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 $2\frac{(y(e))^2}{y(e^2)}$ is ______.

Answer 1

3/4

Answer 2

The given differential equation is $x^2\frac{dy}{dx}+xy = x^2+y^2$. Divide by $x^2$ (since $x > 1/e > 0$, $x^2 \neq 0$): $\frac{dy}{dx} + \frac{y}{x} = 1 + \frac{y^2}{x^2}$ $\frac{dy}{dx} = 1 - \frac{y}{x} + \left(\frac{y}{x}\right)^2$. This is a homogeneous differential equation. Let $v = \frac{y}{x}$. Then $y = vx$, and $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting this into the equation: $v + x\frac{dv}{dx} = 1 - v + v^2$ $x\frac{dv}{dx} = v^2 - 2v + 1$ $x\frac{dv}{dx} = (v-1)^2$

This is a separable differential equation. $\frac{dv}{(v-1)^2} = \frac{dx}{x}$ (assuming $v \neq 1$)

Integrate both sides: $\int \frac{dv}{(v-1)^2} = \int \frac{dx}{x}$ $-\frac{1}{v-1} = \ln|x| + C$ Since $x > 1/e > 0$, $|x| = x$. $-\frac{1}{v-1} = \ln x + C$

Substitute back $v = y/x$: $-\frac{1}{y/x - 1} = \ln x + C$ $-\frac{1}{(y-x)/x} = \ln x + C$ $-\frac{x}{y-x} = \ln x + C$ $\frac{x}{x-y} = \ln x + C$

Use the initial condition $y(1)=0$: Substitute $x=1$ and $y=0$: $\frac{1}{1-0} = \ln 1 + C$ $1 = 0 + C \implies C = 1$

The particular solution is: $\frac{x}{x-y} = \ln x + 1$

We need to find $y(e)$ and $y(e^2)$. For $y(e)$, substitute $x=e$: $\frac{e}{e-y(e)} = \ln e + 1 = 1 + 1 = 2$ $e = 2(e-y(e))$ $e = 2e - 2y(e)$ $2y(e) = 2e - e = e$ $y(e) = e/2$

For $y(e^2)$, substitute $x=e^2$: $\frac{e^2}{e^2-y(e^2)} = \ln(e^2) + 1 = 2\ln e + 1 = 2(1) + 1 = 3$ $\frac{e^2}{e^2-y(e^2)} = 3$ $e^2 = 3(e^2-y(e^2))$ $e^2 = 3e^2 - 3y(e^2)$ $3y(e^2) = 3e^2 - e^2 = 2e^2$ $y(e^2) = \frac{2e^2}{3}$

Now calculate the expression $2\frac{(y(e))^2}{y(e^2)}$: $(y(e))^2 = (e/2)^2 = e^2/4$ $2\frac{(y(e))^2}{y(e^2)} = 2 \times \frac{e^2/4}{2e^2/3}$ $= 2 \times \frac{e^2}{4} \times \frac{3}{2e^2}$ $= 2 \times \frac{3}{8}$ $= \frac{6}{8} = \frac{3}{4}$

The value of $2\frac{(y(e))^2}{y(e^2)}$ is $3/4$.