Solution of First Order Linear Differential Equation

To solve the given first-order linear differential equation:

$\frac{dy}{dx} + \left(\frac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3 + 2)(2 + e^{-2x})}\right)y = 2 + e^{-2x}$

we first simplify the coefficient of $y$, denoted by $P(x)$.

Step 1: Simplify $P(x)$

The coefficient $P(x)$ can be rewritten by splitting the numerator:

$P(x) = \frac{6x^2 + 3x^2 e^{-2x} + 2x^3 e^{-2x} + 4e^{-2x}}{(x^3 + 2)(2 + e^{-2x})}$

Regrouping terms in the numerator: $P(x) = \frac{3x^2(2 + e^{-2x}) + 2e^{-2x}(x^3 + 2)}{(x^3 + 2)(2 + e^{-2x})}$

Dividing each term in the numerator by the denominator, we get: $P(x) = \frac{3x^2}{x^3 + 2} + \frac{2e^{-2x}}{2 + e^{-2x}}$

Step 2: Find the Integrating Factor ($\text{I.F.}$)

The integrating factor is given by $\text{I.F.} = e^{\int P(x) \, dx}$.

First, evaluate the integral $\int P(x) \, dx$: $\int P(x) \, dx = \int \frac{3x^2}{x^3 + 2} \, dx + \int \frac{2e^{-2x}}{2 + e^{-2x}} \, dx$

For the first integral: $\int \frac{3x^2}{x^3 + 2} \, dx = \ln(x^3 + 2)$

For the second integral, substituting $u = 2 + e^{-2x} \implies du = -2e^{-2x} dx$: $\int \frac{2e^{-2x}}{2 + e^{-2x}} \, dx = -\int \frac{du}{u} = -\ln(2 + e^{-2x})$

Combining these results: $\int P(x) \, dx = \ln(x^3 + 2) - \ln(2 + e^{-2x}) = \ln\left(\frac{x^3 + 2}{2 + e^{-2x}}\right)$

Thus, the Integrating Factor is: $\text{I.F.} = e^{\ln\left(\frac{x^3 + 2}{2 + e^{-2x}}\right)} = \frac{x^3 + 2}{2 + e^{-2x}}$

Step 3: Solve the Differential Equation

The general solution of a linear differential equation $\frac{dy}{dx} + P(x)y = Q(x)$ is given by: $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$

Substituting $Q(x) = 2 + e^{-2x}$ and $\text{I.F.} = \frac{x^3 + 2}{2 + e^{-2x}}$: $y \cdot \left(\frac{x^3 + 2}{2 + e^{-2x}}\right) = \int \left(2 + e^{-2x}\right) \cdot \left(\frac{x^3 + 2}{2 + e^{-2x}}\right) dx + C$

$y \cdot \left(\frac{x^3 + 2}{2 + e^{-2x}}\right) = \int (x^3 + 2) \, dx + C$

$y \cdot \left(\frac{x^3 + 2}{2 + e^{-2x}}\right) = \frac{x^4}{4} + 2x + C$

Step 4: Apply the Initial Condition

Using the condition $y(0) = \frac{3}{2}$, substitute $x = 0$ into the equation:

$\frac{3}{2} \cdot \left(\frac{0^3 + 2}{2 + e^0}\right) = \frac{0^4}{4} + 2(0) + C$

$\frac{3}{2} \cdot \left(\frac{2}{3}\right) = C \implies C = 1$

Thus, the particular solution is: $y \cdot \left(\frac{x^3 + 2}{2 + e^{-2x}}\right) = \frac{x^4}{4} + 2x + 1$

Step 5: Calculate $y(1)$ and Find $\alpha$

Substitute $x = 1$ into the solution:

$y(1) \cdot \left(\frac{1^3 + 2}{2 + e^{-2}}\right) = \frac{1^4}{4} + 2(1) + 1$

$y(1) \cdot \left(\frac{3}{2 + e^{-2}}\right) = \frac{1}{4} + 3 = \frac{13}{4}$

Solving for $y(1)$: $y(1) = \frac{13}{4} \cdot \left(\frac{2 + e^{-2}}{3}\right) = \frac{13}{12}(2 + e^{-2})$

Comparing this with $y(1) = \alpha(2 + e^{-2})$, we get: $\alpha = \frac{13}{12}$

This matches Option D.