To solve the differential equation [tex]\( \mathbf{r}''(t) = \left\langle e^{6t-6}, t^2 - 1, 1 \right\rangle \)[/tex] with the given initial conditions [tex]\( \mathbf{r}(1) = \langle 0, 0, 11 \rangle \)[/tex] and [tex]\( \mathbf{r}'(1) = \langle 7, 0, 0 \rangle \)[/tex], follow these steps:
1. Integrate [tex]\( \mathbf{r}''(t) \)[/tex] with respect to [tex]\( t \)[/tex] to find [tex]\( \mathbf{r}'(t) \)[/tex]:
Given:
[tex]\[
\mathbf{r}''(t) = \left\langle e^{6t - 6}, t^2 - 1, 1 \right\rangle
\][/tex]
Integrate each component with respect to [tex]\( t \)[/tex]:
[tex]\[
r''_1(t) = e^{6t - 6} \implies r'_1(t) = \int e^{6t - 6} \, dt = \frac{e^{6t - 6}}{6} + C_1
\][/tex]
[tex]\[
r''_2(t) = t^2 - 1 \implies r'_2(t) = \int (t^2 - 1) \, dt = \frac{t^3}{3} - t + C_2
\][/tex]
[tex]\[
r''_3(t) = 1 \implies r'_3(t) = \int 1 \, dt = t + C_3
\][/tex]
So,
[tex]\[
\mathbf{r}'(t) = \left\langle \frac{e^{6t - 6}}{6} + C_1, \frac{t^3}{3} - t + C_2, t + C_3 \right\rangle
\][/tex]
2. Use the initial condition [tex]\( \mathbf{r}'(1) = \langle 7, 0, 0 \rangle \)[/tex] to solve for [tex]\( C_1, C_2, \)[/tex] and [tex]\( C_3 \)[/tex]:
Plug in [tex]\( t = 1 \)[/tex] into [tex]\( \mathbf{r}'(t) \)[/tex]:
[tex]\[
\mathbf{r}'(1) = \left\langle \frac{e^{6 \cdot 1 - 6}}{6} + C_1, \frac{1^3}{3} - 1 + C_2, 1 + C_3 \right\rangle = \langle 7, 0, 0 \rangle
\][/tex]
Solving each component:
[tex]\[
\frac{e^0}{6} + C_1 = 7 \implies C_1 = 7 - \frac{1}{6} = \frac{42}{6} - \frac{1}{6} = \frac{41}{6}
\][/tex]
[tex]\[
\frac{1}{3} - 1 + C_2 = 0 \implies C_2 = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}
\][/tex]
[tex]\[
1 + C_3 = 0 \implies C_3 = -1
\][/tex]
Therefore:
[tex]\[
\mathbf{r}'(t) = \left\langle \frac{e^{6t - 6}}{6} + \frac{41}{6}, \frac{t^3}{3} - t + \frac{2}{3}, t - 1 \right\rangle
\][/tex]
3. Integrate [tex]\( \mathbf{r}'(t) \)[/tex] with respect to [tex]\( t \)[/tex] to find [tex]\( \mathbf{r}(t) \)[/tex]:
Integrate each component of [tex]\( \mathbf{r}'(t) \)[/tex]:
[tex]\[
r'_1(t) = \frac{e^{6t - 6}}{6} + \frac{41}{6} \implies r_1(t) = \int \left( \frac{e^{6t - 6}}{6} + \frac{41}{6} \right) \, dt = \frac{e^{6t - 6}}{36} + \frac{41t}{6} + C_4
\][/tex]
[tex]\[
r'_2(t) = \frac{t^3}{3} - t + \frac{2}{3} \implies r_2(t) = \int \left( \frac{t^3}{3} - t + \frac{2}{3} \right) \, dt = \frac{t^4}{12} - \frac{t^2}{2} + \frac{2t}{3} + C_5
\][/tex]
[tex]\[
r'_3(t) = t - 1 \implies r_3(t) = \int (t - 1) \, dt = \frac{t^2}{2} - t + C_6
\][/tex]
Thus:
[tex]\[
\mathbf{r}(t) = \left\langle \frac{e^{6t - 6}}{36} + \frac{41t}{6} + C_4, \frac{t^4}{12} - \frac{t^2}{2} + \frac{2t}{3} + C_5, \frac{t^2}{2} - t + C_6 \right\rangle
\][/tex]
4. Use the initial condition [tex]\( \mathbf{r}(1) = \langle 0, 0, 11 \rangle \)[/tex] to solve for [tex]\( C_4, C_5, \)[/tex] and [tex]\( C_6 \)[/tex]:
Plug in [tex]\( t = 1 \)[/tex] into [tex]\( \mathbf{r}(t) \)[/tex]:
[tex]\[
\mathbf{r}(1) = \left\langle \frac{e^0}{36} + \frac{41 \cdot 1}{6} + C_4, \frac{1^4}{12} - \frac{1^2}{2} + \frac{2 \cdot 1}{3} + C_5, \frac{1^2}{2} - 1 + C_6 \right\rangle = \langle 0, 0, 11 \rangle
\][/tex]
Solving each component:
[tex]\[
\frac{1}{36} + \frac{41}{6} + C_4 = 0 \implies C_4 = -\frac{41}{6} - \frac{1}{36} = -\frac{246}{36} - \frac{1}{36} = -\frac{247}{36}
\][/tex]
[tex]\[
\frac{1}{12} - \frac{1}{2} + \frac{2}{3} + C_5 = 0 \implies C_5 = -\frac{1}{12} + \frac{6}{12} - \frac{2}{12} = -\frac{5}{12}
\][/tex]
[tex]\[
\frac{1}{2} - 1 + C_6 = 11 \implies C_6 = 11 - \frac{1}{2} + 1 = \frac{21}{2}
\][/tex]
Therefore:
[tex]\[
\mathbf{r}(t) = \left\langle \frac{e^{6t - 6}}{36} + \frac{41t}{6} - \frac{247}{36}, \frac{t^4}{12} - \frac{t^2}{2} + \frac{2t}{3} - \frac{5}{12}, \frac{t^2}{2} - t + \frac{21}{2} \right\rangle
\][/tex]
Putting everything together, the solution is:
[tex]\[
\mathbf{r}(t) = \left\langle 7t + \frac{e^{6t-6}}{36}, \frac{t^4}{12} - \frac{t^2}{2}, \frac{t^2}{2} + 11 \right\rangle
\][/tex]
and the position vector at [tex]\( t = 1 \)[/tex] is:
[tex]\[
\mathbf{r}(1) = \left\langle \frac{253}{36}, -\frac{5}{12}, \frac{23}{2} \right\rangle
\][/tex]
