Let Y = y(x) be the solution of the differential equation `( dy ) /( dx) + 2y = f(x) ` , where ` f(x) = { {:(1, x in [0,1]),( 0, " Otherwise "):}`
IF ` y(0) =0 ` then ` y ((3)/(2)) ` is :

Options

A. ` ( e^2-1)/( 2e ^3)`
B. ` ( e^2-1)/(e^3)`
C. `(e ^2 +1)/( 2e ^4)`
D. `(1)/(2e)`

Answer

To solve the differential equation \( \frac{dy}{dx} + 2y = f(x) \) where \( f(x) = 1 \) for \( x \in [0, 1] \) and \( f(x) = 0 \) otherwise, and given that \( y(0) = 0 \), we will follow these steps:

### Step 1: Solve the differential equation for \( x \in [0, 1] \)

For \( x \in [0, 1] \), we have:
\[ \frac{dy}{dx} + 2y = 1 \]

To solve this, we will find the integrating factor \( e^{\int 2 \, dx} = e^{2x} \).

Multiplying through by the integrating factor:
\[ e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{2x} \]

This can be rewritten as:
\[ \frac{d}{dx}(e^{2x} y) = e^{2x} \]

### Step 2: Integrate both sides

Integrating both sides with respect to \( x \):
\[ \int \frac{d}{dx}(e^{2x} y) \, dx = \int e^{2x} \, dx \]

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

### Step 3: Solve for \( y \)

Now, we can solve for \( y \):
\[ y = \frac{1}{2} + Ce^{-2x} \]

### Step 4: Apply the initial condition \( y(0) = 0 \)

Using the initial condition \( y(0) = 0 \):
\[ 0 = \frac{1}{2} + C \cdot e^{0} \]
\[ 0 = \frac{1}{2} + C \implies C = -\frac{1}{2} \]

### Step 5: Substitute \( C \) back into the equation

Thus, the solution for \( y \) when \( x \in [0, 1] \) is:
\[ y = \frac{1}{2} - \frac{1}{2} e^{-2x} \]

### Step 6: Solve for \( x > 1 \)

For \( x > 1 \), \( f(x) = 0 \):
\[ \frac{dy}{dx} + 2y = 0 \]

This simplifies to:
\[ \frac{dy}{dx} = -2y \]

The integrating factor is still \( e^{-2x} \), leading to:
\[ \frac{d}{dx}(y) = -2y \]

Integrating gives:
\[ y = Ce^{-2x} \]

### Step 7: Ensure continuity at \( x = 1 \)

To ensure continuity at \( x = 1 \):
\[ y(1) = \frac{1}{2} - \frac{1}{2} e^{-2} = Ce^{-2} \]

Calculating \( y(1) \):
\[ y(1) = \frac{1}{2}(1 - e^{-2}) \]

Setting the two expressions equal:
\[ C e^{-2} = \frac{1}{2}(1 - e^{-2}) \]
\[ C = \frac{1}{2} e^{2}(1 - e^{-2}) = \frac{1}{2} (e^{2} - 1) \]

### Step 8: Final expression for \( y \)

Thus, for \( x > 1 \):
\[ y = \left(\frac{1}{2}(e^{2} - 1)\right)e^{-2x} \]

### Step 9: Calculate \( y\left(\frac{3}{2}\right) \)

Now we can find \( y\left(\frac{3}{2}\right) \):
\[ y\left(\frac{3}{2}\right) = \left(\frac{1}{2}(e^{2} - 1)\right)e^{-3} \]

### Final Answer

Thus, the value of \( y\left(\frac{3}{2}\right) \) is:
\[ y\left(\frac{3}{2}\right) = \frac{1}{2}(e^{2} - 1)e^{-3} \]