The solution of the differential equation 2 f | vedclass

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Question

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

Vedclass · Accepted Answer

$y = \frac{2x}{2 - \sqrt{x}}$

Vedclass · Answer

(B) Given the differential equation: $2 \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^2}$.Divide by $y^2$: $2 y^{-2} \frac{dy}{dx} - \frac{1}{xy} = \frac{1}{x^2}$.Let $v = y^{-1}$,then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$,so $y^{-2} \frac{dy}{dx} = -\frac{dv}{dx}$.Substituting into the equation: $-2 \frac{dv}{dx} - \frac{v}{x} = \frac{1}{x^2}$,which simplifies to $\frac{dv}{dx} + \frac{v}{2x} = -\frac{1}{2x^2}$.This is a linear differential equation with integrating factor $IF = e^{\int \frac{1}{2x} dx} = e^{\frac{1}{2} \ln x} = \sqrt{x}$.Multiplying by $IF$: $\sqrt{x} \frac{dv}{dx} + \frac{v}{2\sqrt{x}} = -\frac{1}{2x^{3/2}}$,so $\frac{d}{dx}(v \sqrt{x}) = -\frac{1}{2} x^{-3/2}$.Integrating both sides: $v \sqrt{x} = -\frac{1}{2} \int x^{-3/2} dx = -\frac{1}{2} \frac{x^{-1/2}}{-1/2} + C = \frac{1}{\sqrt{x}} + C$.Thus,$v = \frac{1}{x} + \frac{C}{\sqrt{x}} = \frac{1 + C\sqrt{x}}{x}$.Since $v = 1/y$,we have $y = \frac{x}{1 + C\sqrt{x}}$.Given $y = 2$ when $x = 1$: $2 = \frac{1}{1 + C}$,so $2 + 2C = 1$,which means $2C = -1$ or $C = -1/2$.Substituting $C$: $y = \frac{x}{1 - \frac{1}{2}\sqrt{x}} = \frac{2x}{2 - \sqrt{x}}$.Therefore,the correct option is $B$.

The solution of the differential equation $2 \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^2}$,given that $y = 2$ when $x = 1$,is

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