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I need to solve the following initial value problem via Laplace transform \begin{align*} \dot{\mathbf{x}} = \begin{pmatrix} 2 & -5 \\ 1 & -2 \end{pmatrix} \mathbf{x} + \begin{pmatrix} \sin t \\ \tan t \end{pmatrix} , \: \mathbf{x}(0) = \begin{pmatrix} -1 \\ 0 \end{pmatrix} \end{align*} but I don't know how to do that since there's no Laplace transform for the tangent function.

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  • $\begingroup$ Are u sure about using L.T.? $\endgroup$
    – Mikasa
    Commented May 30, 2013 at 5:16
  • $\begingroup$ Well, the problem appears in the L.T. section of a book on Diferential Equations so I'm assuming there must be a solution using L.T. $\endgroup$
    – Edward
    Commented May 30, 2013 at 5:18
  • $\begingroup$ Which book is this from? $\endgroup$
    – Amzoti
    Commented May 30, 2013 at 5:22
  • $\begingroup$ It's exercise 6 on page 370 in Braun's Differential Equations and Their Applications. You can read it here: books.google.com.mx/… $\endgroup$
    – Edward
    Commented May 30, 2013 at 5:25

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There is Laplace transform for tangent function, but it's complicated and using digamma function. Laplace transform

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