Solve the initial value problem.
-Differential Equation: $\frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } = \frac { 3 } { 2 } ( \mathrm { t } + 3 ) ^ { 1 / 2 } \mathbf { i } + \mathrm { e } ^ { \mathrm { t } } \mathbf { j }$
Initial Condition: $\mathbf { r } ( 0 ) = 0$
A) $r ( t ) = \left[ ( t + 3 ) ^ { 3 / 2 } - 3 ^ { 3 / 2 } \right] \mathbf { i } + \left( e ^ { t } - 1 \right) \mathbf { j }$
B) $r ( t ) = \left[ ( t + 3 ) ^ { 5 / 2 } - 3 ^ { 5 / 2 } \right] \mathbf { i } + \left( e ^ { t } - 1 \right) \mathbf { j }$
C) $\mathbf { r } ( \mathrm { t } ) = \left[ ( \mathrm { t } + 3 ) ^ { 3 / 2 } + 3 ^ { 3 / 2 } \right] \mathbf { i } + \left( \mathrm { e } ^ { \mathrm { t } } + 1 \right) \mathbf { j }$
D) $\mathbf { r } ( \mathrm { t } ) = ( \mathrm { t } + 3 ) ^ { 3 / 2 } \mathbf { i } + \mathrm { e } _ { \mathbf { j } } \mathbf { j }$

Question

Quizplus · Accepted Answer

To solve the differential equation, we integrate each component with respect to time:
$$\int \frac { \mathrm { d } x } { ( t + 3 ) ^ { 1 / 2 } }= \int \frac { 3 } { 2 } \mathrm { dt }$$
$$2 \left. ( t + 3 ) ^ { 1 / 2 } \right| _ { 0 } ^ { t }= 3 t + \mathrm { C }$$
$$( t + 3 ) ^ { 1 / 2 } - 3 = \frac { 3 } { 2 } t + \frac { 3 } { 2 } \mathrm { C }$$
Using the initial condition $\mathbf { r } ( 0 ) = 0$, we find that $\mathrm { C } = - 3$. So the expression for $\mathbf { r } ( t )$ becomes:
$$r ( t ) = \left[ ( t + 3 ) ^ { 3 / 2 } - 3 ^ { 3 / 2 } \right] \mathbf { i } + \left( e ^ { t } - 1 \right) \mathbf { j }$$

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

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