Question

Calcule o Wronskiano dos seguintes pares de função :

a) e^at
, b^at
b) t, t ln t

c) Sen at , cos bt

Answer 1

Definição:

O Wronskiano de funções  $f_1(x), f_2(x), \dots, f_n(x)$ é dada por

            $W(f_1, f_2,\dots,f_n) = \det \begin{vmatrix}f_1(x) &f_2(x)&\dots&f__n(x)\\f_1'(x)&f_2'(x)&\dots&f_n'(x)\\\vdots&\vdots&\ddots&\vdots\\f_1^{(n)}(x)&f_2^{(n)}(x)&\dots &f_n^{(n)}(x)\end{vmatrix}$

Resposta:

a) $W(\sin(at),\cos(bt)) = \begin{vmatrix}\sin(at)&\cos(bt)\\a\cos(at)&-b\sin(bt) \end{vmatrix} = -b\sin(at)\sin(bt) - a\cos(at)\cos(bt)$

b)    $W = \begin{vmatrix}\sin^2t&1-\cos(2t)\\2\cos(t)\sin(t)&2\sin(2t)\end{vmatrix} =$                      $2\sin^2(t)\sin(2t)+2\cos(2t)\cos(t)\sin(t) - 2\cos(t)\sin(t)$

c)         $W = \begin{vmatrix}e^{at}&e^{bt}\\ae^{at}&be^{bt}\end{vmatrix} = be^{t(b+a)} - ae^{t(b+a)}$

d) $W = \begin{vmatrix}e^{at}&te^{at}\\ae^{at}&e^{at} + ate^{at}\end{vmatrix} = e^{2at} + ate^{2at} - ate^{2at} = e^{2at}$

e)  $W = \begin{vmatrix}t&t\ln(t)\\1&\ln(t) + 1\end{vmatrix} = t$

f)             $W = \begin{vmatrix}e^{at}\sin(bt)&e^{at}\cos(bt)\\ae^{at}\sin(bt) +be^{at}\cos(bt)&ae^{at}\cos(bt) - be^{at}\sin(bt)\end{vmatrix}$

$W = e^{2at}(a\sin(bt)\cos(bt) - b\sin^2(bt)) - e^{2at}(a\sin(bt)\cos(bt) + b\cos^2(bt))$

                       $W = -be^{2at}(\sin^2(bt)+\cos^2(bt)) = -be^{2at}$