Question:

Divide the equation by $x^2$: $$\dfrac{dy}{dx} + \dfrac{y}{x} = 1 + \left(\dfrac{y}{x}\right)^2$$ Substitute $y = vx \implies \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$: $$v + x\dfrac{dv}{dx} + v = 1 + v^2 \implies x\dfrac{dv}{dx} = (v-1)^2$$ Separate variables and integrate: $$\int \\dfrac{dv}{(v-1)^2} = \int \dfrac{dx}{x} \implies -\dfrac{1}{v-1} = \log_e x + C$$ Substitute $v = y/x$: $$\dfrac{x}{x - y} = \log_e x + C$$ Using $y(1) = 0$: $$\dfrac{1}{1-0} = \log_e 1 + C \implies C = 1$$ Thus, the solution is: $$\dfrac{x}{x-y} = 1 + \log_e x \implies y = \dfrac{x \log_e x}{1 + \log_e x}$$ Calculate $y(e)$ and $y(e^2)$: - $y(e) = \dfrac{e \log_e e}{1 + \log_e e} = \dfrac{e}{2}$ - $y(e^2) = \dfrac{e^2 \log_e (e^2)}{1 + \log_e (e^2)} = \dfrac{2e^2}{3}$ Substitute these into the expression: $$2 \dfrac{(y(e))^2}{y(e^2)} = 2 \dfrac{e^2/4}{2e^2/3} = 2 \cdot \dfrac{3}{8} = 0.75$$