General solution of `y(dy)/(dx) + by^(2)=a cos x, 0 lt x lt 1` is

Options

A. `y^(2)= 2a(2b sin x+ cos x)+ ce^(-2bx)`
B. `(4b^(2) +1) y^(2)=2a (sin x+2b cos x) + ce^(-2bx)`
C. `(4b^(2) +1) y^(2)=2a (sin x + 2b cos x) + ce^(2bx)`
D. `y^(2)=2a (2 b sin x+ cos x) + ce^(-2bx)`

Answer

To solve the differential equation \( y \frac{dy}{dx} + by^2 = a \cos x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ y \frac{dy}{dx} + by^2 = a \cos x \]
We can rewrite \( y \frac{dy}{dx} \) as \( \frac{1}{2} \frac{d}{dx}(y^2) \) because \( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \).

### Step 2: Substitute and simplify
Substituting this into the equation gives:
\[ \frac{1}{2} \frac{d}{dx}(y^2) + by^2 = a \cos x \]
Multiplying through by 2 to eliminate the fraction:
\[ \frac{d}{dx}(y^2) + 2by^2 = 2a \cos x \]

### Step 3: Identify the integrating factor
This is a first-order linear differential equation in the form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \( P(x) = 2b \) and \( Q(x) = 2a \cos x \). The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int 2b \, dx} = e^{2bx} \]

### Step 4: Multiply through by the integrating factor
We multiply the entire equation by the integrating factor:
\[ e^{2bx} \frac{d}{dx}(y^2) + 2b e^{2bx} y^2 = 2a e^{2bx} \cos x \]

### Step 5: Integrate both sides
The left-hand side can be rewritten as:
\[ \frac{d}{dx}(e^{2bx} y^2) \]
Thus, we have:
\[ \frac{d}{dx}(e^{2bx} y^2) = 2a e^{2bx} \cos x \]
Integrating both sides with respect to \( x \):
\[ e^{2bx} y^2 = \int 2a e^{2bx} \cos x \, dx \]

### Step 6: Solve the integral
Using integration by parts or a standard integral formula, we find:
\[ \int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) \]
For our case, \( a = 2b \) and \( b = 1 \):
\[ \int 2a e^{2bx} \cos x \, dx = \frac{2a e^{2bx}}{(2b)^2 + 1^2} (2b \cos x + \sin x) \]
Substituting this back, we get:
\[ e^{2bx} y^2 = \frac{2a e^{2bx}}{4b^2 + 1} (2b \cos x + \sin x) + C \]

### Step 7: Solve for \( y^2 \)
Dividing both sides by \( e^{2bx} \):
\[ y^2 = \frac{2a (2b \cos x + \sin x)}{4b^2 + 1} + Ce^{-2bx} \]

### Step 8: General solution
Thus, the general solution of the differential equation is:
\[ y^2 = \frac{2a (2b \cos x + \sin x)}{4b^2 + 1} + Ce^{-2bx} \]