Question

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

Function y = sin x + cos x

Answer 1

We have,
$y=\sin x+\cos x$ .....(1)
Differentiating both sides of (1) with respect to x, we get
$\frac{dy}{dx}=\cos x-\sin x$ .....(2)
Differentiating both sides of (2) with respect to x, we get
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=-\sin x-\cos x \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=-\left(\sin x+\cos x\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=-y\left[\text{Using}\left(1\right)\right] \end{gathered}$$

$\Rightarrow$$\frac{d^{2}y}{d{x}^2}+y=0$
It is the given differential equation.
Therefore, $y=\sin x+\cos x$ satisfies the given differential equation.
Also, when $x=0;y=\sin 0+\cos 0=1,\text{i.e}.y\left(0\right)=1.$
And, when $x=0;y\text{'}=\cos 0-\sin 0=1,\text{i.e}.y\text{'}\left(0\right)=1.$
Hence, $y=\sin x+\cos x$ is the solution to the given initial value problem.