The Bernouli's equation
`(dy)/(dx)-ytanx=(sinxcos^(2)x)/(y^(2))` can be transformed to

Options

A. `(dz)/(dx-y tan z=sinz cos^(2)z`
B. `(dz)/(dx+3z tan x=3 sin cos^(2)x`
C. `(dz)/(dx)-3z tan x=3 sin x cos^(2)x`
D. `(dz)/(dx)-z tan x =sin x cos^(2)x`

Explanation

To transform the given Bernoulli's equation

\[ \frac{dy}{dx} - y \tan x = \frac{\sin x \cos^2 x}{y^2} \]

into a linear differential equation, we will follow these steps:

### Step 1: Rewrite the Equation
We start with the original equation:

\[ \frac{dy}{dx} - y \tan x = \frac{\sin x \cos^2 x}{y^2} \]

### Step 2: Multiply Through by \(y^2\)
To eliminate the fraction on the right side, multiply the entire equation by \(y^2\):

\[ y^2 \frac{dy}{dx} - y^3 \tan x = \sin x \cos^2 x \]

### Step 3: Substitute for \(y^3\)
Let’s make the substitution \(t = y^3\). Then, differentiating both sides gives us:

\[ \frac{dt}{dx} = 3y^2 \frac{dy}{dx} \]

### Step 4: Solve for \(\frac{dy}{dx}\)
From the substitution, we can express \(\frac{dy}{dx}\) in terms of \(t\):

\[ \frac{dy}{dx} = \frac{1}{3y^2} \frac{dt}{dx} \]

### Step 5: Substitute Back into the Equation
Now substitute \(\frac{dy}{dx}\) back into the modified equation:

\[ y^2 \left(\frac{1}{3y^2} \frac{dt}{dx}\right) - t \tan x = \sin x \cos^2 x \]

This simplifies to:

\[ \frac{1}{3} \frac{dt}{dx} - t \tan x = \sin x \cos^2 x \]

### Step 6: Multiply by 3
To convert this into a standard linear form, multiply the entire equation by 3:

\[ \frac{dt}{dx} - 3t \tan x = 3 \sin x \cos^2 x \]

### Step 7: Final Form
This is now in the standard linear form:

\[ \frac{dt}{dx} - 3t \tan x = 3 \sin x \cos^2 x \]

### Conclusion
Thus, the transformed equation can be expressed as:

\[ \frac{dz}{dx} - 3z \tan x = 3 \sin x \cos^2 x \]

where \(z\) is used in place of \(t\).