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6.2 Ordinary Differential Equations • 239and> laplace( eqn2, t, s );m (s (s laplace(x 2 (t), t, s) − x 2 (0)) − D(x 2 )(0)) =k (laplace(x 1 (t), t, s) − laplace(x 2 (t), t, s))Evaluate the set consisting of the two transforms at the initial conditions.> eval( {%, %%}, {ini} );{m s 2 laplace(x 2 (t), t, s) =k (laplace(x 1 (t), t, s) − laplace(x 2 (t), t, s)),α m s (s laplace(x 1 (t), t, s) − 2) =k (laplace(x 2 (t), t, s) − laplace(x 1 (t), t, s)) + e(−s)s }You must solve this set of algebraic equations for the Laplace transformsof the two functions x 1 (t) and x 2 (t).> sol := solve( %, { laplace(x[1](t),t,s),> laplace(x[2](t),t,s) } );sol := {laplace(x 1 (t), t, s) = (m s2 + k) (2 α m s 2 e s + 1)e s s 3 m (k + α m s 2 + α k) ,laplace(x 2 (t), t, s) =k (2 α m s 2 e s + 1)e s s 3 m (k + α m s 2 + α k) }Maple has solved the algebraic problem. You must take the inverseLaplace transform to get the functions x 1 (t) and x 2 (t) .> invlaplace( %, s, t );

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