# first order ODE question.

• Sep 7th 2008, 07:43 PM
04viking
first order ODE question.
I have been given an equation which is first order ODE of the velocity v(t).

v'(t) = g-(k/m)v(t)

Where g is 9.8m/s^2, k is a constant of proportionality based on the size and shape of the body falling, and m is the mass.

I am asked to find the general solution of this DE, satisfying the initial value v(o) = v0(initial velocity).

I believe that I am supposed to take g-(k/m) to be a constant. Any suggestions as to how i should go about solving this?
• Sep 7th 2008, 08:16 PM
04viking
Could someone please verify if this is correct. This is the answer I am coming up with.

v(t) = 9.8m/k - (v(0)+9.8m/k)e^kt/m
• Sep 7th 2008, 10:15 PM
mr fantastic
Quote:

Originally Posted by 04viking
Could someone please verify if this is correct. This is the answer I am coming up with.

v(t) = 9.8m/k - (v(0)+9.8m/k)e^kt/m

I get $\displaystyle v = \frac{mg - (mg - k v_0) e^{-kt/m}}{k} = \frac{mg}{k} - \left(\frac{mg}{k} - v_0 \right) e^{-kt/m}$.

The DE is $\displaystyle \frac{dv}{dt} = \frac{mg - kv}{m} \Rightarrow \frac{dt}{dv} = \frac{m}{mg - kv}$.

The DE clearly suggests a terminal velocity $\displaystyle v_t = \frac{mg}{k}$, which is consistent with my solution but obviously not yours (your solution has v --> +oo as t --> +oo).

At t = 0 my solution gives $\displaystyle v = v_0$ whereas your solution gives $\displaystyle v = -v_0$ ....
• Sep 8th 2008, 08:31 AM
04viking
Thank you very much for the help.
• Sep 13th 2008, 11:27 AM
04viking
I have run into a small problem. I am given that v(t) = s'(t) and to solve for s(t) when s(0)=0. I cannot seem to solve it. Could someone please show me a few steps in the right direction?
• Sep 13th 2008, 11:32 AM
Chris L T521
Quote:

Originally Posted by mr fantastic
I get $\displaystyle v = \frac{mg - (mg - k v_0) e^{-kt/m}}{k} = \frac{mg}{k} - \left(\frac{mg}{k} - v_0 \right) e^{-kt/m}$.

The DE is $\displaystyle \frac{dv}{dt} = \frac{mg - kv}{m} \Rightarrow \frac{dt}{dv} = \frac{m}{mg - kv}$.

The DE clearly suggests a terminal velocity $\displaystyle v_t = \frac{mg}{k}$, which is consistent with my solution but obviously not yours (your solution has v --> +oo as t --> +oo).

At t = 0 my solution gives $\displaystyle v = v_0$ whereas your solution gives $\displaystyle v = -v_0$ ....

Quote:

Originally Posted by 04viking
I have run into a small problem. I am given that v(t) = s'(t) and to solve for s(t) when s(0)=0. I cannot seem to solve it. Could someone please show me a few steps in the right direction?

Set up the differential equation.

mr fantastic has found v for you. Since $\displaystyle v=\frac{\,ds}{\,dt}$, it can be easily determined that $\displaystyle \frac{\,ds}{\,dt}=\frac{mg}{k} - \left(\frac{mg}{k} - v_0 \right) e^{-kt/m}$

This implies that $\displaystyle \int\,ds=\int\left[\frac{mg}{k} - \left(\frac{mg}{k} - v_0 \right) e^{-kt/m}\right]\,dt\implies s(t)=\dots$

Then apply the initial condition $\displaystyle s(0)=0$ to find the value of the constant of integration $\displaystyle C$.

--Chris