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To solve the initial value problem for the differential equation:

\[
y' = 2 \cos\left(\frac{x}{3}\right) + 4 \cos(3x),
\]
with the initial condition \( y\left(\frac{\pi}{2}\right) = 5 \).

### Step 1: Integrate \( y' \) to find \( y(x) \)
Since \( y' = \frac{dy}{dx} \), we can integrate both sides with respect to \( x \) to find \( y(x) \):

\[
y(x) = \int \left(2 \cos\left(\frac{x}{3}\right) + 4 \cos(3x)\right) \, dx.
\]

### Step 2: Integrate each term separately

1. **First term:** \( \int 2 \cos\left(\frac{x}{3}\right) \, dx \)
   - Let \( u = \frac{x}{3} \), then \( du = \frac{1}{3} dx \) or \( dx = 3 \, du \).
   - Substitute into the integral:
     \[
     \int 2 \cos\left(\frac{x}{3}\right) \, dx = 2 \int \cos(u) \cdot 3 \, du = 6 \int \cos(u) \, du = 6 \sin(u).
     \]
   - Substitute \( u = \frac{x}{3} \) back:
     \[
     = 6 \sin\left(\frac{x}{3}\right).
     \]

2. **Second term:** \( \int 4 \cos(3x) \, dx \)
   - Let \( v = 3x \), then \( dv = 3 \, dx \) or \( dx = \frac{dv}{3} \).
   - Substitute into the integral:
     \[
     \int 4 \cos(3x) \, dx = 4 \int \cos(v) \cdot \frac{dv}{3} = \frac{4}{3} \int \cos(v) \, dv = \frac{4}{3} \sin(v).
     \]
   - Substitute \( v = 3x \) back:
     \[
     = \frac{4}{3} \sin(3x).
     \]

### Step 3: Combine results
So,
\[
y(x) = 6 \sin\left(\frac{x}{3}\right) + \frac{4}{3} \sin(3x) + C,
\]
where \( C \) is the constant of integration.

### Step 4: Apply the initial condition \( y\left(\frac{\pi}{2}\right) = 5 \)

Substitute \( x = \frac{\pi}{2} \) and \( y = 5 \):
\[
5 = 6 \sin\left(\frac{\pi}{6}\right) + \frac{4}{3} \sin\left(\frac{3\pi}{2}\right) + C.
\]

Calculate each term:
- \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \), so \( 6 \sin\left(\frac{\pi}{6}\right) = 6 \cdot \frac{1}{2} = 3 \).
- \( \sin\left(\frac{3\pi}{2}\right) = -1 \), so \( \frac{4}{3} \sin\left(\frac{3\pi}{2}\right) = \frac{4}{3} \cdot (-1) = -\frac{4}{3} \).

Thus,
\[
5 = 3 - \frac{4}{3} + C.
\]

Simplify:
\[
5 = \frac{9}{3} - \frac{4}{3} + C = \frac{5}{3} + C.
\]

So,
\[
C = 5 - \frac{5}{3} = \frac{15}{3} - \frac{5}{3} = \frac{10}{3}.
\]

### Final Solution
Substitute \( C = \frac{10}{3} \) into the solution:
\[
y(x) = 6 \sin\left(\frac{x}{3}\right) + \frac{4}{3} \sin(3x) + \frac{10}{3}.
\]

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Let me know if you need further explanation on any steps!

### Related Questions

1. How would the solution change if the initial condition were different?
2. Can this problem be solved using a numerical method? If so, which one?
3. How do the solutions behave as \( x \to \infty \) in this equation?
4. What if the cosine terms were replaced by sine terms in the original differential equation?
5. How can we verify this solution by differentiating?

### Tip
Remember that integrating terms with trigonometric functions often requires a substitution to simplify the calculation.

Solve the initial value problem: y' = 2cos(x/3) + 4cos(3x), with y(π/2) = 5. Do not include 'y =' in your answer.

Learn how to solve an initial value problem for the differential equation y' = 2cos(x/3) + 4cos(3x), with the initial condition y(π/2) = 5. This solution covers integration techniques and applies the Fundamental Theorem of Calculus.

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