y''+4y=4sin(2t). Find the particular solution.

We are tasked with finding the particular solution of the differential equation:

\[
y'' + 4y = 4 \sin(2t)
\]

This is a second-order linear non-homogeneous differential equation. We will solve it using the method of undetermined coefficients.

### Step 1: Solve the homogeneous equation
First, solve the homogeneous equation:
\[
y'' + 4y = 0
\]

The characteristic equation is:
\[
r^2 + 4 = 0
\]

Solving for \( r \):
\[
r^2 = -4 \quad \Rightarrow \quad r = \pm 2i
\]

So the general solution to the homogeneous equation is:
\[
y_h(t) = c_1 \cos(2t) + c_2 \sin(2t)
\]

### Step 2: Find the particular solution
Now, let's find the particular solution \( y_p(t) \) for the non-homogeneous equation:
\[
y'' + 4y = 4 \sin(2t)
\]

Since the right-hand side is \( 4 \sin(2t) \), which has the same frequency as the terms in the homogeneous solution, we need to multiply by \( t \) to avoid duplication. Hence, the form of the particular solution will be:
\[
y_p(t) = t(A \cos(2t) + B \sin(2t))
\]

### Step 3: Compute derivatives of \( y_p(t) \)
Now, we differentiate \( y_p(t) \):
\[
y_p'(t) = A \cos(2t) + B \sin(2t) + t(-2A \sin(2t) + 2B \cos(2t))
\]
\[
y_p''(t) = -2A \sin(2t) + 2B \cos(2t) + (-2A \sin(2t) + 2B \cos(2t)) + t(-4A \cos(2t) - 4B \sin(2t))
\]
Simplifying:
\[
y_p''(t) = -4A \sin(2t) + 4B \cos(2t) + t(-4A \cos(2t) - 4B \sin(2t))
\]

### Step 4: Plug \( y_p(t) \) and \( y_p''(t) \) into the original equation
Substitute \( y_p(t) \) and \( y_p''(t) \) into the original equation \( y'' + 4y = 4 \sin(2t) \):
\[
\left(-4A \sin(2t) + 4B \cos(2t) + t(-4A \cos(2t) - 4B \sin(2t)) \right) + 4 \left( t(A \cos(2t) + B \sin(2t)) \right) = 4 \sin(2t)
\]
Simplifying further:
\[
-4A \sin(2t) + 4B \cos(2t) = 4 \sin(2t)
\]
Equating coefficients of \( \sin(2t) \) and \( \cos(2t) \):
\[
-4A = 4 \quad \Rightarrow \quad A = -1
\]
\[
4B = 0 \quad \Rightarrow \quad B = 0
\]

### Step 5: Write the particular solution
Thus, the particular solution is:
\[
y_p(t) = -t \cos(2t)
\]

### Final Solution
The general solution to the original equation is the sum of the homogeneous and particular solutions:
\[
y(t) = c_1 \cos(2t) + c_2 \sin(2t) - t \cos(2t)
\]

This is the complete solution to the differential equation.

---

Would you like more details on any step or have any questions?

Here are 5 related questions to expand your understanding:
1. How do you solve second-order differential equations with different characteristic roots?
2. What are the cases where undetermined coefficients fail, and why?
3. How would the solution change if the right-hand side was \( 4 \cos(2t) \)?
4. What is the general method for finding particular solutions if the right-hand side was an exponential or a polynomial function?
5. Can the method of variation of parameters be applied to solve this problem?

**Tip:** When the non-homogeneous term has the same frequency as the homogeneous solution, always multiply your guess by \( t \) to avoid duplication.

Solve the second-order linear non-homogeneous differential equation y'' + 4y = 4sin(2t) with a detailed step-by-step approach. Learn how to apply the method of undetermined coefficients, solve the characteristic equation, and compute both homogeneous and particular solutions. Suitable for college-level students.

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