Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
• Integrating Factor (I.F.): The integrating factor for a first-order linear differential equation is given by $I.F. = e^{\int P(x) dx}$. Multiplying the differential equation by the integrating factor allows us to solve for $y$.
• Solution: The general solution is given by $y(x) \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$, where C is the constant of integration.

Step-by-Step Solution

Step 1: Identify P(x) and Q(x)

The given differential equation is: $\frac{dy}{dx} + \left( \frac{2x + 1}{x} \right)y = e^{-2x}$ Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we have: $P(x) = \frac{2x + 1}{x} = 2 + \frac{1}{x}$ $Q(x) = e^{-2x}$

Step 2: Calculate the Integrating Factor (I.F.)

The integrating factor is given by: $I.F. = e^{\int P(x) dx} = e^{\int (2 + \frac{1}{x}) dx}$ $I.F. = e^{2x + \ln x} = e^{2x} \cdot e^{\ln x} = xe^{2x}$

Step 3: Find the general solution

The general solution of the differential equation is given by: $y(x) \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$ Substituting the values of I.F. and Q(x), we get: $y(x) \cdot xe^{2x} = \int e^{-2x} \cdot xe^{2x} dx + C$ $y(x) \cdot xe^{2x} = \int x dx + C$ $y(x) \cdot xe^{2x} = \frac{x^2}{2} + C$ Therefore, $y(x) = \frac{x}{2e^{2x}} + \frac{C}{xe^{2x}}$

Step 4: Use the initial condition to find C

We are given that $y(1) = \frac{1}{2}e^{-2}$. Substituting $x = 1$ into the general solution, we get: $y(1) = \frac{1}{2e^2} + \frac{C}{e^2} = \frac{1}{2}e^{-2}$ $\frac{1}{2e^2} + \frac{C}{e^2} = \frac{1}{2e^2}$ $\frac{C}{e^2} = 0$ $C = 0$

Step 5: Find the particular solution

Since $C = 0$, the particular solution is: $y(x) = \frac{x}{2e^{2x}}$

Step 6: Evaluate y(log e 2)

We need to find $y(\ln 2)$: $y(\ln 2) = \frac{\ln 2}{2e^{2\ln 2}} = \frac{\ln 2}{2e^{\ln 2^2}} = \frac{\ln 2}{2(2^2)} = \frac{\ln 2}{2(4)} = \frac{\ln 2}{8}$

Step 7: Check the options

• (A) y(log e 2) = log e 4 $y(\ln 2) = \frac{\ln 2}{8} \neq \ln 4 = 2\ln 2$. So, option (A) is incorrect.

• (C) y(log e 2) = ${{{{\log }_e}2} \over 4}$ $y(\ln 2) = \frac{\ln 2}{8} \neq \frac{\ln 2}{4}$. So, option (C) is incorrect.

• We must have made an error. Let's go back to Step 3. The general solution is $y(x) \cdot xe^{2x} = \int x dx + C$ $y(x) \cdot xe^{2x} = \frac{x^2}{2} + C$ $y(x) = \frac{x}{2e^{2x}} + \frac{C}{xe^{2x}}$ And the initial condition $y(1) = \frac{1}{2}e^{-2}$ gives $y(1) = \frac{1}{2e^2} + \frac{C}{e^2} = \frac{1}{2e^2}$ $\frac{C}{e^2} = 0$ $C = 0$ So, we still have $y(x) = \frac{x}{2e^{2x}}$.

Now we reconsider the question. Let's investigate option (A):

If $y(\ln 2) = \ln 4$, then $\frac{\ln 2}{2 e^{2 \ln 2}} = \ln 4 = 2 \ln 2$. This simplifies to $\frac{\ln 2}{2(4)} = 2 \ln 2$, which implies $\frac{1}{8} = 2$, which is false.

Let's look at option (B) and (D). We want to determine where y(x) is decreasing. We analyze the derivative of y(x): $y(x) = \frac{x}{2e^{2x}}$ $y'(x) = \frac{1}{2} \cdot \frac{e^{2x} - x(2e^{2x})}{e^{4x}} = \frac{1}{2} \cdot \frac{e^{2x}(1 - 2x)}{e^{4x}} = \frac{1 - 2x}{2e^{2x}}$

$y'(x) < 0$ if $1 - 2x < 0$, which means $2x > 1$, so $x > \frac{1}{2}$. $y'(x) > 0$ if $1 - 2x > 0$, which means $2x < 1$, so $x < \frac{1}{2}$. Thus, $y(x)$ is decreasing in $(\frac{1}{2}, \infty)$. This means y(x) is decreasing in $(\frac{1}{2}, 1)$. Thus, (D) is correct.

Let's re-examine the question again. We are told the correct answer is (A). We made an error in our calculation.

Let's verify our derivative $y'(x)$. $y(x) = \frac{x}{2}e^{-2x}$. $y'(x) = \frac{1}{2} e^{-2x} + \frac{x}{2} (-2) e^{-2x} = \frac{1}{2} e^{-2x} - xe^{-2x} = \frac{e^{-2x}}{2}(1 - 2x)$

Now let's re-examine options. We found that $y(x)$ is decreasing when $x > \frac{1}{2}$. So (D) is correct.

The correct answer MUST be (A). We made an error somewhere. Let's rework from the beginning to find our error.

$\frac{dy}{dx} + (\frac{2x+1}{x})y = e^{-2x}$ $P(x) = \frac{2x+1}{x} = 2 + \frac{1}{x}$ $Q(x) = e^{-2x}$ $I.F. = e^{\int (2 + \frac{1}{x}) dx} = e^{2x + \ln x} = e^{2x} e^{\ln x} = xe^{2x}$ $y(x) xe^{2x} = \int e^{-2x} xe^{2x} dx + C = \int x dx + C = \frac{x^2}{2} + C$ $y(x) = \frac{x}{2e^{2x}} + \frac{C}{xe^{2x}}$ $y(1) = \frac{1}{2e^2} + \frac{C}{e^2} = \frac{1}{2e^2}$ $\frac{C}{e^2} = 0 \implies C = 0$ $y(x) = \frac{x}{2e^{2x}}$ We made no errors. However, the question states that y(log e 2) = log e 4. $y(\ln 2) = \frac{\ln 2}{2 e^{2 \ln 2}} = \frac{\ln 2}{2 e^{\ln 4}} = \frac{\ln 2}{2(4)} = \frac{\ln 2}{8}$ But we are told that $y(\ln 2) = \ln 4 = 2 \ln 2$. These do not match.

There must be an error in the given correct answer.

Common Mistakes & Tips

• Be very careful when calculating the integrating factor and solving the integral. A small mistake can lead to a completely wrong answer.
• Double-check your calculations, especially when dealing with exponential and logarithmic functions.
• Always verify your solution by substituting it back into the original differential equation.

Summary

We solved the first-order linear differential equation using the integrating factor method. We found the particular solution using the given initial condition. However, we found that the value of the solution at $x = \ln 2$ does not match the value given in option (A). Our derivative shows that y(x) is decreasing for x > 1/2, so option (D) is correct. Since we must adhere to the given correct answer, it is likely there is an error in the problem statement or the given answer. The problem might intended for $y(2) = \frac{1}{2e^4}$ instead of $y(1) = \frac{1}{2e^2}$.

The final answer is \boxed{D}.