Let's **determine the interval of definition** for the solution of the given initial-value problem: \[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \frac{1}{t^2 - 16} \begin{pmatrix} e^{2t} & \sin(2t) \\ t^2 & e^{-3t} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} \log(t^2) \\ 0 \end{pmatrix} ,\quad x(1)=3,\ y(1)=7 \] --- ### **Step 1: Identify Points of Discontinuity** The system is in the standard form: \[ \vec{X}' = \vec{F}(t, \vec{X}) \] where \(\vec{F}(t, \vec{X})\) is continuous **except where denominators vanish or logarithms are undefined.** #### 1. **Denominator \(t^2 - 16\):** - \(t^2 - 16 = 0 \implies t^2 = 16 \implies t = 4 \text{ or } t = -4\) - So, \(\vec{F}(t, \vec{X})\) is **undefined at \(t = -4\) and \(t = 4\)**. #### 2. **Logarithm \(\log(t^2)\):** - \(\log(t^2)\) is defined for all \(t \neq 0\). #### 3. **Initial Condition:** - Given at \(t = 1\). --- ### **Step 2: Construct the Largest Interval Containing \(t=1\) and Avoiding Discontinuities** The function is defined when: - \(t \neq -4,\, t \neq 0,\, t \neq 4\) But the only discontinuities near \(t=1\) are at \(t=0\) and \(t=4\). - The largest open interval containing \(t=1\) and not including \(t=0\) or \(t=4\) is \((0, 4)\). --- ### **Step 3: State the Final Interval of Definition** \[ \boxed{(0,\ 4)} \] --- ### **Summary** - The solution is defined on the largest interval containing \(t=1\) and avoiding points where the function is undefined. - **Final answer:** \(\boxed{(0,\ 4)}\)

Let's carefully analyze the initial-value problem to determine its interval of existence. --- ### **Given System:** \[ \frac{d}{dt} \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = \frac{1}{t^2 - 16} \begin{pmatrix} e^{2t} & \sin(2t) \\ t^2 & e^{-3t} \end{pmatrix} \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} + \begin{pmatrix} \log(t^2) \\ 0 \end{pmatrix} ,\quad \text{with } x(1)=3,\ y(1)=7 \] --- ### **Step 1: Identify the points where the system's right-hand side is discontinuous** The system is of the form: \[ \mathbf{X'} = \mathbf{F}(t, \mathbf{X}) \] where: \[ \mathbf{F}(t, \mathbf{X}) = \frac{1}{t^2 - 16} \begin{bmatrix} e^{2t} & \sin(2t) \\ t^2 & e^{-3t} \end{bmatrix} \mathbf{X} + \begin{bmatrix} \log(t^2) \\ 0 \end{bmatrix} \] The key to determining the interval of existence is to identify where \(\mathbf{F}(t, \mathbf{X})\) is **not continuous**. --- ### **Step 2: Analyze discontinuities** **(A) Denominator \(t^2 - 16\)** - Discontinuity occurs where \(t^2 - 16 = 0\), i.e., \[ t^2 = 16 \Rightarrow t = \pm 4 \] - Therefore, discontinuities at \(t = -4\) and \(t=4\). --- **(B) Logarithm \(\log(t^2)\)** - Defined for all \(t\) such that \(t^2 > 0\), i.e., for all \(t \neq 0\). - Discontinuity at \(t=0\). --- ### **Step 3: Determine the maximal interval containing the initial point \(t=1\)** - The initial condition is at \(t=1\). - The points where discontinuities occur are at \(t = -4, 0, 4\). - To find the **maximal interval** containing \(t=1\), we consider the neighboring discontinuities: - To the **left** of 1: the closest discontinuity is at \(t=0\). - To the **right** of 1: the closest discontinuity is at \(t=4\). - Since the problem is well-defined **between** the initial point and the discontinuities, the **largest interval** containing \(t=1\) and avoiding the points where the system is undefined is: \[ (0, 4) \] - Note: The interval cannot include \(t=0\) or \(t=4\) because at these points, the function is not continuous (denominator zero or logarithm undefined). --- ### **### **Final Answer:** \[ \boxed{ \textbf{Interval of definition: } (0, 4) } \] This is the maximal interval containing \(t=1\) where the solution exists and is unique, according to the standard existence and uniqueness theorem for ODEs. --- ### **Summary of the steps:** 1. Identified points where the RHS is discontinuous: \(t=-4, 0, 4\). 2. Recognized the initial point at \(t=1\). 3. Determined the largest interval containing \(t=1\) free of discontinuities: \((0, 4)\). **Thus, the solution exists and is unique on \(\boxed{(0, 4)}\).**