Differential equation d^2y dx^2 +y=0,y(0)=0,y'(0)=1Function y=

Question

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

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Answer

We have,$\text{y}=\sin{\text{x}} ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\cos\text{x} ...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\sin{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{y}$ [Using (1)]
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
it is the given differential equation.
Here, $\text{y}=\sin\text{x}$ satisfies the given differential equation; hence, it is a solution.
Also, when $\text{x}=0,\text{y}=\sin0=0,\text{i.e.},\text{y}(0)=0.$
And, when $\text{x}=0,\text{y}'=\cos 0=1,\text{i.e.,}\text{y}'(0)=1.$
Hence, $\text{y}=\sin\text{x}$ is the solution to the given initial value problem.

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