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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi \left( \frac{y}{x} \right)}{\phi' \left( \frac{y}{x} \right)}$.Substitu

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$\phi \left( \frac{y}{x} \right) = kx$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi \left( \frac{y}{x} \right)}{\phi' \left( \frac{y}{x} \right)}$.Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the differential equation,we get:$v + x \frac{dv}{dx} = v + \frac{\phi(v)}{\phi'(v)}$.Subtracting $v$ from both sides,we have:$x \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)}$.Separating the variables,we get:$\frac{\phi'(v)}{\phi(v)} dv = \frac{dx}{x}$.Integrating both sides:$\int \frac{\phi'(v)}{\phi(v)} dv = \int \frac{dx}{x}$.This yields $\log |\phi(v)| = \log |x| + \log |k|$,where $\log |k|$ is the constant of integration.Using the property of logarithms,$\log |\phi(v)| = \log |kx|$.Therefore,$\phi(v) = kx$.Substituting $v = \frac{y}{x}$ back,we obtain the solution:$\phi \left( \frac{y}{x} \right) = kx$.

The solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi \left( \frac{y}{x} \right)}{\phi' \left( \frac{y}{x} \right)}$ is

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