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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 October 2nd, 2019, 08:35 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Differential equation $\displaystyle s^{2}+(\frac{ds}{dt}\cdot s)^{2}=1$ . s=? October 2nd, 2019, 09:36 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 If $s$ and $t$ are real, $s = \sqrt{1 - (t - \text{c})^2}$, where $c$ is a real constant and $\text{c} - 1 < t < \text{c} + 1$, or $s = -\sqrt{1 - (t - \text{c})^2}$, where $c$ is a real constant and $\text{c} - 1 < t < \text{c} + 1$. There are also the solutions $s = 1$ and $s = -1$, which are the only solutions where $t$ can have any real value. Thanks from idontknow October 3rd, 2019, 01:24 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 $\displaystyle \phi (t,s,C) \equiv (t+C)^{2}+s^{2} =1=r$. Which means $\displaystyle s=\pm 1$ are singular solutions. Substitute for $\displaystyle s=z^{1/2}$ then equation is $\displaystyle 1-z=z\cdot (1/(2\sqrt{z})z')^{2}=z\cdot 1/(4z)(z')^{2}=(z')^{2}/4$. $\displaystyle (4-4z)^{1/2}=dz/dt$. Am I correct? Last edited by skipjack; October 3rd, 2019 at 02:16 AM. October 3rd, 2019, 12:30 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 Almost. You should also allow $(4 - 4z)^{1/2} = -dz/dt$. Thanks from idontknow October 4th, 2019, 02:13 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 The substitution $s = z^{1/2}$ implicitly assumes that $s \geqslant 0$. One could consider substituting $s = -z^{1/2}$ as well, but the equation is separable without any substitution. Either $s^2 = 1$ or $\displaystyle \frac{s ds/dt}{\sqrt{1 - s^2}} = \pm1$, which gives $\sqrt{1 - s^2} = \pm(t - \text{C})$, i.e. $s^2 = 1 - (t - \text{C})^2$, etc. Thanks from idontknow October 5th, 2019, 10:24 AM #6 Newbie   Joined: Oct 2019 From: India Posts: 2 Thanks: 2 $\displaystyle S^2 dS/dt + S^2=1$ $\displaystyle dS/dt+1=1/S^2$ (divide the equation with $\displaystyle S^2$) $\displaystyle dS/dt=(1-S^2)/S^2$ $\displaystyle \int S^2/(1-S^2) dS= \int dt$ (after shifting same variables on same side and integrating on both side) $\displaystyle \int -(1-S^2)/(S^2 - 1) + 1/(S^2 - 1) = \int dt$ $\displaystyle S=1/2 \log |(1+S)/(1-S)| - t$ Thanks from idontknow and DarnItJimImAnEngineer Last edited by skipjack; October 5th, 2019 at 12:26 PM. October 5th, 2019, 12:36 PM #7 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 You forgot about a constant of integration. Also, the differential equation you solved isn't the one asked about. Thanks from topsquark Tags differential, equation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Sonprelis Calculus 6 August 6th, 2014 10:07 AM PhizKid Differential Equations 0 February 24th, 2013 10:30 AM cheyb93 Differential Equations 3 February 7th, 2013 09:28 PM kaanmm Differential Equations 5 June 9th, 2008 08:08 AM pepeatienza Differential Equations 1 May 13th, 2008 01:14 PM

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