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!
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...
$\displaystyle v^{'}= -g + k\ v^{2}\ , \\ v(0)=0$ (1)
In (1) You can separate the variables obtaining...
$\displaystyle \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 $
$\displaystyle \displaystyle \implies e^{2\ \sqrt{k\ g}\ t}= c\ \frac{v-\sqrt{\frac{g}{k}}} {v+\sqrt{\frac{g}{k}}} \implies v= -\sqrt{\frac{g}{k}}\ \frac{e^{2\ \sqrt{k\ g}\ t}+c}{e^{2\ \sqrt{k\ g}\ t}-c}$ (2)
If we set in (2) v(0)=0 we obtain c=-1 so that is...
$\displaystyle \displaystyle v= - \sqrt{\frac{g}{k}}\ (1-\frac{2}{1+e^{2\ \sqrt{k\ g}\ t}})$ (3)
If we integrate (3) with the condition y(0)= L we obtain...
$\displaystyle \displaystyle y(t)=L-\sqrt{\frac{g}{k}}\ \int_{0}^{t} (1-\frac{2}{1+e^{2\ \sqrt{k\ g}\ \tau}})\ d \tau = L + \sqrt{\frac{g}{k}}\ t - \frac{1}{k}\ \ln \frac{1+e^{2\ \sqrt{k\ g}\ t}}{2}$ (4)
May be that, sooner or later', I will explain You what has been the 'scope' amd the 'queue' of the 'discussion' ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
$\displaystyle \frac{m}{mg- kv^2}dv= dt$
$\displaystyle \frac{m}{(\sqrt{mg}-\sqrt{k}v()(\sqrt{mg}+ \sqrt{k}v)}= d$
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]