Question 9

Given differential equation: $x^2 \frac{dy}{dx} + xy = x^2 + y^2$.

Divide by $x^2$: $\frac{dy}{dx}$ + $\frac{y}{x}$ = 1 + $\left(\frac{y}{x} \right)^2$.

Let $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$.

Substituting, we get: $v + x\frac{dv}{dx} + v = 1 + v^2 \implies x\frac{dv}{dx} = v^2 - 2v + 1 = (v-1)^2$.

Separating variables: $\int \frac{dv}{(v-1)^2} = \int \frac{dx}{x} \implies -\frac{1}{v-1} = \ln x + C$.

Replacing $v = \frac{y}{x}$: $-\frac{1}{\frac{y}{x}-1} = \ln x + C \implies \frac{x}{x-y} = \ln x + C$.

Using $y(1) = 0$: $\frac{1}{1-0} = \ln 1 + C \implies C = 1$.

Thus, $\frac{x}{x-y} = \ln x + 1 \implies x-y = \frac{x}{1+\ln x} \implies y = x - \frac{x}{1+\ln x} = \frac{x \ln x}{1+\ln x}$. $y(e) = \frac{e \ln e}{1+\ln e} = \frac{e}{2}$. $y(e^2) = \frac{e^2 \ln e^2}{1+\ln e^2} = \frac{2e^2}{3}$. Value $= 2 \frac{(y(e))^2}{y(e^2)} = 2 \frac{(e/2)^2}{2e^2/3} = 2 \cdot \frac{e^2}{4} \cdot \frac{3}{2e^2} = \frac{3}{4} = 0.75$.