If y(x) is the solution of the differential equation `( dy )/( dx) +((2x+1)/(x))y=e^(-2x), x gt 0` , where `y(1) = (1)/(2) e^(-2)`, then

Options

A. `y(x) ` is decreasing in ` ((1)/(2),1)`
B. y(x) is decreasing in (0,1)
C. ` y( log _(e) 2) =( log _(e) 2) /(4)`
D. ` y ( log _(e) 2)) = log _(e) 4`

Answer

To solve the differential equation

\[ \frac{dy}{dx} + \frac{2x + 1}{x} y = e^{-2x}, \quad x > 0 \]

with the initial condition \( y(1) = \frac{1}{2} e^{-2} \), we will follow these steps:

### Step 1: Identify \( p(x) \) and \( q(x) \)
In the given equation, we can identify:
- \( p(x) = \frac{2x + 1}{x} = 2 + \frac{1}{x} \)
- \( q(x) = e^{-2x} \)

### Step 2: Calculate the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int \left(2 + \frac{1}{x}\right) \, dx} \]

Calculating the integral:

\[ \int \left(2 + \frac{1}{x}\right) \, dx = 2x + \ln |x| + C \]

Thus, the integrating factor is:

\[ \mu(x) = e^{2x} \cdot x \]

### Step 3: Multiply the Differential Equation by the Integrating Factor
Now we multiply the entire differential equation by the integrating factor:

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

This simplifies to:

\[ x e^{2x} \frac{dy}{dx} + (2x + 1) e^{2x} y = 1 \]

### Step 4: Rewrite the Left Side
The left side can be rewritten as the derivative of a product:

\[ \frac{d}{dx} \left( x e^{2x} y \right) = 1 \]

### Step 5: Integrate Both Sides
Integrating both sides with respect to \( x \):

\[ \int \frac{d}{dx} \left( x e^{2x} y \right) \, dx = \int 1 \, dx \]

This gives:

\[ x e^{2x} y = x + C \]

### Step 6: Solve for \( y \)
Now, we solve for \( y \):

\[ y = \frac{x + C}{x e^{2x}} = \frac{1}{e^{2x}} + \frac{C}{x e^{2x}} \]

### Step 7: Apply the Initial Condition
We use the initial condition \( y(1) = \frac{1}{2} e^{-2} \):

\[ \frac{1}{e^{2}} + \frac{C}{1 \cdot e^{2}} = \frac{1}{2} e^{-2} \]

This simplifies to:

\[ \frac{1 + C}{e^{2}} = \frac{1}{2 e^{2}} \]

Multiplying both sides by \( e^{2} \):

\[ 1 + C = \frac{1}{2} \]

Thus,

\[ C = \frac{1}{2} - 1 = -\frac{1}{2} \]

### Step 8: Substitute \( C \) Back into the Equation for \( y \)
Substituting \( C \) back, we have:

\[ y = \frac{1}{e^{2}} - \frac{1/2}{x e^{2}} = \frac{1}{e^{2}} \left( 1 - \frac{1}{2x} \right) \]

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

\[ y(x) = \frac{1}{e^{2}} \left( 1 - \frac{1}{2x} \right) \]