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# help plese!! differential equation 4dt = t (t^2 -4)^(1/2) ds?

can somebody help me to solve this differential equation with step by step solution.

### 1 Answer

- schmisoLv 71 decade agoFavorite Answer
First separate the variables (t to the LHS, s to the RHS)

4dt = t·√(t² -4) ds

→

4/[t·√(t² -4)] dt = ds

Next step is to integrate LHS with respect to t and right hand side with respect to s:

∫ 4/[t·√(t² -4)] dt = ∫ ds

An irrational integral of the type √(t² -a²) like on the LHS can be transformed to a rational integral solved by the substitution

t = a·sec(x)

here

t = 2·sec(x)

dt/dx = 2·sec(x)·tan(x) dx (because 1+tan²(x) = sec²(x))

dt = 2·sec(x)·√(sec²(x) - 1)

Substitute in the integral

∫ 4/[2·sec(x)·√(4·sec²(x) -4)] ·2·sec(x)·√(sec²(x) - 1)dx = ∫ ds

→

∫ 4/[2·√(sec²(x) -1)] ·√(sec²(x) - 1)dx = ∫ ds

→

∫ 2dx = ∫ ds

→

2x = s + C

Where C is the constant of integration.

With the reverse substitution t = 2·sec(x) <=> x = arcsec(t/2) = arccos(2/t) you get:

2·arccos(2/t) = s + C

Hence the solution of the DE is

s = 2·arccos(2/t) - C

or

t = 2/cos((s+C)/2)

For the evaluation of the constant C you need an additional condition.

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