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

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

by RoRi

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*}

3 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…

      • Evangelos says:

        Tom, maybe you had your “eureka” moment like I just did. Admittedly before this I was consulting Wolfram for the solution to the integral of 1/sin(x) but RoRi gave me the hint with the half-angle formulas… and it all fell into place perfectly

            \begin{align*} \int\frac{dv}{\sin v} &= \frac{1}{2}\int\frac{dv}{\sin \frac{v}{2}\cos\frac{v}{2}}\\ &= \frac{1}{2}\int\frac{\cos^2\frac{v}{2} + \sin^2\frac{v}{2}}{\sin\frac{v}{2}\cos\frac{v}{2}} dv\\ &= \frac{1}{2}\int\frac{\cos\frac{v}{2}}{\sin\frac{v}{2}} dv + \frac{1}{2}\int\frac{\sin\frac{v}{2}}{\cos\frac{v}{2}} dv\\ &= \log(\sin\frac{v}{2}) - \log(\cos\frac{v}{2})\\ &= \log(\tan\frac{v}{2}) \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):