To solve this initial value problem, let's go through each part:

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

Step 1: Integrate $y'$ to find $y(x)$

To find $y(x)$, we need to integrate the right-hand side of the differential equation.

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

This can be split into two separate integrals:

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

Integral 1: $\int 2 \cos\left(\frac{x}{3}\right) \, dx$

1. Factor out the constant: $= 2 \int \cos\left(\frac{x}{3}\right) \, dx.$

2. Use a substitution: Let $u = \frac{x}{3}$, then $du = \frac{1}{3} dx$ or $dx = 3 \, du$.

3. Substitute and simplify: $= 2 \int \cos(u) \cdot 3 \, du = 6 \int \cos(u) \, du.$

4. Integrate $\cos(u)$: $= 6 \sin(u) = 6 \sin\left(\frac{x}{3}\right).$

So, the first integral gives: $\int 2 \cos\left(\frac{x}{3}\right) \, dx = 6 \sin\left(\frac{x}{3}\right).$

Integral 2: $\int 4 \cos(3x) \, dx$

1. Factor out the constant: $= 4 \int \cos(3x) \, dx.$

2. Use a substitution: Let $v = 3x$, then $dv = 3 \, dx$ or $dx = \frac{dv}{3}$.

3. Substitute and simplify: $= 4 \int \cos(v) \cdot \frac{dv}{3} = \frac{4}{3} \int \cos(v) \, dv.$

4. Integrate $\cos(v)$: $= \frac{4}{3} \sin(v) = \frac{4}{3} \sin(3x).$

So, the second integral gives: $\int 4 \cos(3x) \, dx = \frac{4}{3} \sin(3x).$

Step 2: Combine the results and add the constant of integration

Putting both integrals together, we get: $y = 6 \sin\left(\frac{x}{3}\right) + \frac{4}{3} \sin(3x) + C.$

Step 3: Apply the initial condition $y\left(\frac{\pi}{2}\right) = 5$

Now we substitute $x = \frac{\pi}{2}$ and $y = 5$ to solve for $C$.

$5 = 6 \sin\left(\frac{\pi}{6}\right) + \frac{4}{3} \sin\left(\frac{3\pi}{2}\right) + C.$

Calculating each term:

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

Substitute these values: $5 = 3 - \frac{4}{3} + C.$

To solve for $C$, combine the constants: $5 = \frac{9}{3} - \frac{4}{3} + C = \frac{5}{3} + C.$

Subtract $\frac{5}{3}$ from both sides: $C = 5 - \frac{5}{3} = \frac{15 - 5}{3} = \frac{10}{3}.$

Final Answer

Thus, the solution to the initial value problem is: $y = 6 \sin\left(\frac{x}{3}\right) + \frac{4}{3} \sin(3x) + \frac{10}{3}.$

Would you like further clarification or have any additional questions?

Here are 5 questions related to this type of problem:

1. How would the solution change if the initial condition was different?
2. Can this method be applied to other trigonometric functions, like sine?
3. What happens if we use a different frequency in the cosine terms?
4. How would the solution look if $y'$ involved exponential functions instead?
5. Can this technique be extended to solve systems of differential equations?

Tip: When integrating trigonometric functions with coefficients, always look for substitution to simplify the integration process.