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According to my cosmology book:$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$where $a$ is the scale factor, the dot indicates derivative wrt time and $\varLambda$ and $K$ are constants. The author then says “Introducing a new variable $x$ by $a^{3}=x^{2}$ and integrating once more with the initial condition $a\left(0\right)=0$ we obtain”$$a^{3}=\frac{3K}{\varLambda}\sinh^{2}\left(\frac{t}{t_{\varLambda}}\right),$$where $t_{\varLambda}=\frac{2}{\sqrt{3\varLambda}}$. I've tried (with difficulty) integrating the first equation using WolframAlpha but end up with nothing like the second equation. Any suggestions or advice, please?

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$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$ $$x^2=a^3 \implies 3a^2 \dot a =2\dot x x \implies 9a^4 \dot a ^2=4x^2\dot x ^2 $$ $$\implies 9a^3(a\dot a^2)=4x^2\dot x^2 \implies (a\dot a ^2)=\frac {4\dot x^2}{9}$$ The equation becomes $$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$ $$\frac {4\dot x^2}{9}=\frac{\varLambda}{3}x^2+K$$ $$2\dot x=\pm\sqrt {3\varLambda x^2+9K}$$ I integrate for the positive equation... $$ \int \frac {dx}{\sqrt { x^2+3K/\varLambda }}= \frac{\sqrt {3\varLambda }}{2}\int dt$$ Use a $\sinh(..)$ substitution for the integral and square the result to get $a^3$ $$s^2=3K/\varLambda $$ Substitute $x=s\sinh(m) \implies dx=s\cosh(m)$...for the positive equation we get $$ \int \frac {dx}{\sqrt { x^2+s^2 }}= \frac{\sqrt {3\varLambda }}{2}\int dt$$ $$ m= \frac{\sqrt {3\varLambda }}{2}t+C$$ $$ arcsinh(x/s)= \frac{\sqrt {3\varLambda }}{2}t+C$$ $$ x= s\sinh(\frac{\sqrt {3\varLambda }}{2}t+C)$$ for the initial condition given $C=0$ $$\boxed{ a^3(t)= \frac {3K}{\varLambda}\sinh^2(\frac{\sqrt {3\varLambda }}{2}t)}$$


Edit for the integral substitute $x=s \sinh(m)$ $$I=\int \frac {dx}{\sqrt { x^2+s^2 }}= \int \frac {s\cosh(m)dm}{\sqrt { s^2\sinh^2(m)+s^2 }}$$ $$I= \int \frac {s\cosh(m)dm}{\sqrt { s^2(\sinh^2(m)+1) }}$$ since you have the equality for hyperbolic functions $\cosh^2(m)-\sinh^2(m)=1$ $$I= \int \frac {s\cosh(m)dm}{\sqrt { s^2\cosh^2(m) }}$$ $$I= \int dm=m+K$$ $$I=arcsinh(x/s)+K$$

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  • $\begingroup$ Brilliant. I was eventually able to follow this, but only when I found out that $\int\frac{dx}{\sqrt{x^{2}+s^{2}}}=\sinh^{-1}\frac{x}{s}+constant$ (from en.wikipedia.org/wiki/Trigonometric_substitution). Thank you so much. $\endgroup$ – Peter4075 May 29 '18 at 14:20
  • $\begingroup$ yw @Peter4075 use a sinh substitution for the integral simply..then the integral becomes very simple ... $\endgroup$ – Isham May 29 '18 at 14:38
  • $\begingroup$ Sorry to be so slow, but what do you mean by a sinh substitution? I thought that's what you'd done when you used $\int\frac{dx}{\sqrt{x^{2}+s^{2}}}=\sinh^{-1}\frac{x}{s}+constant$. $\endgroup$ – Peter4075 May 29 '18 at 14:51
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    $\begingroup$ Thanks. I'll need to study that. I think at my basic level it's more straightforward to use the standard integral. Hyperbolic functions we're never part of my high school maths education! $\endgroup$ – Peter4075 May 29 '18 at 15:13
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    $\begingroup$ Yes, just done it. That's really neat. I'm assuming the tan substitution would be difficult because it involves a sec integral. $\endgroup$ – Peter4075 May 29 '18 at 17:05
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Use that $2x\dot x=3a^2\dot a$ so that $4x^2\dot x^2=9a^4\dot a^2=9x^2a\dot a^2$ and then $$ 4\dot x^2=9a\dot a^2=3Λx^2+9K $$ which you can now solve with standard substitutions.

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Using Mathematica:

sol = DSolve[{a[t] a'[t]^2 == Λ/3*a[t]^3 + K, a[0] == 0}, a[t], t] // Quiet

(*{{a[t] -> -(((-3)^(1/3) K^(1/3) (-Tanh[1/2 Sqrt[3] t Sqrt[Λ]])^(
    2/3))/(-Λ (-1 + Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3))},
  {a[t] -> (3^(1/3) K^(1/3) (-Tanh[1/2 Sqrt[3] t Sqrt[Λ]])^(2/3))/(-Λ (-1 + 
   Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3)},
  {a[t] -> ((-1)^(2/3) 3^(1/3) K^(1/3) (-Tanh[1/2 Sqrt[3] t 
   Sqrt[Λ]])^(2/3))/(-Λ (-1 + Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3)}, 
  {a[t] -> -(((-3)^(1/3) K^(1/3)Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^(2/3))/(-Λ (-1 + 
   Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3))}, 
  {a[t] -> (3^(1/3) K^(1/3) Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^(2/3))/(-Λ (-1 + 
  Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3)}, 
  {a[t] -> ((-1)^(2/3) 3^(1/3) K^(1/3)Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^(
   2/3))/(-Λ (-1 + Tanh[1/2 Sqrt[3] t Sqrt[Λ]]^2))^(1/3)} *)

Mathematica found 6 solution.Let's raise the 6 equation to the power of 3,and substitution: $\text{t$\Lambda $}=\frac{2}{\sqrt{3} \sqrt{\Lambda }}$

Table[(a[t] /. sol[[n]])^3 // FullSimplify, {n, 1, 6}] /. 1/2 Sqrt[3] Sqrt[Λ] -> 1/tΛ

(*{(3 k Sinh[t/tΛ]^2)/Λ, (3 k Sinh[t/tΛ]^2)/Λ, (3 k Sinh[t/tΛ]^2)/Λ, 
       3 k Sinh[t/tΛ]^2)/Λ, (3 k Sinh[t/tΛ]^2)/Λ, (3 k Sinh[t/tΛ]^2)/Λ} *)

we have 6 identical solutions.One of them:

$$a(t)^3=\frac{3 k \sinh ^2\left(\frac{t}{\text{t$\Lambda $}}\right)}{\Lambda }$$

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