Let's solve both parts step by step. Theorem 3.1.1 is the **existence and uniqueness theorem** for initial value problems (IVPs) for linear ODEs, which typically states: > If the functions \( a_n(x), a_{n-1}(x), ..., a_0(x), f(x) \) are continuous on an open interval \( I \) containing \( x_0 \), then the IVP has a unique solution on \( I \). Let's analyze each problem: --- ## **(a)** Given the IVP: \[ (x+5)y''' + (x^2-64)y'' + 18y = \frac{1}{x+6},\quad y(0)=5,\ y'(0)=8,\ y''(0)=1 \] ### **Step 1: Write in Standard Form** Divide both sides by \(x+5\) (as long as \(x \neq -5\)): \[ y''' + \frac{x^2-64}{x+5}y'' + \frac{18}{x+5}y = \frac{1}{(x+5)(x+6)} \] ### **Step 2: Identify Points of Discontinuity** - \( \frac{x^2 - 64}{x+5} \): Discontinuous at \(x = -5\) - \( \frac{18}{x+5} \): Discontinuous at \(x = -5\) - \( \frac{1}{(x+5)(x+6)} \): Discontinuous at \(x = -5\) and \(x = -6\) ### **Step 3: Largest Interval around \(x=0\)** - The points of discontinuity are at \(x = -5\) and \(x = -6\). - The initial value \(x_0 = 0\). The largest interval containing \(x = 0\) and not containing \(x = -5\) or \(x = -6\) is \( (-5, \infty) \) since \(-6 < -5 < 0\). **Final Answer for (a):** \[ \boxed{(-5, \infty)} \] --- ## **(b)** Given the IVP: \[ (x-5)y''' + (x^2-64)y'' + 18y = \frac{1}{x+6},\quad y(0)=5,\ y'(0)=8,\ y''(0)=1 \] ### **Step 1: Write in Standard Form** Divide both sides by \(x-5\) (as long as \(x \neq 5\)): \[ y''' + \frac{x^2-64}{x-5}y'' + \frac{18}{x-5}y = \frac{1}{(x-5)(x+6)} \] ### **Step 2: Identify Points of Discontinuity** - \( \frac{x^2 - 64}{x-5} \): Discontinuous at \(x = 5\) - \( \frac{18}{x-5} \): Discontinuous at \(x = 5\) - \( \frac{1}{(x-5)(x+6)} \): Discontinuous at \(x = 5\) and \(x = -6\) ### **Step 3: Largest Interval around \(x=0\)** - The points of discontinuity are at \(x = 5\) and \(x = -6\). - The initial value \(x_0 = 0\). The largest interval containing \(x = 0\) and not containing \(x = 5\) or \(x = -6\) is \( (-6, 5) \). **Final Answer for (b):** \[ \boxed{(-6,\,5)} \] --- ### **Summary Table** | Part | Largest Interval for Unique Solution | |------|---------------------------------------| | (a) | \((-5,\,\infty)\) | | (b) | \((-6,\,5)\) | --- If you need the reasoning behind each step or more details on the theorem, let me know!

Let's analyze both problems carefully, step-by-step, with detailed explanations. --- ### Background: Theorem 3.1.1 (Existence and Uniqueness Theorem) This theorem states that for a linear differential equation of order \( n \): \[ a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + \dots + a_1(x) y' + a_0(x) y = f(x), \] if the functions \(a_i(x)\) and \(f(x)\) are continuous on an open interval \(I\), then the initial value problem (IVP) \[ y^{(k)}(x_0) = y_k, \quad k=0,1,\dots,n-1, \] has a unique solution on some subinterval of \(I\) containing \(x_0\). **Key Point:** The largest such interval is determined by the points where the coefficients \(a_i(x)\) or \(f(x)\) are discontinuous. --- ## **Part (a)** ### Given: \[ ( x+5 ) y''' + ( x^2 - 64 ) y'' + 18 y = \frac{1}{x+6}, \] with initial conditions at \(x=0\): \[ y(0)=5,\quad y'(0)=8,\quad y''(0)=1. \] --- ### **Step 1: Write the differential equation in standard form** Divide through by \(x+5\) (assuming \(x \neq -5\)): \[ y''' + \frac{x^2 - 64}{x+5} y'' + \frac{18}{x+5} y = \frac{1}{(x+5)(x+6)}. \] Now, the equation is in the form: \[ y''' + p_2(x) y'' + p_1(x) y' + p_0(x) y = q(x), \] where: \[ p_2(x) = \frac{x^2 - 64}{x+5}, \quad p_1(x)=0, \quad p_0(x)=\frac{18}{x+5}, \quad q(x)= \frac{1}{(x+5)(x+6)}. \] ### **Step 2: Identify points of discontinuity** - \( p_2(x) \), \( p_0(x) \), and \( q(x) \) are discontinuous where denominators are zero: - At \(x = -5\), - At \(x = -6\). ### **Step 3: Find the largest interval containing \(x=0\) without discontinuities** - The discontinuities are at \(x=-6\) and \(x=-5\). - Since \(x=0\) lies between \(-6\) and \(-5\), the largest interval containing \(0\) **and** avoiding the discontinuities is: \[ (-5, \infty). \] This is because: - To the left of \(x=-6\), the interval ends at \(-6\), - To the right of \(-5\), the interval extends to infinity. But, note that the interval must **not** include any point where discontinuities occur. So, the largest open interval containing \(0\) and not crossing any discontinuity is: \[ \boxed{(-5, \infty)}. \] --- ### **Final Answer for part (a):** \[ \boxed{\textbf{Largest interval: } (-5, \infty).} \] --- ## **Part (b)** ### Given: \[ ( x-5 ) y''' + ( x^2 - 64 ) y'' + 18 y = \frac{1}{x+6}, \] with initial conditions at \(x=0\): \[ y(0)=5,\quad y'(0)=8,\quad y''(0)=1. \] --- ### **Step 1: Write in standard form** Divide through by \(x-5\) (assuming \(x \neq 5\)): \[ y''' + \frac{x^2 - 64}{x-5} y'' + \frac{18}{x-5} y = \frac{1}{(x-5)(x+6)}. \] The coefficients are: \[ p_2(x) = \frac{x^2 - 64}{x-5}, \quad p_1(x) = 0, \quad p_0(x) = \frac{18}{x-5}, \quad q(x) = \frac{1}{(x-5)(x+6)}. \] ### **Step 2: Identify points of discontinuity** Discontinuities occur where denominators are zero: - At \(x=5\), - At \(x=-6\). ### **Step 3: Find the largest interval containing \(x=0\) without discontinuities** - The points are at \(-6\) and \(5\). - The initial point \(x=0\) lies between \(-6\) and \(5\). Therefore, the largest interval containing \(0\) and avoiding the discontinuities is: \[ (-6, 5). \] --- ### **Final answer for part (b):** \[ \boxed{\textbf{Largest interval: } (-6, 5).} \] --- ## **Summary** | Part | Discontinuities | Largest Interval containing \(x=0\) | |--------|-------------------|------------------------------------| | (a) | \(x = -5, -6\) | \((-5, \infty)\) | | (b) | \(x = -6, 5\) | \((-6, 5)\) | --- ## **Key Takeaways:** - The theorem guarantees a unique solution on an interval where all coefficients and the forcing function are continuous. - Discontinuities in coefficients or the forcing function determine the limits of the interval. - The largest such interval around the initial point \(x_0=0\) is the one that avoids these discontinuities. --- If you'd like, I can also explain why these points cause discontinuities or how to verify the continuity of the functions.

Great question! Let's clarify why the two problems, although similar in form, lead to different largest intervals for guaranteed unique solutions. --- ### **Understanding the Differences Between the Two Problems** **1. The Structure of the Differential Equations** - **Part (a):** \[ (x+5) y''' + (x^2 - 64) y'' + 18 y = \frac{1}{x+6} \] - **Part (b):** \[ (x-5) y''' + (x^2 - 64) y'' + 18 y = \frac{1}{x+6} \] **2. Points of Discontinuity** The key to understanding the largest interval of existence and uniqueness is **where the coefficients and forcing functions are discontinuous**. - **In Part (a):** - The coefficients involve denominators \(x+5\), which are discontinuous at \(x=-5\). - The forcing function involves \(\frac{1}{x+6}\), which is discontinuous at \(x=-6\). - **In Part (b):** - The coefficients involve denominators \(x-5\), which are discontinuous at \(x=5\). - The forcing function is the same, \(\frac{1}{x+6}\), discontinuous at \(x=-6\). **3. How These Discontinuities Affect the Largest Interval** - The **initial conditions** are given at \(x=0\) in both cases. - The **interval** where the theorem guarantees a **unique solution** is the **largest interval** containing \(x=0\) that **does not** include any points where the coefficients or forcing function are discontinuous. --- ### **Why are the answers different?** - **In Part (a):** - Discontinuities at **\(x=-6\)** and **\(x=-5\)**. - The initial point is at \(x=0\). - The interval containing \(x=0\) **and avoiding** \(-6\) and \(-5\) is **from \(-5\) to \(\infty\)** because: - To the **left** of \(-6\), the interval would end at \(-6\). - To the **left** of \(-5\), the interval ends at \(-5\). - Since \(x=0\) lies between \(-6\) and \(-5\), the **largest open interval** containing \(0\) **without crossing** any discontinuity is **\((-5, \infty)\)**. - **In Part (b):** - Discontinuities at **\(x=-6\)** and **\(x=5\)**. - The initial point is again at \(x=0\). - The interval must avoid crossing \(x=5\) and \(x=-6\). - The largest such interval containing \(0\) is **\((-6, 5)\)**. --- ### **Summary** | Aspect | Part (a) | Part (b) | |---------|------------|------------| | Discontinuities | \(x=-6, -5\) | \(x=-6, 5\) | | Interval around \(x=0\) | \(\boxed{(-5, \infty)}\) | \(\boxed{(-6, 5)}\) | --- ### **In essence:** Though the equations look similar, the points where coefficients are undefined (discontinuities) differ because of the different factors in the denominators (\(x+5\) vs. \(x-5\)). These differences directly impact the largest interval on which the existence and uniqueness theorem applies, explaining why the answers differ. --- **If you'd like, I can help you analyze the steps further or clarify any specific part!**

Certainly! The final answers are: **(a)** Largest interval: **\((-5, \infty)\)** **(b)** Largest interval: **\((-6, 5)\)**