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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?

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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.

Solve the initial value problem: y' = 2 cos(x/3) + 4 cos(3x), y(π/2) = 5

Learn how to solve the initial value problem for y' = 2 cos(x/3) + 4 cos(3x) with the condition y(π/2) = 5. This solution involves integrating trigonometric functions and applying an initial condition to find the constant of integration. Suitable for advanced high school and undergraduate calculus students.

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