Find the laplace transform of f(t)=t(3sin2t-2cos2t)
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I can't understand your language sorry...
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-2s^2+12s+8/(s^2+4)^2
Step-by-step explanation:
L[3sin2t-2cos2t]
L[sinat]=a/s^2+a^2
L[cosat]=s/s^2+a^2
L[3sin2t]=6/s^2+4
L[2cos2t]=2s/s^2+4
L[t(3sin2t-2cos2t)]=
= -d/ds[(6/s^2+4)-(2s/s^2+4)]
= -d/ds(6-2s/s^2+4)
= -[(s^2+4)(-2)-(6-2s)(2s)/(s^2+4)^2]
=-[2s^2-12s-8/(s^2+4)^2]
= -2s^2+12s+8/(s^2+4)^2
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