Solve: `sin^3x(dx)/(dy)=siny`

Explanation

To solve the differential equation \( \frac{\sin^3 x}{dy} = \sin y \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the given equation:

\[ \sin^3 x \frac{dx}{dy} = \sin y \]

We can rearrange this to separate the variables \(x\) and \(y\):

\[ \sin^3 x \, dx = \sin y \, dy \]

### Step 2: Integrate both sides
Now we will integrate both sides:

\[ \int \sin^3 x \, dx = \int \sin y \, dy \]

### Step 3: Solve the integral on the left side
To solve the integral \( \int \sin^3 x \, dx \), we can use the identity for \(\sin^3 x\):

\[ \sin^3 x = \frac{3 \sin x - \sin 3x}{4} \]

Substituting this into the integral gives:

\[ \int \sin^3 x \, dx = \int \frac{3 \sin x - \sin 3x}{4} \, dx \]

This can be simplified to:

\[ \frac{1}{4} \int (3 \sin x - \sin 3x) \, dx \]

### Step 4: Integrate the right side
Now we integrate both sides:

1. For the left side:

\[ \frac{1}{4} \left( -3 \cos x + \frac{1}{3} \cos 3x \right) + C_1 \]

2. For the right side:

\[ \int \sin y \, dy = -\cos y + C_2 \]

### Step 5: Combine results
Now we can equate the results from both integrals:

\[ \frac{1}{4} \left( -3 \cos x + \frac{1}{3} \cos 3x \right) = -\cos y + C \]

Where \(C = C_2 - C_1\) is a constant.

### Final Solution
Thus, the general solution of the differential equation is:

\[ -\cos y = \frac{1}{4} \left( -3 \cos x + \frac{1}{3} \cos 3x \right) + C \]