Question

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

\[\frac{d^2 y}{d x^2} = - \sin x\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = - y ..........\left[\text{Using (1)}\right]\]

⇒ \[\frac{d^2 y}{d x^2} + y = 0\]

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 \[x = 0, y' = \cos 0 = 1,\text{ i.e. }, y'\left( 0 \right) = 1\]

Hence, \[y = \sin x\] is the solution to the given initial value problem.