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

Solutions

VerifiedIn this case, we have a basis:

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

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

$\implies$ The characteristic equation is

$\begin{align*} (\lambda + 2i \pi)(\lambda - 2i \pi) = 0 &\iff \lambda ^2 + \cancel{2i \pi\lambda} - \cancel{2i \pi\lambda} - (2i \pi)^2 = 0 \\ &\iff \lambda ^2 + 4 \pi ^2 = 0 \end{align*}$

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

$\Rightarrow \textbf{Form :}\;\; \boxed{\;ay''+by'+cy=0\;}$

$Where\; : \;a\;,\;b\;,\;c \;\; \Rightarrow Constants$

$In\; order\; to\; solve\; this\; D.E \;, \;we\; need \;to\; get\; the\; "characterestic\; equation\;" by\; replacing$

$\boxed{y'' \;\; \Rightarrow \;\; m^{2}\;\;\;,\;\;\; y'\;\; \Rightarrow \;\; m\;\;\;,\;\;\;y\;\; \Rightarrow \;\; 1}$

$\textbf{Then Characterestic Eq :}\;\; \Rightarrow \;\;{\color{#c34632} {am^{2}+bm+c=0}}$

$m_{1,2}=\dfrac{-b \;\pm \sqrt{b^{2}-4ac}}{2a}\;\; \Rightarrow \;\; \textbf{Where :}\; \sqrt{b^{2}-4ac}=\;-ve$

$\boxed{\;\color{#19804f} {m_{1,2}=\alpha \pm \beta i}\;}$

$\therefore \;\; \textbf{Solution is : }\; \boxed{\color{#c34632} {y=e^{\alpha x}(C_{1}\cos(\beta x)+C_{2}\sin(\beta x))}}\;\;\; \Rightarrow \;\; \textbf{For Complix Roots}$

$\because \boxed{\;\color{#4257b2} {\cos(2\pi x)\;\;,\;\;\sin(2\pi x)}\;}$

$\therefore {\color{#19804f} {m_{1,2} = \pm 2\pi i}}\;\;\; \Rightarrow \;\; \boxed{\;\alpha =0\;\;,\;\;\beta =2\pi \;}$

$(m+2\pi i)(m-2\pi i)=0\;\;\; \Rightarrow \;\; \boxed{\;m^{2}+4\pi^{2} =0\;}$

$a=1\;\;,\;\;b=0\;\;,\;\;c=4\pi^{2}$

$\boxed{\;\color{#c34632} {y''+4\pi^{2} y=0}\;}$

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