Question

Given the differential equation: \[ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad \text{with } y(1) = 0 \] Find the value of \(\frac{y^2(e)}{y(e^2)}\).

• A. \(\frac{1}{2}\)
• B. \(\frac{3}{8}\)
• C. \(e^3\)
• D. \(1\)

Hint:

When a differential equation includes \(y^2/x^2\) and \(y/x\), try substituting \(v = \frac{y}{x}\) to reduce it to a separable form.

Answer 1

The Correct Option is B

We are given: \[ x^2 \frac{dy}{dx} + xy = x^2 + y^2 \Rightarrow x^2 \frac{dy}{dx} = x^2 + y^2 - xy \] Divide both sides by \(x^2\): \[ \frac{dy}{dx} = 1 + \frac{y^2}{x^2} - \frac{y}{x} \] Step 1: Use substitution \(v = \frac{y}{x} \Rightarrow y = vx\) Differentiate: \[ \frac{dy}{dx} = v + x \frac{dv}{dx} \] Substitute into the DE: \[ v + x \frac{dv}{dx} = 1 + v^2 - v \Rightarrow x \frac{dv}{dx} = v^2 - 2v + 1 = (v - 1)^2 \] Step 2: Separate and integrate: \[ \frac{dv}{(v - 1)^2} = \frac{dx}{x} \Rightarrow \int \frac{dv}{(v - 1)^2} = \int \frac{dx}{x} \Rightarrow -\frac{1}{v - 1} = \ln x + C \] Step 3: Use initial condition \(y(1) = 0 \Rightarrow v = 0\) at \(x = 1\): \[ -\frac{1}{0 - 1} = \ln 1 + C \Rightarrow 1 = 0 + C \Rightarrow C = 1 \] Step 4: Back-substitute for \(v = \frac{y}{x}\): \[ -\frac{1}{\frac{y}{x} - 1} = \ln x + 1 \Rightarrow \frac{1}{\frac{y}{x} - 1} = -(\ln x + 1) \Rightarrow \frac{y}{x} - 1 = \frac{-1}{\ln x + 1} \Rightarrow \frac{y}{x} = 1 - \frac{1}{\ln x + 1} \] So: \[ y = x \left(1 - \frac{1}{\ln x + 1} \right) \] Step 5: Compute the required value: We need: \[ \frac{y^2(e)}{y(e^2)} \] Recall: \[ \ln e = 1, \quad \ln e^2 = 2 \] So: \[ y(e) = e \left(1 - \frac{1}{1 + 1} \right) = e \cdot \frac{1}{2} = \frac{e}{2} \Rightarrow y^2(e) = \left(\frac{e}{2}\right)^2 = \frac{e^2}{4} \] \[ y(e^2) = e^2 \left(1 - \frac{1}{2 + 1} \right) = e^2 \cdot \frac{2}{3} = \frac{2e^2}{3} \] \[ \frac{y^2(e)}{y(e^2)} = \frac{e^2/4}{2e^2/3} = \frac{1}{4} \cdot \frac{3}{2} = \frac{3}{8} \]