Let y(x) be the solution of the differential equation x²dy/dx + xy = x² + y², x > 1/e, satisfying y(1) = 0. Then the value of 2(y(e))²/y(e²) is ________.

Let y(x) be the solution of the differential equation x²dy/dx + xy = x² + y², x > 1/e, satisfying y(1) = 0. Then the value of 2(y(e))²/y(e²) is ________. x²dy/dx+xy = x²+y², x>1/e y(1)=0 To find: 2(y(e))²/y(e²) dy/dx+y/x = 1+(y/x)² 1. v=y/x ⇒ separable 2. integrate 3. y(1)=0 ⇒ C ⇒ value y=vx ⇒ dy/dx=v+xdv/dx v+xv'+v = 1+v² ⇒ x v' = (v-1)² dv/((v-1)²) = dx/x ⇒ -1/(v-1) = ln x + C v = 1 - 1/(ln x + C) y = vx = x - x/(ln x + C) y(1)=0: 0 = 1 - 1/C ⇒ C=1 y(x) = x - x/(ln x + 1) y(e) = e - e/2 = e/2 y(e²) = e² - e²/3 = 2e²/3 2(y(e))²/y(e²) = 2((e/2)²)/(2e²/3) = (e²/2)/(2e²/3) = 3/4 boxed3/4

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

Answer: $0.75$

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Written solution

$x^2\dfrac{dy}{dx}+xy = x^2+y^2,\quad x>\dfrac1e$

$y(1)=0$

$\text{To find: } 2\dfrac{(y(e))^2}{y(e^2)}$

$\dfrac{dy}{dx}+\dfrac{y}{x} = 1+\left(\dfrac{y}{x}\right)^2$ $1.\ v=\dfrac{y}{x} \Rightarrow \text{separable}$ $2.\ \text{integrate}$ $3.\ y(1)=0 \Rightarrow C \Rightarrow \text{value}$ $y=vx \Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ $v+xv'+v = 1+v^2 \Rightarrow x v' = (v-1)^2$ $\dfrac{dv}{(v-1)^2} = \dfrac{dx}{x}$ $\Rightarrow -\dfrac{1}{v-1} = \ln x + C$ $v = 1 - \dfrac{1}{\ln x + C}$ $y = vx = x - \dfrac{x}{\ln x + C}$ $y(1)=0:\ 0 = 1 - \dfrac{1}{C} \Rightarrow C=1$ $y(x) = x - \dfrac{x}{\ln x + 1}$ $y(e) = e - \dfrac{e}{2} = \dfrac{e}{2}$ $y(e^2) = e^2 - \dfrac{e^2}{3} = \dfrac{2e^2}{3}$ $2\dfrac{(y(e))^2}{y(e^2)} = 2\dfrac{(e/2)^2}{2e^2/3} = \dfrac{e^2/2}{2e^2/3} = \dfrac34$ $\boxed{\dfrac34}$

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