Let y(x) be the solution of the differential | vedclass

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(A) Given the differential equation $x^2 \frac{dy}{dx} + xy = x^2 + y^2$. Divide by $x^2$: $\frac{dy}{dx} + \frac{y}{x} = 1 + (\frac{y}{x})^2$. Let $v

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(A) Given the differential equation $x^2 \frac{dy}{dx} + xy = x^2 + y^2$. Divide by $x^2$: $\frac{dy}{dx} + \frac{y}{x} = 1 + (\frac{y}{x})^2$. Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting into the equation: $v + x \frac{dv}{dx} + v = 1 + v^2$. $x \frac{dv}{dx} = 1 + v^2 - 2v = (v - 1)^2$. Separating variables: $\int \frac{dv}{(v - 1)^2} = \int \frac{dx}{x}$. $-\frac{1}{v - 1} = \ln|x| + C$. Substituting $v = \frac{y}{x}$: $-\frac{x}{y - x} = \ln x + C$,which simplifies to $\frac{x}{x - y} = \ln x + C$. Given $y(1) = 0$,we have $\frac{1}{1 - 0} = \ln(1) + C \Rightarrow C = 1$. Thus,$\frac{x}{x - y} = \ln x + 1 = \ln(ex)$. For $y(e)$: $\frac{e}{e - y(e)} = \ln(e^2) = 2 \Rightarrow e = 2e - 2y(e) \Rightarrow y(e) = \frac{e}{2}$. For $y(e^2)$: $\frac{e^2}{e^2 - y(e^2)} = \ln(e^3) = 3 \Rightarrow e^2 = 3e^2 - 3y(e^2) \Rightarrow 3y(e^2) = 2e^2 \Rightarrow y(e^2) = \frac{2e^2}{3}$. Finally,$2 \frac{(y(e))^2}{y(e^2)} = 2 \frac{(e/2)^2}{2e^2/3} = 2 \cdot \frac{e^2/4}{2e^2/3} = 2 \cdot \frac{3}{8} = \frac{3}{4} = 0.75$.

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 $....$

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