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y''+ay'+by=0 for the given basis. cos 2πx, sin 2πx.

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In this case, we have a basis:

{ϕ1(x)=cos(2πx)=e0xcos(2πx)ϕ2(x)=sin(2πx)=e0xsin(2πx)\begin{align*} \begin{cases} & \phi _1 (x) = \cos (2\pi x) = e^{0 \cdot x} \cos (2\pi x)\\ &\phi _2 (x) =\sin (2\pi x) = e^{0 \cdot x} \sin (2\pi x) \end{cases} \end{align*}

so λ1/2=0±2πi=±2πi\boxed{\lambda_{1/2} = 0 \pm 2\pi i = \pm 2\pi i} are the complex conjugate roots of the characteristic equation.

    \implies The characteristic equation is

(λ+2iπ)(λ2iπ)=0    λ2+2iπλ2iπλ(2iπ)2=0    λ2+4π2=0\begin{align*} (\lambda + 2i \pi)(\lambda - 2i \pi) = 0 &\iff \lambda ^2 + \cancel{2i \pi\lambda} - \cancel{2i \pi\lambda} - (2i \pi)^2 = 0 \\ &\iff \lambda ^2 + 4 \pi ^2 = 0 \end{align*}

    \implies The equation is y+4π2y=0\boxed{{\color{Fuchsia}{ y'' + 4 \pi ^2 y= 0}}}

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