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(A) Given differential equation: $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$.Rearranging,we get $2x^2 \frac{dy}{dx} = 2xy + y^2$,so $\frac{dy}{dx} = \frac{2x

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$y = \frac{2x}{1 - \log |x|}, (x 
eq 0, x 
eq e)$

Vedclass · Answer

(A) Given differential equation: $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$.Rearranging,we get $2x^2 \frac{dy}{dx} = 2xy + y^2$,so $\frac{dy}{dx} = \frac{2xy + y^2}{2x^2} \dots (1)$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting into $(1)$: $v + x \frac{dv}{dx} = \frac{2x(vx) + (vx)^2}{2x^2} = \frac{2vx^2 + v^2x^2}{2x^2} = v + \frac{v^2}{2}$.Thus,$x \frac{dv}{dx} = \frac{v^2}{2}$,which implies $\frac{2}{v^2} dv = \frac{dx}{x}$.Integrating both sides: $\int 2v^{-2} dv = \int \frac{1}{x} dx \Rightarrow -\frac{2}{v} = \log |x| + C$.Substituting $v = \frac{y}{x}$,we get $-\frac{2x}{y} = \log |x| + C \dots (2)$.Given $y = 2$ when $x = 1$: $-\frac{2(1)}{2} = \log |1| + C \Rightarrow -1 = 0 + C \Rightarrow C = -1$.Substituting $C = -1$ into $(2)$: $-\frac{2x}{y} = \log |x| - 1 \Rightarrow \frac{2x}{y} = 1 - \log |x|$.Therefore,$y = \frac{2x}{1 - \log |x|}$ for $x 
eq 0, x 
eq e$.

Find the particular solution satisfying the given condition: $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$; $y = 2$ when $x = 1$.

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