3
$\begingroup$

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!

$\endgroup$
  • 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
4
$\begingroup$

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

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

Hence

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

$\endgroup$
2
$\begingroup$

Hint.

Try to solve for $y'$ giving

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

with solution

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.