VIPSolutions4. The solution in the notes for periodic forcing of a system is correct if the system is damped and the forcing frequency is not equal to the natural frequency w. Consider the remaining case of an undamped oscillator forced at its natural frequency: (1) + y(t) = sin(20). [) lof2 f I li 0 Mime in 0 Figure 1: sin(107t) + sin(107t + 0), so wi = 10m, A = 1. The envelope appears to have frequency 2 = 2r. 0) Show that for some value of b the particular solution is bt cos(2t). What is b7 (b) Show then that for certain values of A, B y(t) = Asin(20) + Beos(21) + bt cos(2t) 0) solves (3) with initial conditions y(0) = 1, (0) = —1.
Question:
4. The solution in the notes for periodic forcing of a system is correct if the system is
damped and the forcing frequency is not equal to the natural frequency w. Consider the
remaining case of an undamped oscillator forced at its natural frequency:
(1) + y(t) = sin(20). [)
lof2
f
I li
0 Mime in 0
Figure 1: sin(107t) + sin(107t + 0), so wi = 10m, A = 1. The envelope appears to have
frequency 2 = 2r.
0) Show that for some value of b the particular solution is bt cos(2t). What
is b7
(b) Show then that for certain values of A, B
y(t) = Asin(20) + Beos(21) + bt cos(2t) 0)
solves (3) with initial conditions y(0) = 1, (0) = —1.
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Asked by: Adapaka Jagadeesh
Created at: 2025-08-30 20:28:21
Asked by: Adapaka Jagadeesh
Created at: 2025-08-30 20:28:21
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### Given Differential Equation
\[
\ddot{y}(t) + 4y(t) = \sin(2t).
\]
---
### Part (a)
**Question:**
Show that for some value of \( b \), the particular solution is \( bt \cos(2t) \). What is \( b \)?
#### **Solution:**
Since the forcing term \(\sin(2t)\) resonates with the natural frequency of the system (since the homogeneous solution has terms with frequency 2), we look for a particular solution of the form:
\[
y_p(t) = bt \cos(2t)
\]
Calculate the first and second derivatives:
1. **First derivative:**
\[
\dot{y}_p(t) = b[\cos(2t) - 2t\sin(2t)]
\]
2. **Second derivative:**
\[
\ddot{y}_p(t) = b[-2\sin(2t) - 2\sin(2t) - 4t\cos(2t)] = b[-4\sin(2t) - 4t\cos(2t)]
\]
Plug into the original equation:
\[
\ddot{y}_p + 4y_p = [-4b\sin(2t) - 4b t \cos(2t)] + 4 (b t \cos(2t))
= -4b\sin(2t)
\]
Set this equal to the right-hand side, \(\sin(2t)\):
\[
-4b\sin(2t) = \sin(2t)
\implies b = -\frac{1}{4}
\]
---
### Part (b)
Show that for certain values of \(A, B\):
\[
y(t) = A\sin(2t) + B\cos(2t) + bt\cos(2t)
\]
solves the equation with initial conditions \( y(0) = 1,\, \dot{y}(0) = -1 \).
#### **Solution:**
Given \( b = -\frac{1}{4} \) from above,
\[
y(t) = A \sin(2t) + B \cos(2t) - \frac{1}{4} t \cos(2t)
\]
**Initial conditions:**
- \( y(0) = 1 \)
- \( \dot{y}(0) = -1 \)
**Calculate \( y(0) \):**
\[
y(0) = A \sin(0) + B \cos(0) - \frac{1}{4} \cdot 0 \cdot \cos(0) = 0 + B \cdot 1 + 0 = B
\]
So, \( B = 1 \).
**Calculate \( \dot{y}(0) \):**
First derivative:
\[
\dot{y}(t) = 2A \cos(2t) - 2B \sin(2t) - \frac{1}{4} [\cos(2t) - 2t \sin(2t)]
\]
At \( t = 0 \):
\[
\dot{y}(0) = 2A \cos(0) - 2B \sin(0) - \frac{1}{4} [\cos(0) - 0]
= 2A \cdot 1 - 0 - \frac{1}{4} \cdot 1
= 2A - \frac{1}{4}
\]
Set equal to \(-1\):
\[
2A - \frac{1}{4} = -1
\implies 2A = -1 + \frac{1}{4} = -\frac{3}{4}
\implies A = -\frac{3}{8}
\]
---
### **Summary of Answers**
**(a)** The particular solution is \( y_p(t) = -\frac{1}{4} t \cos(2t) \), so \( b = -\frac{1}{4} \).
**(b)** The general solution is
\[
y(t) = -\frac{3}{8} \sin(2t) + \cos(2t) - \frac{1}{4} t \cos(2t)
\]
with \( A = -\frac{3}{8} \), \( B = 1 \), and \( b = -\frac{1}{4} \).
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