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October 2nd, 2019, 08:35 AM   #1
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Differential equation

$\displaystyle s^{2}+(\frac{ds}{dt}\cdot s)^{2}=1$ .
s=?
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October 2nd, 2019, 09:36 AM   #2
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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.
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October 3rd, 2019, 01:24 AM   #3
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$\displaystyle \phi (t,s,C) \equiv (t+C)^{2}+s^{2} =1=r$.
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.
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October 3rd, 2019, 12:30 PM   #4
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Almost. You should also allow $(4 - 4z)^{1/2} = -dz/dt$.
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October 4th, 2019, 02:13 PM   #5
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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.
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October 5th, 2019, 10:24 AM   #6
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$\displaystyle S^2 dS/dt + S^2=1$

$\displaystyle dS/dt+1=1/S^2$ (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 skipjack; October 5th, 2019 at 12:26 PM.
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October 5th, 2019, 12:36 PM   #7
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You forgot about a constant of integration. Also, the differential equation you solved isn't the one asked about.
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