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Solve the Initial Value Problem Initial Condition r(0)=0\mathbf { r } ( 0 ) = 0

Question 96

Multiple Choice

Solve the initial value problem.
-Differential Equation: drdt=32(t+3) 1/2i+etj\frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } = \frac { 3 } { 2 } ( \mathrm { t } + 3 ) ^ { 1 / 2 } \mathbf { i } + \mathrm { e } ^ { \mathrm { t } } \mathbf { j }
Initial Condition: r(0) =0\mathbf { r } ( 0 ) = 0


A) r(t) =[(t+3) 3/233/2]i+(et1) jr ( t ) = \left[ ( t + 3 ) ^ { 3 / 2 } - 3 ^ { 3 / 2 } \right] \mathbf { i } + \left( e ^ { t } - 1 \right) \mathbf { j }
B) r(t) =[(t+3) 5/235/2]i+(et1) jr ( t ) = \left[ ( t + 3 ) ^ { 5 / 2 } - 3 ^ { 5 / 2 } \right] \mathbf { i } + \left( e ^ { t } - 1 \right) \mathbf { j }
C) r(t) =[(t+3) 3/2+33/2]i+(et+1) j\mathbf { r } ( \mathrm { t } ) = \left[ ( \mathrm { t } + 3 ) ^ { 3 / 2 } + 3 ^ { 3 / 2 } \right] \mathbf { i } + \left( \mathrm { e } ^ { \mathrm { t } } + 1 \right) \mathbf { j }
D) r(t) =(t+3) 3/2i+ejj\mathbf { r } ( \mathrm { t } ) = ( \mathrm { t } + 3 ) ^ { 3 / 2 } \mathbf { i } + \mathrm { e } _ { \mathbf { j } } \mathbf { j }

Correct Answer:

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