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équation différentielles du second ordre


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Posted

Coucou @jlr.10!

 

Dans l'item B la fonction qu'on te donne c'est f(t) = sin(3t) + 1/9.

Si on la dérive ça donne f'(t) = 3 * cos(3t).

Si on calcule sa dérivée seconde ça donne f''(t) =-9 * sin(3t).

 

On a donc bien:

f''(t) + 9 * f(t)-9 * sin(3t) + 9 * [sin(3t) + 1/9] = 9 (sin(3t) - sin(3t) + 1/ 9) = 1

 

On teste les valeurs de l'énoncé soient f(0) et f'(0) :

f(0) = sin(0) + 1/9

f(0) = 0 +1/9

f(0) = 1/9

 

f'(0) = 3 * cos(0)

f'(0) = 3 * 1

f'(0) = 3

 

Ca colle bien avec les résultats qu'on nous propose donc l'item est vrai.

 

C'est plus clair ?

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