation is of the form $y=c_1\cos x+c_2\sin x$, where $c_1 \& c_2$ can be found from the initial conditions.

Now, we see that Only option $1$ is of this form.