2. (Section 5.6, problem 11) Find the solution of the following IVP: y" + 2y' + 10y h(t – 2), y(0) y'(0) 0, 0.

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### Section 5.6, Problem 11: Solving the Initial Value Problem (IVP)

**Problem Statement:**

Find the solution of the following IVP:

\[
y'' + 2y' + 10y = h(t - 2),
\]

with the initial conditions:

\[
y(0) = 0, \quad y'(0) = 0.
\]

**Solution Approach:**

The problem mentions that the Laplace transform of the solution is given by:

\[
Y(s) = \frac{e^{-2s}}{s(s^2 + 2s + 10)}.
\]

**Additional Information:**

Note that the source reference uses \(u(t)\) in place of \(h(t)\).

### Explanation:

- **Equation:** The given differential equation is a second-order linear ordinary differential equation with constant coefficients.
  
- **Initial Conditions:** The conditions \(y(0) = 0\) and \(y'(0) = 0\) imply that initially, both the function and its first derivative are zero.

- **Laplace Transform:** The Laplace transform is used to convert differential equations into algebraic equations in the complex frequency domain, which are often easier to solve.

- **Expression for \(Y(s)\):** The solution in the s-domain shows the impact of a delayed impulse, represented by the exponential term \(e^{-2s}\), and a denominator indicative of a second-order system.

### Key Concepts:

- **Laplace Transform in Control Systems:** This approach is widely used in control systems to handle initial conditions and delay terms efficiently.
  
- **Interpretation of Poles:** The denominator \(s(s^2 + 2s + 10)\) suggests a system with one real pole at \(s=0\) and two complex poles, which determine the system's transient response.
Transcribed Image Text:### Section 5.6, Problem 11: Solving the Initial Value Problem (IVP) **Problem Statement:** Find the solution of the following IVP: \[ y'' + 2y' + 10y = h(t - 2), \] with the initial conditions: \[ y(0) = 0, \quad y'(0) = 0. \] **Solution Approach:** The problem mentions that the Laplace transform of the solution is given by: \[ Y(s) = \frac{e^{-2s}}{s(s^2 + 2s + 10)}. \] **Additional Information:** Note that the source reference uses \(u(t)\) in place of \(h(t)\). ### Explanation: - **Equation:** The given differential equation is a second-order linear ordinary differential equation with constant coefficients. - **Initial Conditions:** The conditions \(y(0) = 0\) and \(y'(0) = 0\) imply that initially, both the function and its first derivative are zero. - **Laplace Transform:** The Laplace transform is used to convert differential equations into algebraic equations in the complex frequency domain, which are often easier to solve. - **Expression for \(Y(s)\):** The solution in the s-domain shows the impact of a delayed impulse, represented by the exponential term \(e^{-2s}\), and a denominator indicative of a second-order system. ### Key Concepts: - **Laplace Transform in Control Systems:** This approach is widely used in control systems to handle initial conditions and delay terms efficiently. - **Interpretation of Poles:** The denominator \(s(s^2 + 2s + 10)\) suggests a system with one real pole at \(s=0\) and two complex poles, which determine the system's transient response.
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