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College
Algebra
Linear algebra
Forms of linear equations

cistG

Answered question

2020-11-02

The coefficient matrix for a system of linear differential equations of the form

y_{1} = A y

has the given eigenvalues and eigenspace bases. Find the general solution for the system

λ_{1} = 3 + i \Rightarrow {[\begin{matrix} 2 i \\ i \end{matrix}]}, λ_{2} = 3 - i \Rightarrow {[\begin{matrix} - 2 i \\ - i \end{matrix}]}

Answer & Explanation

SoosteethicU

Skilled2020-11-03Added 102 answers

The general solution is
$y = c_{1} y_{1} + c_{2} y_{2}$
$for y_{1} = e^{a t} (\cos (b t) R e (u) - \sin (b t) I m (u))$
$y_{2} = e^{a t} (\sin (b t) R e (u) + \cos (b t) I m (u))$
We have
$λ_{1} = 3 + i \Rightarrow {[\begin{matrix} 2 i \\ i \end{matrix}]}, λ_{2} = 3 - i \Rightarrow {[\begin{matrix} - 2 i \\ - i \end{matrix}]}$
Then
$y_{1} = e^{3 t} (\cos (t) [\begin{matrix} 0 \\ 0 \end{matrix}] - \sin (t) [\begin{matrix} 2 \\ 1 \end{matrix}])$
$y_{2} = e^{t} (\sin (- t) [\begin{matrix} 0 \\ 0 \end{matrix}] + \cos (- t) [\begin{matrix} - 2 \\ - 1 \end{matrix}])$
Hence the general solution is
$y = c_{1} y_{1} + c_{2} y_{2}$
$= c_{1} e^{3 t} (\cos (t) [\begin{matrix} 0 \\ 0 \end{matrix}] - \sin (t) [\begin{matrix} 2 \\ 1 \end{matrix}]) + c_{2} e^{t} (\sin (- t) [\begin{matrix} 0 \\ 0 \end{matrix}] + \cos (- t) [\begin{matrix} - 2 \\ - 1 \end{matrix}])$
The individual functions are
$y_{1} = - 2 c_{1} e^{3 t} (\sin (t) + \cos (- t))$
$y_{2} = - c_{2} e^{3 t} (\sin (t) + \cos (- t))$

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