Question:

Step 1: Rewrite the given differential equation in standard linear form. The given differential equation is $x\log_e x\,\dfrac{dy}{dx}+y=x^2\log_e x$. To convert it into the standard linear first-order differential equation form $\dfrac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $x\log_e x$. $$ \dfrac{dy}{dx} + \dfrac{1}{x\log_e x}y = \dfrac{x^2\log_e x}{x\log_e x} $$ $$ \dfrac{dy}{dx} + \left(\dfrac{1}{x\log_e x}\right)y = x $$ Here, we identify $P(x) = \dfrac{1}{x\log_e x}$ and $Q(x) = x$. Step 2: Calculate the Integrating Factor (IF). The integrating factor for a linear first-order differential equation is given by $\text{IF} = e^{\int P(x)dx}$. First, let's find $\int P(x)dx$: $$ \int P(x)dx = \int \dfrac{1}{x\log_e x}dx $$ Let $u = \log_e x$. Then $du = \dfrac{1}{x}dx$. $$ \int \dfrac{1}{u}du = \log_e |u| $$ Since $x > 1$, $\log_e x > 0$, so $|u| = \log_e x$. Thus, $\int P(x)dx = \log_e (\log_e x)$. Now, substitute this into the formula for the Integrating Factor: $$ \text{IF} = e^{\log_e (\log_e x)} = \log_e x $$ Step 3: Solve the differential equation using the integrating factor. The general solution of a linear first-order differential equation is given by $y \cdot \text{IF} = \int (Q(x) \cdot \text{IF}) dx + C$. Substitute the values of $\text{IF}$ and $Q(x)$: $$ y \cdot \log_e x = \int \left(x \cdot \log_e x\right) dx + C $$ To evaluate the integral $\int x \log_e x \, dx$, we use integration by parts, $\int u \, dv = uv - \int v \, du$. Let $u = \log_e x \implies du = \dfrac{1}{x} dx$. Let $dv = x \, dx \implies v = \dfrac{x^2}{2}$. $$ \int x \log_e x \, dx = \log_e x \cdot \dfrac{x^2}{2} - \int \dfrac{x^2}{2} \cdot \dfrac{1}{x} dx $$ $$ = \dfrac{x^2}{2}\log_e x - \int \dfrac{x}{2} dx $$ $$ = \dfrac{x^2}{2}\log_e x - \dfrac{x^2}{4} $$ Now substitute this back into the solution equation: $$ y \log_e x = \dfrac{x^2}{2}\log_e x - \dfrac{x^2}{4} + C $$ Divide by $\log_e x$ to find $y(x)$: $$ y(x) = \dfrac{x^2}{2} - \dfrac{x^2}{4\log_e x} + \dfrac{C}{\log_e x} $$ Step 4: Use the initial condition to find the constant $C$. We are given the initial condition $y(2)=2$. Substitute $x=2$ and $y=2$ into the general solution: $$ 2 = \dfrac{2^2}{2} - \dfrac{2^2}{4\log_e 2} + \dfrac{C}{\log_e 2} $$ $$ 2 = \dfrac{4}{2} - \dfrac{4}{4\log_e 2} + \dfrac{C}{\log_e 2} $$ $$ 2 = 2 - \dfrac{1}{\log_e 2} + \dfrac{C}{\log_e 2} $$ Subtracting 2 from both sides: $$ 0 = -\dfrac{1}{\log_e 2} + \dfrac{C}{\log_e 2} $$ $$ 0 = \dfrac{C-1}{\log_e 2} $$ Since $\log_e 2 \neq 0$, we must have $C-1=0$, which implies $C=1$. Step 5: Write the particular solution. Substitute the value of $C=1$ back into the general solution: $$ y(x) = \dfrac{x^2}{2} - \dfrac{x^2}{4\log_e x} + \dfrac{1}{\log_e x} $$ Step 6: Evaluate $y(e)$. To find $y(e)$, substitute $x=e$ into the particular solution. Recall that $\log_e e = 1$. $$ y(e) = \dfrac{e^2}{2} - \dfrac{e^2}{4\log_e e} + \dfrac{1}{\log_e e} $$ $$ y(e) = \dfrac{e^2}{2} - \dfrac{e^2}{4(1)} + \dfrac{1}{1} $$ $$ y(e) = \dfrac{e^2}{2} - \dfrac{e^2}{4} + 1 $$ To combine these terms, find a common denominator, which is 4: $$ y(e) = \dfrac{2e^2}{4} - \dfrac{e^2}{4} + \dfrac{4}{4} $$ $$ y(e) = \dfrac{2e^2 - e^2 + 4}{4} $$ $$ y(e) = \dfrac{e^2 + 4}{4} $$ Step 7: Conclude the final answer. The value of $y(e)$ is $\dfrac{4+e^2}{4}$. This matches Option 1. The final answer is $\boxed{\dfrac{4+e^2}{4}}$.