5. Solve the second order Initial Value Problem (IVP): y" + 4y' + 5y = 10e' - 10u2(t)e,y(0) = 0, y'(0) = 10 The form of the solution should be: v(t) = ce + de"[cos(at - p)] for 0st< 2 he' + keb [cos(at – q)] for 2st Try not to round very much until the final answer is achieved. Give the value of p in radians (Round to 3 decimal places, 0
5. Solve the second order Initial Value Problem (IVP): y" + 4y' + 5y = 10e' - 10u2(t)e,y(0) = 0, y'(0) = 10 The form of the solution should be: v(t) = ce + de"[cos(at - p)] for 0st< 2 he' + keb [cos(at – q)] for 2st Try not to round very much until the final answer is achieved. Give the value of p in radians (Round to 3 decimal places, 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Value of P
![### Problem 5: Solving a Second Order Initial Value Problem (IVP)
Given the second order Initial Value Problem (IVP):
\[ y'' + 4y' + 5y = 10e^t - 10u_2(t)e^t, \quad y(0) = 0, \quad y'(0) = 10 \]
### Form of the Solution:
The form of the solution should be:
\[
y(t) =
\begin{cases}
ce^t + de^{bt}[\cos(at - p)] & \text{for } 0 \le t \le 2 \\
he^t + ke^{bt}[\cos(at - q)] & \text{for } 2 \le t
\end{cases}
\]
### Instructions:
- Try not to round very much until the final answer is achieved.
- Give the value of \( p \) in radians (Round to 3 decimal places, \( 0 < p < 6.28 \))
### Detailed Explanation:
1. **Equation Components:**
- \( y'' \) refers to the second derivative of \( y \) with respect to \( t \).
- \( y' \) is the first derivative of \( y \) with respect to \( t \).
- \( y \) is the function of \( t \) we need to solve for.
- \( e^t \) is the exponential function.
- \( u_2(t) \) is the Heaviside step function which is 0 for \( t < 2 \) and 1 for \( t \ge 2 \).
2. **Initial Conditions:**
- \( y(0) = 0 \): At \( t = 0 \), the function \( y \) is 0.
- \( y'(0) = 10 \): At \( t = 0 \), the first derivative of \( y \) is 10.
3. **Solution Form:**
- The solution is piecewise, meaning it has different forms based on the interval of \( t \).
- For \( 0 \le t \le 2 \):
\[
y(t) = ce^t + de^{bt}[\cos(at - p)]
\]
where \( c \), \( d \), \( b \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd8339843-a6c0-40e6-979c-08fc9d6ca9a2%2Ffeec6b1b-2138-451d-a6c1-3686214b48d0%2Fgqlfnt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 5: Solving a Second Order Initial Value Problem (IVP)
Given the second order Initial Value Problem (IVP):
\[ y'' + 4y' + 5y = 10e^t - 10u_2(t)e^t, \quad y(0) = 0, \quad y'(0) = 10 \]
### Form of the Solution:
The form of the solution should be:
\[
y(t) =
\begin{cases}
ce^t + de^{bt}[\cos(at - p)] & \text{for } 0 \le t \le 2 \\
he^t + ke^{bt}[\cos(at - q)] & \text{for } 2 \le t
\end{cases}
\]
### Instructions:
- Try not to round very much until the final answer is achieved.
- Give the value of \( p \) in radians (Round to 3 decimal places, \( 0 < p < 6.28 \))
### Detailed Explanation:
1. **Equation Components:**
- \( y'' \) refers to the second derivative of \( y \) with respect to \( t \).
- \( y' \) is the first derivative of \( y \) with respect to \( t \).
- \( y \) is the function of \( t \) we need to solve for.
- \( e^t \) is the exponential function.
- \( u_2(t) \) is the Heaviside step function which is 0 for \( t < 2 \) and 1 for \( t \ge 2 \).
2. **Initial Conditions:**
- \( y(0) = 0 \): At \( t = 0 \), the function \( y \) is 0.
- \( y'(0) = 10 \): At \( t = 0 \), the first derivative of \( y \) is 10.
3. **Solution Form:**
- The solution is piecewise, meaning it has different forms based on the interval of \( t \).
- For \( 0 \le t \le 2 \):
\[
y(t) = ce^t + de^{bt}[\cos(at - p)]
\]
where \( c \), \( d \), \( b \
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