Question

Differential equation $\frac{d^{2}y}{d{x}^2}+y=0,y\left(0\right)=0,y\text{'}\left(0\right)=1$

Function y = sin x

Answer 1

We have,
$y=\sin x$ ...(1)
Differentiating both sides of (1) with respect to $x$, we get
$\frac{dy}{dx}=\cos x$ ...(2)
Differentiating both sides of (2) with respect to $x$, we get
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=-\sin x \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=-y\left[\text{Using}\left(1\right)\right] \end{gathered}$$
$\Rightarrow$$\frac{d^{2}y}{d{x}^2}+y=0$
It is the given differential equation.
Here, $y=\sin x$ satisfies the given differential equation; hence, it is a solution.
Also, when $x=0,y=\sin 0=0,\text{i.e}.,y\left(0\right)=0.$
And, when $x=0,y\text{'}=\cos 0=1,\text{i.e}.,y\text{'}\left(0\right)=1.$
Hence, $y=\sin x$ is the solution to the given initial value problem.