I need help to solve a differential equation?

Question

4dt = t (t^2 - 4 )^ 1/2 ds
i will really apreciate if somebody can help me to solve this separable differential equation step bystep

Answer 1

ds = ∫ 4 dt / t* (t^2 - 4 )^1/2 +c

It is separated. For integrating use t= 2 sec (x)
t^2 - 4 )^1/2 = 2 tan (x)
dt = 2 sec (x) tan (x) dx

ds = 4 ∫ 2 sec (x) tan (x) dx / 2 sec (x) 2 tan (x) + c

ds = 2 ∫ dx +c
ds = 2 x +c -----(a) put back t >>> t =2 sec (x)

x= sec^-1(t/2)

s = 2 sec^-1 (t /2) + c ===== (G) answer

Also t >>> t =2 sec (x) gives x = 2 tan^-1{[sqrt(t^2 - 4] / 2} by pythagorous in right angled triangle of (x).

S = 2 tan^-1{ [sqrt (t^2 - 4)] /2} +c another form......(M)

S = - 2 tan^-1{2 / [sqrt (t^2 - 4)] } +c' another form...... (N)

c' = (2+Pi)+ c
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You will find different forms of final answer in "s" and "t" without considering the "c" constant. But if you have a boundary condition (say at t=2, s=2 then c=0 for (G) and (M) - derived from same substitution). But (N), will c' = 2+pi
when put in (N) (let us see if (M) can be had

S = - 2 tan^-1{2 / [sqrt (t^2 - 4)] } +(2+pi) >>>(N)
S - 2 = - 2 tan^-1{2 / [sqrt (t^2 - 4)] } + pi
(S - 2)/2 = - tan^-1{2 / [sqrt (t^2 - 4)] } + (pi/2)

(pi/2)-(S- 2)/2 = tan^-1{2 / [sqrt (t^2 - 4)] }
take tangent on both sides

Tan {(pi/2)-(S- 2)/2} = {2 / [sqrt (t^2 - 4)] }
COT {(S- 2)/2} = {2 / [sqrt (t^2 - 4)] } invert
TAN {(S- 2)/2} = {[sqrt (t^2 - 4)] / 2 }

{(S- 2)/2} = TAN^-1 {[sqrt (t^2 - 4)] / 2 }

S = 2 TAN^-1 {[sqrt (t^2 - 4)] / 2 } + 2

That is same as (M) with c=2 as assumed.

You can easily leave answer of diff. equ as (G)

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If you want direct this form
S = - 2 tan^-1{2 / [sqrt (t^2 - 4)] } +c' take t^2 common from sqrt

= ∫ 4 dt / t^2 (1 - 4/t^2)^1/2 +c' -(P)
use substitution in integral ds
2 / t = sin x thus (1 - 4/t^2)^1/2 = cos x
-2/t^2 . dt = cos x dx put in (P)

s= - 2x +c' put back value of x >>>>

from sin x = 2/t so tan x = 2 / sqrt(t^2 - 4)
x = tan^-1 [2 / sqrt(t^2 - 4)]

S= - 2 tan^-1 [2 / sqrt(t^2 - 4)] +c' (it is as good as c) now

Answer 2

Then ds = 4/[t (t^2 - 4 )^ 1/2] dt. Just integrate.