Question

Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.

Hint:

In such differential equations, try using substitution like \( y = xz \) when you see expressions like \( \frac{y}{x} \). This often reduces the equation to a separable form.

Answer 1

Step 1: Given differential equation: \[ x^2 \frac{dy}{dx} + xy = x^2 + y^2 \Rightarrow x^2 \frac{dy}{dx} = x^2 + y^2 - xy \] Divide both sides by \( x^2 \) (since \( x>\frac{1}{e}>0 \)): \[ \frac{dy}{dx} = 1 + \frac{y^2}{x^2} - \frac{y}{x} \Rightarrow \frac{dy}{dx} = \left( \frac{y}{x} \right)^2 - \frac{y}{x} + 1 \] Let \( z = \frac{y}{x} \Rightarrow y = xz \) Differentiate both sides: \[ \frac{dy}{dx} = z + x \frac{dz}{dx} \] Substitute in the original equation: \[ z + x \frac{dz}{dx} = z^2 - z + 1 \Rightarrow x \frac{dz}{dx} = z^2 - 2z + 1 = (z - 1)^2 \]
Step 2: Separate variables and integrate: \[ \frac{dz}{(z - 1)^2} = \frac{dx}{x} \Rightarrow \int \frac{1}{(z - 1)^2} dz = \int \frac{1}{x} dx \] Integrate: \[ - \frac{1}{z - 1} = \ln x + C \Rightarrow \frac{1}{1 - z} = \ln x + C \Rightarrow z = 1 - \frac{1}{\ln x + C} \] Recall \( z = \frac{y}{x} \Rightarrow y = xz = x\left(1 - \frac{1}{\ln x + C}\right) \)
Step 3: Apply initial condition \( y(1) = 0 \) At \( x = 1 \), \( \ln 1 = 0 \Rightarrow \) \[ y(1) = 1 \cdot \left(1 - \frac{1}{C} \right) = 0 \Rightarrow 1 - \frac{1}{C} = 0 \Rightarrow C = 1 \] So the solution is: \[ y(x) = x \left(1 - \frac{1}{\ln x + 1} \right) = x \left( \frac{\ln x}{\ln x + 1} \right) \]
Step 4: Compute \( y(e) \) and \( y(e^2) \) - \( \ln e = 1 \Rightarrow y(e) = e \cdot \frac{1}{1 + 1} = \frac{e}{2} \) - \( \ln(e^2) = 2 \Rightarrow y(e^2) = e^2 \cdot \frac{2}{2 + 1} = e^2 \cdot \frac{2}{3} \) Now: \[ 2 \cdot \frac{(y(e))^2}{y(e^2)} = 2 \cdot \frac{\left( \frac{e}{2} \right)^2}{\frac{2e^2}{3}} = 2 \cdot \frac{e^2 / 4}{2e^2 / 3} = 2 \cdot \frac{1}{4} \cdot \frac{3}{2} = \frac{3}{4} \Rightarrow \boxed{3} \]