Key Concepts and Formulas

• Homogeneous Differential Equation: A differential equation of the form $\frac{dy}{dx} = f(x,y)$ is homogeneous if $f(\lambda x, \lambda y) = f(x,y)$ for any constant $\lambda$.
• Substitution for Homogeneous Equations: To solve a homogeneous differential equation, use the substitution $y = vx$, which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
• Separable Differential Equations: A differential equation is separable if it can be written in the form $g(y)dy = h(x)dx$.

Step-by-Step Solution

Step 1: Identify the Differential Equation as Homogeneous

The given differential equation is $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$ To confirm it is homogeneous, we test the function $f(x,y) = \frac{xy}{x^2 + y^2}$: $f(\lambda x, \lambda y) = \frac{(\lambda x)(\lambda y)}{(\lambda x)^2 + (\lambda y)^2} = \frac{\lambda^2 xy}{\lambda^2 x^2 + \lambda^2 y^2} = \frac{\lambda^2 xy}{\lambda^2 (x^2 + y^2)} = \frac{xy}{x^2 + y^2} = f(x,y)$ Since $f(\lambda x, \lambda y) = f(x,y)$, the differential equation is homogeneous.

Step 2: Apply the Substitution $y = vx$

Let $y = vx$. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the original equation, we get $v + x\frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} = \frac{vx^2}{x^2 + v^2x^2} = \frac{vx^2}{x^2(1 + v^2)} = \frac{v}{1 + v^2}$

Step 3: Separate Variables

We now have $v + x\frac{dv}{dx} = \frac{v}{1 + v^2}$ Subtract $v$ from both sides: $x\frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v(1 + v^2)}{1 + v^2} = \frac{v - v - v^3}{1 + v^2} = \frac{-v^3}{1 + v^2}$ Separate the variables $x$ and $v$: $\frac{1 + v^2}{v^3} dv = -\frac{1}{x} dx$

Step 4: Integrate Both Sides

Integrate both sides of the equation: $\int \frac{1 + v^2}{v^3} dv = \int -\frac{1}{x} dx$ $\int \left(\frac{1}{v^3} + \frac{1}{v}\right) dv = -\int \frac{1}{x} dx$ $\int (v^{-3} + v^{-1}) dv = -\int \frac{1}{x} dx$ $\frac{v^{-2}}{-2} + \ln|v| = -\ln|x| + C$ $-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$

Step 5: Substitute Back $v = \frac{y}{x}$

Substitute $v = \frac{y}{x}$ back into the equation: $-\frac{1}{2(\frac{y}{x})^2} + \ln\left|\frac{y}{x}\right| = -\ln|x| + C$ $-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C$ $-\frac{x^2}{2y^2} + \ln|y| = C$ Since $y(1)=1$, $y$ is positive. $-\frac{x^2}{2y^2} + \ln y = C$

Step 6: Apply the Initial Condition $y(1) = 1$

Substitute $x = 1$ and $y = 1$ into the equation to find $C$: $-\frac{1^2}{2(1)^2} + \ln(1) = C$ $-\frac{1}{2} + 0 = C$ $C = -\frac{1}{2}$ So the particular solution is $-\frac{x^2}{2y^2} + \ln y = -\frac{1}{2}$

Step 7: Find $x$ when $y = e$

Substitute $y = e$ into the particular solution: $-\frac{x^2}{2e^2} + \ln e = -\frac{1}{2}$ $-\frac{x^2}{2e^2} + 1 = -\frac{1}{2}$ $-\frac{x^2}{2e^2} = -\frac{3}{2}$ $\frac{x^2}{2e^2} = \frac{3}{2}$ $x^2 = 3e^2$ $x = \sqrt{3e^2} = e\sqrt{3}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs during integration and algebraic manipulations.
• Logarithm Properties: Remember that $\ln(a/b) = \ln(a) - \ln(b)$ and $\ln(e) = 1$.
• Substitution: Ensure you substitute back $v = y/x$ after integration to get the solution in terms of the original variables.

Summary

We identified the given differential equation as homogeneous and used the substitution $y = vx$ to transform it into a separable equation. After integrating and applying the initial condition $y(1) = 1$, we obtained the particular solution. Finally, we substituted $y = e$ into the particular solution and solved for $x$, finding $x = e\sqrt{3}$.

The final answer is \boxed{\sqrt 3 e}, which corresponds to option (D).