# Thread: Integrating after seperating vars

1. ## Integrating after seperating vars

Haven't done much integration for a couple of years and am a bit rusty,
have this differential equation and have to integrate but can't get my head around the suggested substitution any help would be great, cheers.

After seperating vars i get;

(1/γv^2 - g).dv/dt = 1

then it asks me to integrate both sides given v(0) = 0 , using the substitution;

u= tanh^-1(sqrt(γ/g).v))

can someone please explain next step for me, would be greatly appreciated, thanks.
(γ is just a constant as is g for gravity)

2. ## Re: Integrating after seperating vars

Is it $\displaystyle \displaystyle \left(\frac{1}{\gamma v^2 - g}\right)\,\frac{dv}{dt} = 1$ or $\displaystyle \displaystyle \left(\frac{1}{\gamma v^2} - g\right)\,\frac{dv}{dt} = 1$?

3. ## Re: Integrating after seperating vars

sorrys its the first of your 2 options with -g on the denominator, i had trouble with the latex commands.
cheers

4. ## Re: Integrating after seperating vars

Originally Posted by Prove It
Is it $\displaystyle \displaystyle \left(\frac{1}{\gamma v^2 - g}\right)\,\frac{dv}{dt} = 1$ or $\displaystyle \displaystyle \left(\frac{1}{\gamma v^2} - g\right)\,\frac{dv}{dt} = 1$?
Originally Posted by monster
sorrys its the first of your 2 options with -g on the denominator, i had trouble with the latex commands.
cheers
Having trouble with latex commands is no excuse for not using brackets correctly.

Anyway

\displaystyle \displaystyle \begin{align*} \left(\frac{1}{\gamma v^2 - g}\right)\frac{dv}{dt} &= 1 \\ \int{\left(\frac{1}{\gamma v^2 - g}\right)\frac{dv}{dt}\,dt} &= \int{1\,dt} \\ \int{\frac{1}{\gamma v^2 - g}\,dv} &= t + C_1 \\ -\frac{1}{\gamma}\int{\frac{1}{\frac{g}{\gamma} - v^2}\,dv} &= t + C_1\end{align*}

Now make the substitution $\displaystyle \displaystyle v = \sqrt{\frac{g}{\gamma}}\tanh{u} \implies dv = \sqrt{\frac{g}{\gamma}}\,\textrm{sech}^2\,{u}\,du$ and the DE becomes

\displaystyle \displaystyle \begin{align*} -\frac{1}{\gamma}\int{\frac{\sqrt{\frac{g}{\gamma}} \,\textrm{sech}^2\,{u}}{\frac{g}{\gamma} - \left(\sqrt{\frac{g}{\gamma}}\tanh{u}\right)^2}\,d u} &= t + C_1 \\ -\frac{\sqrt{\frac{g}{\gamma}}}{\gamma}\int{\frac{ \textrm{sech}^2\,{u}}{\frac{g}{\gamma} - \frac{g}{\gamma}\tanh^2{u}}\,du}&= t + C_1 \\ -\frac{\sqrt{\frac{g}{\gamma}}}{g}\int{\frac{ \textrm{sech}^2\,{u}}{1 - \tanh^2{u}}\,du} &= t + C_1 \\ -\frac{\sqrt{\frac{g}{\gamma}}}{g} \int{\frac{ \textrm{sech}^2\,{u}}{ \textrm{sech}^2\,{u}}\,du} &= t + C_1 \\ -\frac{\sqrt{\frac{g}{\gamma}}}{g} \int{1\,du} &= t + C_1\end{align*}

Go from here. Make sure to convert your answer back to a function of $\displaystyle \displaystyle v$ and write it with a single constant.

5. ## Re: Integrating after seperating vars

Alternatively, integrate $\displaystyle \int \frac{1}{\frac{g}{\gamma}- v^2} dv$ using "partial fractions":
$\displaystyle \frac{1}{\frac{g}{\gamma}- v^2}$$\displaystyle = \frac{A}{\sqrt{\frac{g}{\gamma}}- v}+ \frac{B}{\sqrt{\frac{g}{\gamma}}+ v}$.

That will give you a solution in terms of logarithms rather than inverse hyperbolic functions.