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Solve the Initial Value Problem A) s=sin(3t2+3t)4s = \sin \left( 3 t ^ { 2 } + 3 t \right) - 4

question 120

Multiple Choice

Solve the initial value problem.
- dsdt=(6t+3) sin(3t2+3t) ,s(0) =4\frac { \mathrm { ds } } { \mathrm { dt } } = ( 6 \mathrm { t } + 3 ) \sin \left( 3 \mathrm { t } ^ { 2 } + 3 \mathrm { t } \right) , \quad \mathrm { s } ( 0 ) = - 4

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