The solution of `(x^2-1)dy / dx*siny-2x*cosy=2x-2x^3` is

Options

A. `(x^(2)-1)cosy=(x^(4))/(2)-x^(2)+C`
B. `(x^(2)-1)siny=(x^(4))/(2)-x^(2)+C`
C. `(x^(2)-1)cosy=(x^(4))/(2)-(x^(2))/(2)+C`
D. `(x^(2)-1)siny=(x^(4))/(2)-(x^(2))/(2)+C`

Answer

To solve the differential equation

\[ (x^2 - 1) \frac{dy}{dx} \sin y - 2x \cos y = 2x - 2x^3, \]

we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate the derivative term:

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

### Step 2: Introducing a New Variable
Let’s define a new variable \( t \) such that:

\[ t = (x^2 - 1) \cos y. \]

### Step 3: Differentiate \( t \)
Now, we differentiate \( t \) with respect to \( x \) using the product rule:

\[ \frac{dt}{dx} = (x^2 - 1) \frac{d}{dx}(\cos y) + \cos y \frac{d}{dx}(x^2 - 1). \]

Calculating the derivatives, we have:

\[ \frac{d}{dx}(\cos y) = -\sin y \frac{dy}{dx}, \]
and
\[ \frac{d}{dx}(x^2 - 1) = 2x. \]

Thus, we get:

\[ \frac{dt}{dx} = (x^2 - 1)(-\sin y \frac{dy}{dx}) + 2x \cos y. \]

### Step 4: Substitute Back into the Equation
From the rearranged equation, we can express \( (x^2 - 1) \frac{dy}{dx} \sin y \) in terms of \( \frac{dt}{dx} \):

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

This gives us:

\[ -\frac{dt}{dx} = 2x - 2x^3. \]

### Step 5: Multiply by \( dx \)
We can now multiply both sides by \( dx \):

\[ -dt = (2x - 2x^3) dx. \]

### Step 6: Integrate Both Sides
Integrating both sides:

\[ \int -dt = \int (2x - 2x^3) dx. \]

The left side integrates to:

\[ -t + C, \]

and the right side integrates to:

\[ x^2 - \frac{1}{2} x^4 + C. \]

### Step 7: Substitute Back for \( t \)
Substituting back for \( t \):

\[ -(x^2 - 1) \cos y = x^2 - \frac{1}{2} x^4 + C. \]

### Step 8: Rearranging the Final Solution
Rearranging gives us:

\[ (x^2 - 1) \cos y = -x^2 + \frac{1}{2} x^4 - C. \]

### Final Solution
Thus, the solution to the differential equation is:

\[ (x^2 - 1) \cos y = -x^2 + \frac{1}{2} x^4 + C. \]

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