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Step-by-Step Solution
Step 1: Rewrite the Differential Equation
The given differential equation is:
$y^2 \, dx \;+\;\bigl(x^2 - x\,y + y^2\bigr)\, dy \;=\; 0.$
We can express this in the form $\,\dfrac{dy}{dx}=\,\dots$ by moving all terms properly:
$\displaystyle \dfrac{dy}{dx} \;=\; \dfrac{y^2}{\,x\,y - x^2 - y^2\,}.$
Step 2: Use the Substitution $y = v\,x$
We let $y = v\,x$, which implies $y' = \dfrac{dy}{dx} = v + x\,\dfrac{dv}{dx}$. Substitute these into the differential equation:
$v + x \dfrac{dv}{dx} \;=\; \dfrac{v^2}{\,v - 1 - v^2\,}.
Rearrange to isolate $\dfrac{dv}{dx}$:
$\displaystyle x \,\dfrac{dv}{dx} \;=\; \dfrac{v^2}{\,v - 1 - v^2\,} \;-\; v
\;=\; \dfrac{v^2 - v(v - 1 - v^2)}{\,v - 1 - v^2\,}.
Simplify the numerator as needed. The result will allow us to separate variables in $v$ and $x$.
Step 3: Separate the Variables and Integrate
We aim to form integrals of the type $\int f(v)\,dv = \int \dfrac{dx}{x}$. Through some algebraic manipulation, it turns out that one obtains an integral of the form:
$\displaystyle \int \dfrac{v - 1 - v^2}{\,v\,(1 + v^2)\!}\,dv
\;=\; \int \dfrac{dx}{\,x\,}.
Carrying out the left-hand integral typically yields:
$\tan^{-1}\Bigl(\dfrac{y}{x}\Bigr)
\;-\;
\ln\Bigl(\dfrac{y}{x}\Bigr)
\;=\;
\ln(x) + C,
where $C$ is the constant of integration.
Step 4: Use the Condition Through $(1,1)$ to Find $C$
The solution curve passes through the point $(1, 1)$. At this point, substitute $x=1$ and $y=1$ into
$\tan^{-1}\bigl(\tfrac{y}{x}\bigr) - \ln\bigl(\tfrac{y}{x}\bigr)=\ln(x)+C$:
$\tan^{-1}(1) \;-\; \ln(1) \;=\; \ln(1) \;+\; C
\;\;\Longrightarrow\;\;
\dfrac{\pi}{4} - 0 \;=\; 0 + C
\;\;\Longrightarrow\;\;
C = \dfrac{\pi}{4}.
Thus, the solution is:
$\displaystyle \tan^{-1}\Bigl(\dfrac{y}{x}\Bigr)
\;-\;
\ln\Bigl(\dfrac{y}{x}\Bigr)
\;=\;
\ln(x)
\;+\;
\dfrac{\pi}{4}.
Step 5: Find the Intersection with $y = \sqrt{3}\,x$
We next impose the additional condition that the solution intersects the line $y = \sqrt{3}\,x$ at $(\alpha,\sqrt{3}\,\alpha)$. Substitute $y = \sqrt{3}\,x$ into the general solution:
$\tan^{-1}\Bigl(\sqrt{3}\Bigr)
\;-\;
\ln\Bigl(\sqrt{3}\Bigr)
\;=\;
\ln(x)
\;+\;
\dfrac{\pi}{4}.
We know $\tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}$. Therefore:
$\dfrac{\pi}{3}
\;-\;
\ln\bigl(\sqrt{3}\bigr)
\;=\;
\ln(x)
\;+\;
\dfrac{\pi}{4}.
Hence,
$\ln(x) = \dfrac{\pi}{3}
\;-\;
\ln\bigl(\sqrt{3}\bigr)
\;-\;
\dfrac{\pi}{4}
\;=\;
\dfrac{\pi}{12}
\;-\;
\ln\bigl(\sqrt{3}\bigr).
So, if $x = \alpha$, then:
$\ln(\alpha) = \dfrac{\pi}{12}
\;-\;
\ln\bigl(\sqrt{3}\bigr).
Step 6: Compute $\ln(\sqrt{3}\,\alpha)$
The problem asks for $\ln\bigl(\sqrt{3}\,\alpha\bigr)$. Observe:
$\ln(\sqrt{3}\,\alpha)
\;=\;
\ln(\sqrt{3})
\;+\;
\ln(\alpha).
Substitute $\ln(\alpha)$ from above:
$\ln(\sqrt{3}\,\alpha)
\;=\;
\ln(\sqrt{3})
\;+\;
\Bigl(\dfrac{\pi}{12}
\;-\;
\ln(\sqrt{3})\Bigr)
\;=\;
\dfrac{\pi}{12}.
Thus, the value of $\ln\bigl(\sqrt{3}\,\alpha\bigr)$ is:
$\boxed{\dfrac{\pi}{12}}.
