Please I need it done today and can you help me ### Solving Differential Equations - ExampleConsider 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 Process1. **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

r''(t) =<e6t-6, t2-1, 1>r(1)=<0,0,11>r'(1)=<7,0,0>

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