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Integrate the differential equation y′ = 1 + y/x

Integrate the differential equation:

    \[ y' = 1 + \frac{y}{x}. \]


We use the substitution y = vx (used in the examples of Section 8.25 of Apostol). So we have

    \[ y = vx \quad \implies \quad y' = v + v'x. \]

(Where the derivative v' is taken with respect to y.) Making this substitution in the given equation (and noting that y = vx implies \frac{y}{x} = v) we have

    \begin{align*}  y' = 1 + \frac{y}{x} && \implies && v + v'x &= 1 + v \\  && \implies && v' &= \frac{1}{x} \\[9pt]  && \implies && \int dv &= \int \frac{1}{x} \, dx \\[9pt]  && \implies && v &= \log |x| + C \\  && \implies && y &= x \log |x| + Cx. \end{align*}

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