1) I'm wondering how one might solve this differential equation: (here $c$ is a real constant)

$$ c = y - \frac{x}{2} \left( y' - \frac{1}{y'}\right).$$

WolframAlpha says it is $y = \frac{1}{2}\left( 2c \pm e^{-k}x^2 \mp e^{k}\right)$, and I expected to find an $x^2$ in there based off of where the equation comes from, but I'm not super experienced with nonlinear DE's. (For those that are curious: lets say that a ray comes from above $x_0$ parallel to the $y$-axis, hits the graph of $y = f(x)$, and the angle between the ray and the tangent line at $x_0$ was equal to the angle between the reflected line and the tangent line---as if the ray were a pool ball and the tangent line were the side of the table. Then the reflected line has a $y$-intercept of $c$.)

2) I'm curious if anyone might have some advice for solving differential equations of this form?

$$ g(x) = y - \frac{x}{2} \left( y' - \frac{1}{y'}\right).$$

Thank you!

  • 1
    $\begingroup$ This is quite a difficult question to do without a computer. I looked, and couldn’t really find much theory on questions of this form. I guess that’s kind of what you start to expect when dealing with non linear DE’s as opposed to all the theory there is on linear DEs. $\endgroup$ – Colin Hicks Jul 6 '18 at 23:38

Multiply through by $y'$ and rearrange to obtain a quadratic in $y'$

$$ y'^2 - 2\frac{y-c}{x}y' - 1 = 0 $$

Then, complete the square

$$ \left(y' - \frac{y-c}{x} \right)^2 = 1 + \left(\frac{y-c}{x}\right)^2 $$

The substitution $u = \frac{y-c}{x}$ is useful here. Then

$$ y = c + xu \implies y' =u + xu' $$

$$ \implies (xu')^2 = 1 + u^2 $$

You can solve for $u'$ here and separate the equation

$$ \frac{u'}{\sqrt{1+u^2}} = \pm\frac{1}{x} $$

The solution to this is

$$ u(x) = \sinh\big(\ln(\pm ax)\big) = \pm\frac{1}{2}\left(ax - \frac{1}{ax}\right) $$


$$ y(x) = c \pm \frac{1}{2}\left(ax^2 -\frac{1}{a}\right) $$



Try to solve for $y'$ giving

$$ y' = \frac{y-c\pm\sqrt{(y-c)^2+x^2}}{x} $$

with solution

$$ y = \frac 12\left(2c\pm C_1\mp\frac{x^2}{C_1}\right) $$


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