Find the solution of the differential equation r' (t) = (et-6, 2-1, 1) with the initial conditions r(1) = (0, 0, 11), r' (1) = (7, 0, 0). (Use symbolic notation and fractions where needed. Give your answer in vector form.) r(t) = Incorrect 1 61-6 41 36 + t+ 245 - 2 + ¹ -1.2- A 36 12 -t+ 23 2
Find the solution of the differential equation r' (t) = (et-6, 2-1, 1) with the initial conditions r(1) = (0, 0, 11), r' (1) = (7, 0, 0). (Use symbolic notation and fractions where needed. Give your answer in vector form.) r(t) = Incorrect 1 61-6 41 36 + t+ 245 - 2 + ¹ -1.2- A 36 12 -t+ 23 2
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Solving Differential Equations - Example
Consider the differential equation given by:
\[ \mathbf{r}''(t) = \left( e^{6t} - t^2 - 1, t, 1 \right) \]
with the initial conditions:
\[ \mathbf{r}(1) = \langle 0, 0, 11 \rangle \]
\[ \mathbf{r}'(1) = \langle 7, 0, 0 \rangle \]
We need to find the solution \( \mathbf{r}(t) \) using symbolic notation and fractions where needed, and giving the answer in vector form.
In the provided image, the attempted solution is:
\[ \mathbf{r}(t) = \left\langle \frac{1}{36} e^{6t} - 6 + \frac{41}{6} + \frac{245}{36}, \frac{t^2}{12} - \frac{2}{3} t + 1, \frac{t^2}{4} - \frac{23}{2} \right\rangle \]
However, this solution was marked as incorrect.
### Explanation of Notation and Process
1. **Given Differential Equation**:
\[ \mathbf{r}''(t) = \left( e^{6t} - t^2 - 1, t, 1 \right) \]
This is a second-order vector differential equation, indicating we deal with the second derivative of a vector function.
2. **Initial Conditions**:
\[
\mathbf{r}(1) = \langle 0, 0, 11 \rangle \quad \text{and} \quad \mathbf{r}'(1) = \langle 7, 0, 0 \rangle
\]
3. **Integration**:
To find \( \mathbf{r}(t) \), start by integrating \( \mathbf{r}''(t) \):
\[
\mathbf{r}'(t) = \int \mathbf{r}''(t) \, dt + \mathbf{C_1}
\]
Integrating each component:
\[
r_1''(t) = e^{6t} - t^2 - 1 \Rightarrow r](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe732258a-c25c-432c-8516-ff507048c7b7%2F627ea0d4-9a1d-4f4b-bba0-55fcaf8d5b42%2Fz8b4dt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Differential Equations - Example
Consider the differential equation given by:
\[ \mathbf{r}''(t) = \left( e^{6t} - t^2 - 1, t, 1 \right) \]
with the initial conditions:
\[ \mathbf{r}(1) = \langle 0, 0, 11 \rangle \]
\[ \mathbf{r}'(1) = \langle 7, 0, 0 \rangle \]
We need to find the solution \( \mathbf{r}(t) \) using symbolic notation and fractions where needed, and giving the answer in vector form.
In the provided image, the attempted solution is:
\[ \mathbf{r}(t) = \left\langle \frac{1}{36} e^{6t} - 6 + \frac{41}{6} + \frac{245}{36}, \frac{t^2}{12} - \frac{2}{3} t + 1, \frac{t^2}{4} - \frac{23}{2} \right\rangle \]
However, this solution was marked as incorrect.
### Explanation of Notation and Process
1. **Given Differential Equation**:
\[ \mathbf{r}''(t) = \left( e^{6t} - t^2 - 1, t, 1 \right) \]
This is a second-order vector differential equation, indicating we deal with the second derivative of a vector function.
2. **Initial Conditions**:
\[
\mathbf{r}(1) = \langle 0, 0, 11 \rangle \quad \text{and} \quad \mathbf{r}'(1) = \langle 7, 0, 0 \rangle
\]
3. **Integration**:
To find \( \mathbf{r}(t) \), start by integrating \( \mathbf{r}''(t) \):
\[
\mathbf{r}'(t) = \int \mathbf{r}''(t) \, dt + \mathbf{C_1}
\]
Integrating each component:
\[
r_1''(t) = e^{6t} - t^2 - 1 \Rightarrow r
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