# Thread: Ordinary Differential Equation Help

1. ## Ordinary Differential Equation Help

Hi, can anyone help me with the following question:

Find the solution of the ODE:

m(dv/dt)=mg-kv^2

with the initial condition v=v0 at t=0

Is it integration by parts that should be used?
Thanks in advance for any help!

2. $kv^2 + m\dfrac{dv}{dt} = mg$

Separate the variables: $\dfrac{k}{m}v^2\, dv = g\, dt$

If, as I assume m is mass then it is non-zero and can be cancelled. I have also assumed that v << c so we can ignore relativity

3. Originally Posted by e^(i*pi)
$\displaystyle kv^2 \cdot \frac{dv}{dt} = m(g-1)$
This is not the same DE as $\displaystyle m\,\frac{dv}{dt} = m\,g - k\,v^2$...

4. Originally Posted by AlanC877
Hi, can anyone help me with the following question:

Find the solution of the ODE:

m(dv/dt)=mg-kv^2

with the initial condition v=v0 at t=0

Is it integration by parts that should be used?
Thanks in advance for any help!
Several years ago, during an interesting discussion in an argument [apparently...] 'far from math', I imposted the problem of the fall of an object of mass m with initial speed 0 and initial position L under action of gravity and 'aerodinamic resistance'. The DE that describes the problem is...

$v^{'}= -g + k\ v^{2}\ , \\ v(0)=0$ (1)

In (1) You can separate the variables obtaining...

$\displaystyle t= \frac{1}{k}\ \int \frac{dv}{v^{2}-\frac{g}{k}} + \ln c = \frac{1}{2\ \sqrt{k\ g}}\ \ln \frac{v-\sqrt{\frac{g}{k}}} {v+\sqrt{\frac{g}{k}}} + \ln c \implies$

$\displaystyle \implies e^{2\ \sqrt{k\ g}\ t}= c\ \frac{v-\sqrt{\frac{g}{k}}} {v+\sqrt{\frac{g}{k}}} \implies v= -\sqrt{\frac{g}{k}}\ \frac{e^{2\ \sqrt{k\ g}\ t}+c}{e^{2\ \sqrt{k\ g}\ t}-c}$ (2)

If we set in (2) v(0)=0 we obtain c=-1 so that is...

$\displaystyle v= - \sqrt{\frac{g}{k}}\ (1-\frac{2}{1+e^{2\ \sqrt{k\ g}\ t}})$ (3)

If we integrate (3) with the condition y(0)= L we obtain...

$\displaystyle y(t)=L-\sqrt{\frac{g}{k}}\ \int_{0}^{t} (1-\frac{2}{1+e^{2\ \sqrt{k\ g}\ \tau}})\ d \tau = L + \sqrt{\frac{g}{k}}\ t - \frac{1}{k}\ \ln \frac{1+e^{2\ \sqrt{k\ g}\ t}}{2}$ (4)

May be that, sooner or later', I will explain You what has been the 'scope' amd the 'queue' of the 'discussion' ...

Kind regards

$\chi$ $\sigma$

5. Originally Posted by e^(i*pi)
$kv^2 + m\dfrac{dv}{dt} = mg$

Separate the variables: $\dfrac{k}{m}v^2\, dv = g\, dt$

If, as I assume m is mass then it is non-zero and can be cancelled. I have also assumed that v << c so we can ignore relativity
Thats kinda what I was thinking, rearranging and dividing throught by m gives (k/m)v^2+(dv/dt)=g, but multiplying the dt on each side would then give (k/m)v^2+dv=g dt, can you really just get rid of the plus?

6. No, that was a mistake on my part. Please disregard all of my posts in this thread

7. Originally Posted by AlanC877
Hi, can anyone help me with the following question:

Find the solution of the ODE:

m(dv/dt)=mg-kv^2
$\frac{m}{mg- kv^2}dv= dt$

$\frac{m}{(\sqrt{mg}-\sqrt{k}v()(\sqrt{mg}+ \sqrt{k}v)}= d$

Use "partial fractions" to do the integral on the left.

with the initial condition v=v0 at t=0

Is it integration by parts that should be used?
Thanks in advance for any help![/QUOTE]