Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y)$ and $Q(y)$ are functions of $y$ only.
• Integrating Factor (IF): $IF = e^{\int P(y) dy}$. Multiplying the differential equation by the IF makes the left-hand side a perfect derivative.
• General Solution: $x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C$, where C is the constant of integration.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$

Step-by-Step Solution

Step 1: Rewrite the Differential Equation in Standard Linear Form

We are given the differential equation $y\frac{dx}{dy} = 2x + y^3(y + 1){e^y}$. Our goal is to rewrite it in the standard form $\frac{dx}{dy} + P(y)x = Q(y)$. First, rearrange the equation to isolate the $\frac{dx}{dy}$ term and group the $x$ term: $y\frac{dx}{dy} - 2x = y^3(y + 1){e^y}$ Now, divide both sides by $y$ (assuming $y \neq 0$, which is valid since we are evaluating at $y=1$ and $y=e$): $\frac{dx}{dy} - \frac{2}{y}x = \frac{y^3(y + 1){e^y}}{y}$ Simplify: $\frac{dx}{dy} - \frac{2}{y}x = y^2(y + 1){e^y}$ Now, we can identify $P(y)$ and $Q(y)$: $P(y) = -\frac{2}{y}$ $Q(y) = y^2(y + 1){e^y}$

Step 2: Calculate the Integrating Factor (IF)

The Integrating Factor is given by $IF = e^{\int P(y) dy}$. We have $P(y) = -\frac{2}{y}$, so we need to calculate $\int P(y) dy$: $\int P(y) dy = \int -\frac{2}{y} dy = -2 \int \frac{1}{y} dy = -2 \ln|y|$ Since $y$ is positive in our domain (1 and e), we can drop the absolute value: $-2 \ln y = \ln(y^{-2}) = \ln\left(\frac{1}{y^2}\right)$ Now, calculate the Integrating Factor: $IF = e^{\ln\left(\frac{1}{y^2}\right)} = \frac{1}{y^2}$

Step 3: Find the General Solution

The general solution is given by $x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C$. Substitute $IF = \frac{1}{y^2}$ and $Q(y) = y^2(y + 1){e^y}$: $x \cdot \frac{1}{y^2} = \int y^2(y + 1){e^y} \cdot \frac{1}{y^2} \, dy + C$ Simplify: $\frac{x}{y^2} = \int (y + 1){e^y} \, dy + C$ Now, we need to evaluate $\int (y + 1){e^y} \, dy$. We use integration by parts with $u = y + 1$ and $dv = {e^y} \, dy$. Then $du = dy$ and $v = {e^y}$. $\int (y + 1){e^y} \, dy = (y + 1){e^y} - \int {e^y} \, dy = (y + 1){e^y} - {e^y} = y{e^y} + {e^y} - {e^y} = y{e^y}$ Substitute this back into the general solution: $\frac{x}{y^2} = y{e^y} + C$ Multiply both sides by $y^2$ to solve for $x$: $x(y) = y^2(y{e^y} + C)$

Step 4: Apply the Initial Condition to Find the Particular Solution

We are given $x(1) = 0$. Substitute $y = 1$ and $x = 0$ into the general solution: $0 = (1)^2(1 \cdot {e^1} + C)$ $0 = 1(e + C)$ $C = -e$ Now, substitute $C = -e$ back into the general solution to get the particular solution: $x(y) = y^2(y{e^y} - e)$

Step 5: Evaluate x(e)

We want to find $x(e)$. Substitute $y = e$ into the particular solution: $x(e) = (e)^2(e \cdot {e^e} - e)$ $x(e) = {e^2}({e^{e+1}} - e)$ Factor out an $e$ from the parenthesis: $x(e) = {e^2} \cdot e({e^e} - 1)$ $x(e) = {e^3}({e^e} - 1)$

Common Mistakes & Tips

• Sign Errors: Be extra careful with the negative signs when calculating $P(y)$ and integrating.
• Integration by Parts: Remember the correct formula for integration by parts and carefully choose $u$ and $dv$. A good choice makes the integral simpler.
• Constant of Integration: Don't forget the constant of integration, $C$, when finding the general solution.

Summary

We solved the given first-order linear differential equation by first rewriting it in standard form, calculating the integrating factor, finding the general solution, and then using the initial condition to find the particular solution. Finally, we evaluated the particular solution at $y=e$ to find the value of $x(e)$. The value of $x(e)$ is ${e^3}({e^e} - 1)$.

The final answer is $\boxed{{e^3}({e^e} - 1)}$, which corresponds to option (A).