`y(dy)/(dx)sinx=cosx(sinx-(y^2)/2);` where at `x=pi/2,`

Options

A. `y^2 = sin x`
B. `y^2 = 2 sin x`
C. `x^2 = sin Y`
D. `x^2 = 2 sin Y`

Answer

To solve the differential equation

\[ y \frac{dy}{dx} \sin x = \cos x \left( \sin x - \frac{y^2}{2} \right) \]

with the initial condition \( y\left(\frac{\pi}{2}\right) = 1 \), we can follow these steps:

### Step 1: Rearranging the Equation

First, we divide both sides by \(\sin x\) (assuming \(\sin x \neq 0\)):

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

### Step 2: Simplifying the Equation

Rearranging gives us:

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

### Step 3: Substituting \(y^2\)

Let \(y^2 = t\). Then, differentiating gives us:

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

Thus, we can express \(\frac{dy}{dx}\) as:

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

### Step 4: Substituting Back into the Equation

Substituting this back into our equation gives:

\[ \frac{1}{2} \frac{dt}{dx} + \frac{t}{2} \cos x = \cos x \sin x \]

Multiplying through by 2 to eliminate the fractions:

\[ \frac{dt}{dx} + t \cos x = 2 \cos x \sin x \]

### Step 5: Identifying the Integrating Factor

This is a linear first-order differential equation in standard form:

\[ \frac{dt}{dx} + P(x)t = Q(x) \]

where \(P(x) = \cos x\) and \(Q(x) = 2 \cos x \sin x\).

The integrating factor \(I(x)\) is given by:

\[ I(x) = e^{\int P(x) \, dx} = e^{\int \cos x \, dx} = e^{\sin x} \]

### Step 6: Multiplying by the Integrating Factor

Multiplying the entire equation by the integrating factor:

\[ e^{\sin x} \frac{dt}{dx} + e^{\sin x} t \cos x = 2 e^{\sin x} \cos x \sin x \]

### Step 7: Recognizing the Left Side as a Derivative

The left-hand side can be expressed as:

\[ \frac{d}{dx} \left( e^{\sin x} t \right) = 2 e^{\sin x} \cos x \sin x \]

### Step 8: Integrating Both Sides

Integrating both sides gives:

\[ e^{\sin x} t = \int 2 e^{\sin x} \cos x \sin x \, dx \]

Using the substitution \(u = \sin x\), \(du = \cos x \, dx\):

\[ = 2 \int u e^u \, du \]

Using integration by parts:

Let \(v = e^u\) and \(dw = u \, du\):

\[ = 2 \left( u e^u - \int e^u \, du \right) = 2 \left( u e^u - e^u \right) = 2 e^{\sin x} \left( \sin x - 1 \right) \]

Thus:

\[ e^{\sin x} t = 2 e^{\sin x} \left( \sin x - 1 \right) + C \]

### Step 9: Solving for \(t\)

Dividing by \(e^{\sin x}\):

\[ t = 2(\sin x - 1) + Ce^{-\sin x} \]

Recalling that \(t = y^2\):

\[ y^2 = 2(\sin x - 1) + Ce^{-\sin x} \]

### Step 10: Applying Initial Condition

Using the initial condition \(y\left(\frac{\pi}{2}\right) = 1\):

\[ 1^2 = 2(1 - 1) + Ce^{-1} \implies 1 = 0 + \frac{C}{e} \implies C = e \]

### Final Solution

Substituting \(C\) back into the equation:

\[ y^2 = 2(\sin x - 1) + e e^{-\sin x} \]

Thus, the final solution is:

\[ y^2 = 2(\sin x - 1) + e^{1 - \sin x} \]