Differential equation

Dec 2015
1,085
169
Earth
\(\displaystyle s^{2}+(\frac{ds}{dt}\cdot s)^{2}=1\) .
s=?
 

skipjack

Forum Staff
Dec 2006
21,482
2,472
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.
 
  • Like
Reactions: 1 person
Dec 2015
1,085
169
Earth
\(\displaystyle \phi (t,s,C) \equiv (t+C)^{2}+s^{2} =1=r\).
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 a moderator:

skipjack

Forum Staff
Dec 2006
21,482
2,472
Almost. You should also allow $(4 - 4z)^{1/2} = -dz/dt$.
 
  • Like
Reactions: 1 person

skipjack

Forum Staff
Dec 2006
21,482
2,472
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.
 
  • Like
Reactions: 1 person
Oct 2019
2
2
India
\(\displaystyle S^2 dS/dt + S^2=1\)

\(\displaystyle dS/dt+1=1/S^2\) (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)

\(\displaystyle \int -(1-S^2)/(S^2 - 1) + 1/(S^2 - 1) = \int dt\)

\(\displaystyle S=1/2 \log |(1+S)/(1-S)| - t \)
 
Last edited by a moderator:
  • Like
Reactions: 2 people

skipjack

Forum Staff
Dec 2006
21,482
2,472
You forgot about a constant of integration. Also, the differential equation you solved isn't the one asked about.
 
  • Like
Reactions: 1 person