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Step-by-Step Solution
Step 1: Rewrite the Given Differential Equation
The given differential equation is
(y - 2 \log x)\,dx + \bigl(x \log x^2\bigr)\,dy = 0, \quad x > 1.
First, rewrite it in the form \frac{dy}{dx} = f(x,y) . To do this, rearrange as follows:
(y - 2 \log x)\,dx + (x \log x^2)\,dy = 0
\;\;\;\Longrightarrow\;\;\; (x \log x^2)\,\frac{dy}{dx} = -\bigl(y - 2 \log x\bigr).
Then,
\frac{dy}{dx}
= \frac{- (y - 2 \log x)}{x \log x^2}
= \frac{2 \log x - y}{x \,\bigl(2 \log x\bigr)}.
We can simplify the denominator x\cdot \log x^2 = x\cdot(2 \log x) . Thus,
\frac{dy}{dx}
= \frac{2\log x - y}{2x \log x}.
Step 2: Express the Equation in Standard Linear Form
A linear differential equation in y has the standard form
\frac{dy}{dx} + P(x)\,y = Q(x).
Comparing, we get
\frac{dy}{dx} + \frac{y}{2x \log x}
= \frac{1}{x}.
Hence, here
P(x) = \frac{1}{2x \log x},
\quad Q(x) = \frac{1}{x}.
Step 3: Find the Integrating Factor (IF)
The integrating factor (IF) for a linear differential equation
\frac{dy}{dx} + P(x)\,y = Q(x)
is given by
\text{IF} = e^{\int P(x)\,dx}.
In our case,
P(x) = \frac{1}{2x \log x}.
We compute
\int P(x)\,dx = \int \frac{1}{2x \log x}\,dx.
Let t = \log x . Then dt = \frac{1}{x}\,dx . This transforms the integral into
\int \frac{1}{2x \log x}\,dx = \int \frac{1}{2\,t}\,dt
= \frac{1}{2}\,\log t
= \frac{1}{2}\,\log(\log x).
Therefore,
\text{IF} = e^{\frac{1}{2}\,\log(\log x)} = \bigl(\log x\bigr)^{\frac{1}{2}} = \sqrt{\log x}.
Step 4: Multiply the Differential Equation by the IF
Multiply the entire linear differential equation
\frac{dy}{dx} + \frac{y}{2x \log x} = \frac{1}{x}
by \sqrt{\log x} :
\sqrt{\log x}\,\frac{dy}{dx} + \frac{y\sqrt{\log x}}{2x \log x} = \frac{\sqrt{\log x}}{x}.
This can be recognized as the derivative of the product y\sqrt{\log x} with respect to x . Indeed,
\frac{d}{dx}\Bigl[y\sqrt{\log x}\Bigr]
= \sqrt{\log x}\,\frac{dy}{dx} + y \cdot \frac{1}{2\sqrt{\log x}} \cdot \frac{1}{x},
and simplifying the second term yields \frac{y\sqrt{\log x}}{2x \log x} , matching the left-hand side. Hence the equation becomes
\frac{d}{dx} \bigl[y \sqrt{\log x}\bigr] = \frac{\sqrt{\log x}}{x}.
Step 5: Integrate Both Sides
Integrate with respect to x :
y \sqrt{\log x}
= \int \frac{\sqrt{\log x}}{x}\,dx + C.
Use the substitution v = \log x , so dv = \frac{1}{x}\,dx . Then
y \sqrt{\log x}
= \int \sqrt{v}\,dv + C
= \frac{2}{3}v^{\frac{3}{2}} + C
= \frac{2}{3}\bigl(\log x\bigr)^{\frac{3}{2}} + C.
Step 6: Apply the First Initial Condition
We know the curve passes through \bigl(e, \frac{4}{3}\bigr) . For x = e , \log e = 1 . Substitute x = e and y = \frac{4}{3} into
y \sqrt{\log x}
= \frac{2}{3}\bigl(\log x\bigr)^{\frac{3}{2}} + C.
We get
\frac{4}{3} \cdot \sqrt{1}
= \frac{2}{3}\cdot (1)^{\frac{3}{2}} + C
\quad\Longrightarrow\quad
\frac{4}{3} = \frac{2}{3} + C
\quad\Longrightarrow\quad
C = \frac{4}{3} - \frac{2}{3} = \frac{2}{3}.
Step 7: Apply the Second Initial Condition
The curve also passes through \bigl(e^4, \alpha\bigr) . For x = e^4 , \log \bigl(e^4\bigr) = 4 . Substitute x = e^4 and y = \alpha into
y \sqrt{\log x}
= \frac{2}{3}\bigl(\log x\bigr)^{\frac{3}{2}} + \frac{2}{3}.
Hence,
\alpha \sqrt{4}
= \frac{2}{3}\bigl(4\bigr)^{\frac{3}{2}} + \frac{2}{3}.
Note that \sqrt{4} = 2 and (4)^{\frac{3}{2}} = 4^{1.5} = 8. Thus,
2\alpha
= \frac{2}{3}\cdot 8 + \frac{2}{3}
= \frac{16}{3} + \frac{2}{3}
= \frac{18}{3} = 6.
Therefore,
\alpha = 3.
Step 8: State the Final Answer
Thus, the value of \alpha that satisfies the given conditions is
\alpha = 3.
