- Thread starter
- #1

Thanks

Søren, Denmark

- Thread starter U477
- Start date

- Thread starter
- #1

Thanks

Søren, Denmark

- Feb 13, 2012

- 1,704

The ODE can be written with the variables v and t separated as...I have been unable to solve this problem featured in "Elementary Differential Equations" by Edwards and Penney. The problem is this: I do not know how to integrate the function, and get the desired result. I would very much appreciated any help and instructions on how to get the two air resistance equations by integrating the function.

Thanks

Søren, Denmark

View attachment 615

$\displaystyle \frac{d v}{1 - \frac{k}{m g}\ v^{2}} = g\ d t$ (1)

... and remembering that is...

$\displaystyle \int \frac{d u}{1-a\ u^{2}} = \frac{\tanh^{-1} (\sqrt{a}\ u)}{\sqrt{a}} + c$ (2)

... You obtain...

$\displaystyle \tanh^{-1} (\sqrt{\frac{k}{m\ g}}\ v) = \sqrt{\frac{k\ g}{m}}\ t + c $ (3)

... and from (3)...

$\displaystyle v = \sqrt{\frac{m\ g}{k}} \tanh (\sqrt{\frac{k\ g}{m}}\ t + c)$ (4)

Now You can proceed if You are able to find the constant c and that is possible if the initial speed v(0) is given...

Kind regards

$\chi$ $\sigma$

- Thread starter
- #3

I have a few questions for your solution -

1) Why are the position of the constants changed inside the radical in (3) – I am thinking they should be the same.

I'm still new to all this thanks for helping me out

- Feb 13, 2012

- 1,704

I apologize to have omitted some intermediate stage and I suggest You, as useful exercise, to try Yourself to complete my work...

I have a few questions for your solution -

1) Why are the position of the constants changed inside the radical in (3) – I am thinking they should be the same.

2) When I try to isolate ”v” I end up with a fraction - how did you go from (3) to (4)

I'm still new to all this thanks for helping me out

Kind regards

$\chi$ $\sigma$

- Thread starter
- #5

- Feb 13, 2012

- 1,704

Unfortunately I don't have available such a powerful math tool......for your information, I use a N-spire calculator...

Kind regards

$\chi$ $\sigma$

- Admin
- #7

- Jan 26, 2012

- 4,202

So you have

$$\frac{1}{g} \, \frac{dv}{dt}=1- \frac{k}{mg}\,v^{2}$$

$$\int \frac{dv}{1- \frac{k}{mg} \,v^{2}}=g \int dt$$

$$ \frac{ \tanh^{-1} \left( \sqrt{ \frac{k}{mg}} \, v \right)}{ \sqrt{ \frac{k}{mg}} }=gt+c$$

$$\tanh^{-1} \left( \sqrt{ \frac{k}{mg}} \, v \right)= \sqrt{ \frac{k}{mg}}\, (gt+c)$$

$$\tanh^{-1} \left( \sqrt{ \frac{k}{mg}} \, v \right)= \sqrt{ \frac{k g^{2}}{mg}}\,t+\sqrt{ \frac{k}{mg}} \, c= \sqrt{ \frac{kg}{m}}\,t+c'.$$

Can you continue from here? Note that a constant times an arbitrary constant is just another arbitrary constant - in DE's, you often absorb known constants into arbitrary constants, and relabel as the original arbitrary constant. So the $c'$ above becomes $c$ again.

$$\frac{1}{g} \, \frac{dv}{dt}=1- \frac{k}{mg}\,v^{2}$$

$$\int \frac{dv}{1- \frac{k}{mg} \,v^{2}}=g \int dt$$

$$ \frac{ \tanh^{-1} \left( \sqrt{ \frac{k}{mg}} \, v \right)}{ \sqrt{ \frac{k}{mg}} }=gt+c$$

$$\tanh^{-1} \left( \sqrt{ \frac{k}{mg}} \, v \right)= \sqrt{ \frac{k}{mg}}\, (gt+c)$$

$$\tanh^{-1} \left( \sqrt{ \frac{k}{mg}} \, v \right)= \sqrt{ \frac{k g^{2}}{mg}}\,t+\sqrt{ \frac{k}{mg}} \, c= \sqrt{ \frac{kg}{m}}\,t+c'.$$

Can you continue from here? Note that a constant times an arbitrary constant is just another arbitrary constant - in DE's, you often absorb known constants into arbitrary constants, and relabel as the original arbitrary constant. So the $c'$ above becomes $c$ again.

Last edited:

- Thread starter
- #8