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Integrate the differential equation y′ = (y / x) + sin (y / x)

Integrate the following differential equation:

    \[ y' = \frac{y}{x} + \sin \left( \frac{y}{x} \right). \]


Make the substitution v = \frac{y}{x} which gives us y' = v + v'x and so

    \begin{align*}  && y' &= \frac{y}{x} + \sin \left( \frac{y}{x} \right) \\[9pt]  \implies && v + v'x &= v + \sin v \\[9pt]  \implies && v'x &= \sin v \\[9pt]  \implies && \int \frac{1}{\sin v} \, dv &= \int \frac{1}{x} \, dx \\[9pt]  \implies && \log \left( \sin \frac{v}{2} \right) - \log \left( \cos \frac{v}{2} \right) &= \log |x| + C \\[9pt]  \implies && \log \left( \tan \frac{y}{2x} \right) &= \log |x| + C \\[9pt]  \implies && \tan \frac{y}{2x} &= Cx. \end{align*}

2 comments

  1. tom says:

    Wondering about the integral for csc(v) – tanx/2 substitution works but you seem to have an intermediate result involving sin and cos, leading one to think a partial fraction method with sin2x identity was used. Curious…

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