# More Help yet again with Solving differential equations

1. Mar 22, 2005

### $id Hi i posted last time about solving an equation regarding air resistance being proportional to the speed This time i need it to be proportional to the speed squared I have managed to get an equation but havent managed to solve it yet dv/dt -(k/m)v^2 = g Any ideas on how to solve this. I have tried all the methods i know, I think i need to substitute in. I will need more than just a hint here guys Thanks a lot sid 2. Mar 22, 2005 ### Crosson The equation is seperable (you can seperate v and t). Any first order seperable equation can be reduced to a problem of indefinite integration: $$\int \frac{dv}{g - \frac{k}{m} v^2} = t + Const.$$ 3. Mar 22, 2005 ### Data $$v^\prime = g + \frac{k}{m}v^2 \Longrightarrow \frac{v^\prime}{\frac{k}{m}v^2 + g} = 1$$ $$\Longrightarrow \int \frac{dv}{\frac{k}{m}v^2 + g} = \frac{m}{k} \int \frac{dv}{v^2 + \frac{gm}{k}} = \sqrt{\frac{m}{kg}} \arctan \left(v \sqrt{\frac{k}{mg} \right) = t + C$$ $$\Longrightarrow \arctan \left(v \sqrt{\frac{k}{mg}} \right) = \sqrt{\frac{kg}{m}}(t+C) \Longrightarrow v \sqrt{\frac{k}{mg}} = \tan \left( \sqrt{\frac{kg}{m}}(t+C)\right)$$ $$\Longrightarrow v = \sqrt{\frac{mg}{k}} \tan \left( \sqrt{\frac{kg}{m}}(t+C)\right)$$ 4. Mar 22, 2005 ### dextercioby A bit weird not to include among "all the methods" the separation of variables,which is the simplest possible. Anyway,i hope you understood the solution Data gave you. Daniel. 5. Mar 22, 2005 ###$id

Hmm i can follow Data logic despite never have seen anything like that before,

I guess now the main problem is that equation gives me the velocity. for the model i am making the data readings are distances at a given time. Hence I will most probably to integrate that beast.

All i know is that integral of tanx = - ln(cos x) + c or - ln (sec^2 x) or something like that

sid

6. Mar 22, 2005

### dextercioby

$$\int \tan x \ dx=-\ln|\cos x|+C$$ it can be proven using the definition of tangent & the substitution $\cos x=u$.

Daniel.

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