Question

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

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

\[ \Rightarrow \frac{d^2 y}{d x^2} = - \left( \sin x + \cos x \right)\]

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

⇒ \[\frac{d^2 y}{d x^2} + y = 0\]

It is the given differential equation.

Therefore, \[y = \sin x + \cos x\]  satisfies the given differential equation.
Also, when \[x = 0; y = \sin 0 + \cos 0 = 1,\text{ i.e. }y(0) = 1\]

And, when \[x = 0; y' = \cos 0 - \sin 0 = 1,\text{ i.e. }y'(0) = 1\]

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