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theme 4 QCM 5


Go to solution Solved by l'addictologue,

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  • Solution
Posted (edited)

Tu transformes ton second membre ( 2cos(2t) ) en signe sinusoïdal complexe donc ça donne 2e2it et donc tu cherches une solution complexe de la forme fp(t) = C e2it tu dérives ça donne fp'(t) = 2iC e2it et tu redérives ça donne fp''(t) = 2i*2iC e2it = 4i²C e2it = -4C e2it t'injecte tout ça dans ton équation : 

-4C e2it - 3C e2it = 2e2it soit -4C - 3C = 2

Edited by l'addictologue

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