# Differential equation

#### idontknow

$$\displaystyle s^{2}+(\frac{ds}{dt}\cdot s)^{2}=1$$ .
s=?

#### skipjack

Forum Staff
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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#### idontknow

$$\displaystyle \phi (t,s,C) \equiv (t+C)^{2}+s^{2} =1=r$$.
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?

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#### skipjack

Forum Staff
Almost. You should also allow $(4 - 4z)^{1/2} = -dz/dt$.

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#### skipjack

Forum Staff
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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#### jinxrifle

$$\displaystyle S^2 dS/dt + S^2=1$$

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

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

$$\displaystyle S=1/2 \log |(1+S)/(1-S)| - t$$

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#### skipjack

Forum Staff
You forgot about a constant of integration. Also, the differential equation you solved isn't the one asked about.

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