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(D) Given the differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$.Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}

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$-\frac{1}{4}$

Vedclass · Answer

(D) Given the differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$.Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting this into the differential equation,we get $v + x \frac{dv}{dx} = v + \phi \left( \frac{1}{v} \right)$,which simplifies to $x \frac{dv}{dx} = \phi \left( \frac{1}{v} \right)$.Rearranging the terms,we have $\frac{dv}{\phi(1/v)} = \frac{dx}{x}$.Integrating both sides,$\int \frac{dv}{\phi(1/v)} = \ln |x| + C_1$.Given the general solution $y \ln |cx| = x$,we can rewrite this as $\ln |cx| = \frac{x}{y} = \frac{1}{v}$.Thus,$\ln |x| + \ln |c| = \frac{1}{v}$.Differentiating both sides with respect to $v$,we get $\frac{1}{x} \frac{dx}{dv} = -\frac{1}{v^2}$.From our earlier equation $x \frac{dv}{dx} = \phi(1/v)$,we have $\frac{dx}{dv} = \frac{x}{\phi(1/v)}$.Substituting this into the derivative result: $\frac{1}{x} \cdot \frac{x}{\phi(1/v)} = -\frac{1}{v^2}$,which implies $\phi(1/v) = -v^2$.We want to find $\phi(2)$. Let $\frac{1}{v} = 2$,so $v = \frac{1}{2}$.Then $\phi(2) = -(\frac{1}{2})^2 = -\frac{1}{4}$.

If the general solution of the differential equation $y' = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$,for some function $\phi$,is given by $y \ln |cx| = x$,where $c$ is an arbitrary constant,then $\phi(2)$ is equal to:

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