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Step-by-Step Solution
Step 1: Rewrite the Differential Equation in Standard Form
The given differential equation is
$x \frac{dy}{dx} + y = x \ln x,$ for $x > 1.$
Divide both sides by $x$ (assuming $x \neq 0$):
$\frac{dy}{dx} + \frac{y}{x} = \ln x.$
Step 2: Identify the Integrating Factor
A linear differential equation of the form
$\frac{dy}{dx} + P(x)\,y = Q(x)$
has the integrating factor
$\mu(x) = e^{\int P(x)\,dx}.$
Here, $P(x) = \frac{1}{x}$. So
$\mu(x) = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x.$
Step 3: Multiply the Differential Equation by the Integrating Factor
Multiply both sides of
$\frac{dy}{dx} + \frac{y}{x} = \ln x$
by $x$:
$x \frac{dy}{dx} + y = x \ln x.$
Notice that the left-hand side is the derivative of $xy$:
$\frac{d}{dx} (x\,y) = x \ln x.$
Step 4: Integrate Both Sides
Integrate with respect to $x$:
$x\,y = \int x \ln x \, dx + C,$
We use integration by parts on $\int x \ln x \, dx$. Let
$u = \ln x \implies du = \frac{1}{x} dx,$
and
$dv = x\,dx \implies v = \frac{x^2}{2}.$
Then:
$\int x \ln x \, dx
= \ln x \cdot \frac{x^2}{2} \;-\; \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx
= \frac{x^2}{2} \ln x \;-\; \frac{1}{2}\int x \, dx
= \frac{x^2}{2} \ln x \;-\; \frac{x^2}{4}.$
Hence,
$x\,y
= \frac{x^2}{2} \ln x - \frac{x^2}{4} + C.$
Divide by $x$ (for $x \neq 0$):
$y = \frac{x}{2} \ln x - \frac{x}{4} + \frac{C}{x}.$
Step 5: Use the Initial Condition to Find the Constant C
We are given
$2\,y(2) = \ln(4) - 1.$
Thus,
$y(2) = \frac{\ln(4) - 1}{2} = \frac{2\ln(2) - 1}{2} = \ln(2) - \frac{1}{2}.$
Substitute $x = 2$ into the general solution:
$y(2) = \frac{2}{2} \ln(2) - \frac{2}{4} + \frac{C}{2}
= \ln(2) - \frac{1}{2} + \frac{C}{2}.$
This must equal $\ln(2) - \frac{1}{2}$. Hence,
$\ln(2) - \frac{1}{2} + \frac{C}{2} = \ln(2) - \frac{1}{2}
\implies \frac{C}{2} = 0
\implies C = 0.$
Step 6: Write the Particular Solution and Find y(e)
With $C = 0$, the particular solution becomes:
$y = \frac{x}{2} \ln x - \frac{x}{4}.$
Finally, evaluate at $x = e$:
$y(e) = \frac{e}{2} \ln(e) - \frac{e}{4} = \frac{e}{2}\cdot 1 - \frac{e}{4} = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}.$
Therefore, the value of $y(e)$ is $\frac{e}{4}.$
