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The given initial value problem is:

\[
\cos(x) y' + \sin(x) y = 2 \cos^3(x) \sin(x) - 1
\]

with the initial condition:

\[
y\left(\frac{\pi}{4}\right) = 3 \sqrt{2}.
\]

### Step 1: Simplify the differential equation
First, divide through by \(\cos(x)\) (assuming \(\cos(x) \neq 0\)) to make the equation simpler:

\[
y' + \frac{\sin(x)}{\cos(x)} y = \frac{2 \cos^3(x) \sin(x) - 1}{\cos(x)}.
\]

This becomes:

\[
y' + \tan(x) y = 2 \cos^2(x) \sin(x) - \sec(x).
\]

### Step 2: Solve the homogeneous equation
The corresponding homogeneous equation is:

\[
y' + \tan(x) y = 0.
\]

This is a first-order linear differential equation. We solve it using an integrating factor:

\[
\mu(x) = e^{\int \tan(x) \, dx} = e^{-\ln|\cos(x)|} = \frac{1}{\cos(x)}.
\]

Thus, multiplying the entire homogeneous equation by \(\frac{1}{\cos(x)}\) gives:

\[
\frac{d}{dx} \left( \frac{y}{\cos(x)} \right) = 0.
\]

Integrating both sides, we get:

\[
\frac{y}{\cos(x)} = C,
\]

where \(C\) is a constant. Therefore, the general solution to the homogeneous equation is:

\[
y_h(x) = C \cos(x).
\]

### Step 3: Solve the particular solution
Now, we find a particular solution to the non-homogeneous equation:

\[
y' + \tan(x) y = 2 \cos^2(x) \sin(x) - \sec(x).
\]

We try to find the particular solution by inspection or substitution method, but this part involves integrating the right-hand side.

Let me calculate the particular solution step.

Solve the initial value problem cos(x)y' + sin(x)y = 2cos^3(x)sin(x) - 1 with the condition y(π/4) = 3√2.

Learn how to solve a linear first-order differential equation with trigonometric functions. This step-by-step solution involves simplifying the equation, finding the homogeneous solution, and solving an initial value problem with y(π/4) = 3√2. Suitable for advanced calculus students.

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