Show that the function `varphi` defined by `varphi(x)=cos x\ (x in RR)` ; satisfies the initial value problem `(d^2y)/(dx^2)+y=0, y(0)=1,\ y^(prime)(0)=0`.

Explanation

`phi(x)=cos x `

` = phi^{prime}(x)=-sin x`

`
= phi^{prime prime}(x)=-cos x `

`
phi rightarrow y . `

`
phi^{prime prime}(x)=-phi(x) .
Rightarrow frac{d^{2} y}{d x^{2}}=-y
`

`Rightarrow d^{2} y+y=0 . `


Now,

`
y(0)= at x=0 . `

`
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