Key Concepts and Formulas

• Homogeneous Differential Equations: A differential equation of the form $M(x, y)dx + N(x, y)dy = 0$ is homogeneous if $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree. This means $M(tx, ty) = t^n M(x, y)$ and $N(tx, ty) = t^n N(x, y)$ for some $n$.
• Solution Method for Homogeneous Equations: Substitute $x = vy$ or $y = vx$. This transforms the equation into a separable differential equation.
• Integration: Basic integration techniques and knowledge of standard integrals.

Step-by-Step Solution

Step 1: Check for Homogeneity and Rearrange the Equation

The given differential equation is $y^2 dx + (x^2 - xy + y^2) dy = 0$. We can rewrite it as: $\frac{dx}{dy} = \frac{-(x^2 - xy + y^2)}{y^2} = -\frac{x^2}{y^2} + \frac{x}{y} - 1$ Let $f(x,y) = -\frac{x^2}{y^2} + \frac{x}{y} - 1$. Then $f(tx, ty) = -\frac{(tx)^2}{(ty)^2} + \frac{tx}{ty} - 1 = -\frac{x^2}{y^2} + \frac{x}{y} - 1 = f(x, y)$. So, the equation is homogeneous.

Step 2: Substitute $x = vy$

Let $x = vy$. Then $\frac{dx}{dy} = v + y \frac{dv}{dy}$. Substituting into the differential equation: $v + y \frac{dv}{dy} = -v^2 + v - 1$

Step 3: Separate Variables

$y \frac{dv}{dy} = -v^2 - 1$ $\frac{dv}{v^2 + 1} = -\frac{dy}{y}$

Step 4: Integrate Both Sides

$\int \frac{dv}{v^2 + 1} = \int -\frac{dy}{y}$ $\arctan(v) = -\ln|y| + C$

Step 5: Substitute Back $v = x/y$

$\arctan\left(\frac{x}{y}\right) = -\ln|y| + C$

Step 6: Apply the Initial Condition (1, 1)

The solution curve passes through (1, 1). Substituting $x = 1$ and $y = 1$: $\arctan\left(\frac{1}{1}\right) = -\ln|1| + C$ $\arctan(1) = -\ln(1) + C$ $\frac{\pi}{4} = 0 + C$ So, $C = \frac{\pi}{4}$.

Step 7: Write the Particular Solution

The particular solution is: $\arctan\left(\frac{x}{y}\right) = -\ln|y| + \frac{\pi}{4}$

Step 8: Apply the Intersection Condition

The curve intersects the line $y = \sqrt{3}x$ at $(\alpha, \sqrt{3}\alpha)$. Substituting $x = \alpha$ and $y = \sqrt{3}\alpha$: $\arctan\left(\frac{\alpha}{\sqrt{3}\alpha}\right) = -\ln|\sqrt{3}\alpha| + \frac{\pi}{4}$ $\arctan\left(\frac{1}{\sqrt{3}}\right) = -\ln(\sqrt{3}\alpha) + \frac{\pi}{4}$ $\frac{\pi}{6} = -\ln(\sqrt{3}\alpha) + \frac{\pi}{4}$

Step 9: Solve for $\ln(\sqrt{3}\alpha)$

$\ln(\sqrt{3}\alpha) = \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi - 2\pi}{12} = \frac{\pi}{12}$ We want to find the value of $\log_e(\sqrt{3}\alpha)$, which is the same as $\ln(\sqrt{3}\alpha)$. So, $\ln(\sqrt{3}\alpha) = \frac{\pi}{12}$.

Step 10: Find the Value of $\alpha$ Since $\ln(\sqrt{3}\alpha) = \frac{\pi}{12}$, we have $\sqrt{3}\alpha = e^{\frac{\pi}{12}}$. Thus, $\alpha = \frac{1}{\sqrt{3}}e^{\frac{\pi}{12}}$. However, we are looking for the value of $\ln(\sqrt{3}\alpha)$, which is $\frac{\pi}{12}$.

Step 11: Find the value of ${\log _e}(\sqrt 3 \alpha )$ We have already calculated $\ln(\sqrt{3}\alpha) = \frac{\pi}{12}$ in Step 9.

Common Mistakes & Tips

• Sign Errors: Be very careful with signs when rearranging and integrating the differential equation.
• Correct Substitution: Ensure you substitute back to the original variables after integration.
• Logarithm Properties: Remember logarithm properties when simplifying expressions.

Summary

We solved the homogeneous differential equation by using the substitution $x = vy$. After separating variables and integrating, we applied the initial condition (1, 1) to find the particular solution. Then, we used the intersection condition with the line $y = \sqrt{3}x$ to find the value of $\ln(\sqrt{3}\alpha)$, which is $\frac{\pi}{12}$.

The final answer is \boxed{\frac{\pi}{12}}, which corresponds to option (C).