
Differential Equations Ordinary and Partial Differential Equations Math Forum 
 LinkBack  Thread Tools  Display Modes 
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. 
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 $\displaystyle 1z=z\cdot (1/(2\sqrt{z})z')^{2}=z\cdot 1/(4z)(z')^{2}=(z')^{2}/4$. $\displaystyle (44z)^{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 $(4  4z)^{1/2} = dz/dt$.

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. 
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 $\displaystyle S^2$) $\displaystyle dS/dt=(1S^2)/S^2$ $\displaystyle \int S^2/(1S^2) dS= \int dt $ (after shifting same variables on same side and integrating on both side) $\displaystyle \int (1S^2)/(S^2  1) + 1/(S^2  1) = \int dt$ $\displaystyle S=1/2 \log (1+S)/(1S)  t $ 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.


Tags 
differential, equation 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Gompertz equation  differential equation  Sonprelis  Calculus  6  August 6th, 2014 10:07 AM 
Show that an equation satisfies a differential equation  PhizKid  Differential Equations  0  February 24th, 2013 10:30 AM 
differential equation  cheyb93  Differential Equations  3  February 7th, 2013 09:28 PM 
differential equation???  kaanmm  Differential Equations  5  June 9th, 2008 08:08 AM 
Help with a differential equation  pepeatienza  Differential Equations  1  May 13th, 2008 01:14 PM 