Solve the initial value problem.
-Differential Equation: $\frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } = \left( \mathrm { t } ^ { 4 } + 5 \mathrm { t } ^ { 2 } \right) \mathbf { i } + 4 \mathrm { t } \mathbf { j }$
Initial Condition: $\mathbf { r } ( 0 ) = 5 \mathbf { i } - 6 \mathbf { j }$
A) $r ( t ) = \left( \frac { t ^ { 5 } } { 5 } + \frac { 5 t ^ { 3 } } { 3 } - 6 \right) i + \left( 2 t ^ { 2 } + 5 \right) \mathbf { j }$
B) $\mathbf { r } ( \mathrm { t } ) = \left( \frac { \mathrm { t } ^ { 5 } } { 5 } + \frac { 5 \mathrm { t } ^ { 3 } } { 3 } \right) \mathbf { i } + 2 \mathrm { t } ^ { 2 } \mathbf { j }$
C) $r ( t ) = \left( \frac { t ^ { 5 } } { 5 } + \frac { 5 t ^ { 3 } } { 3 } + 5 \right) i + \left( 2 t ^ { 2 } - 6 \right) \mathbf { j }$
D) $r ( t ) = \left\{ \frac { t ^ { 5 } } { 5 } + \frac { 5 t ^ { 3 } } { 3 } + 5 \right) i + \left( t ^ { 2 } - 6 \right) j$

Question

Quizplus · Accepted Answer

Integrating the given differential equation with respect to $t$ gives the components of $\mathbf{r}(t)$. For the $\mathbf{i}$ component, $\int (t^4 + 5t^2) dt = \frac{t^5}{5} + \frac{5t^3}{3} + C$. For the $\mathbf{j}$ component, $\int 4t dt = 2t^2 + C$. Using the initial condition $\mathbf{r}(0) = 5\mathbf{i} - 6\mathbf{j}$ to solve for $C$, we find $C = 5$ for the $\mathbf{i}$ component and $C = -6$ for the $\mathbf{j}$ component, leading to the correct expression for $\mathbf{r}(t)$.

Solve the Initial Value Problem Initial Condition r(0)=5i−6j\mathbf { r } ( 0 ) = 5 \mathbf { i } - 6 \mathbf { j }r(0)=5i−6j

Solve the Initial Value Problem Initial Condition $\mathbf { r } ( 0 ) = 5 \mathbf { i } - 6 \mathbf { j }$