A Second Degree Differential Equation

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[@mcwulf25]

Easy to separate and integrate using *substitution*
Write y' = +/- sqrt(1-y^2)
dy / sqrt(1-y^2) = +/- dx
y = sin(t)
dy = cos(t) dt
1- y^2 = cos(t)
Integrate
Int{ cos(t)dt / cos(t) } = t + c = x + k
a = k-c
t = x + a
sin(t) = y = sin(x+a)

[@lawrencejelsma8118]

As an electrical engineer this introduces mathematics to understanding sinusoidal responses and the idea of "phase shifts". In mathematics it was translational function notions not centered around an origin by considering sin(x + ). I think we are getting closer to Fourier Series Calculus understandings here, moreso.

[@jejnsndn]

Back to geometry puzzles

[@user-ud6ui7zt3r]

Polynomials have *degree.*
Differential Equations have *order.*

[@DonEnsley]

y=sin ±(x+c),
or y=cos ± (x+c)

[@martinplesinger]

You can also glue those two solutions together in many ways, e.g., f(x) = 1 for x ≤ 0 and f(x) = cos(x) for x > 0...

[@sabe607]

y = cos theta
y' = - sin theta theta'
cos^2 theta + sin^2 theta * theta'^2 = 1
theta'^2 = 1
theta' = +/- 1
theta = +/- x + C
y = cos(+/- x + C)
y = cos(x+C)

[@aminzqrti7672]

Why don’t you just use Laplace transform?!

[@FisicTrapella]

3rd method
y^2 + (y')^2 = 1
(y')^2 = 1 - y^2 -> y' = dy/dx = sqr(1 - y^2)
dy/sqr(1 - y^2) = dx
x + c = arcsin y -> y = sin (x+c)
y = cons. is a particular solution (x=0)

[@williamperez-hernandez3968]

Though your method is valid, it is incorrect to claim y=constant as a special case (x=0). Y = constant is a solution for all values of x. The key point is: when doing separation of variables we take y' = dy/dx, so dx = dy/y'. This is only valid if y' isn't zero! Therefore we must separate the two cases: y' = 0, and y' ≠ 0. Understandable, we usually jump into the y' ≠ 0 case because the solution y=constant is "boring".

[@moeberry8226]

Y does not equal any constant for all values of x. Y=-1 or 1 for all values of x. Those are the only two constants that work. And of course the non trivial solution of the combination of sin(x) and cos(x). When solving a differential equation using separation of variables you have to make sure the denominators are not 0. In this case 1-y^2 cannot be 0 unles y=1 or -1 and those are your missing solutions.

[@FisicTrapella]

You're right 👍