first order ODE question.

**04viking** · September 7th 2008, 07:43 PM

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?

**04viking** · September 7th 2008, 08:16 PM

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

**mr fantastic** · September 7th 2008, 10:15 PM

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

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

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

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

**04viking** · September 8th 2008, 08:31 AM

Thank you very much for the help.

**04viking** · September 13th 2008, 11:27 AM

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?

**Chris L T521** · September 13th 2008, 11:32 AM

Originally Posted by mr fantastic

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

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

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

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

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 $v=\frac{\,ds}{\,dt}$ , it can be easily determined that $\frac{\,ds}{\,dt}=\frac{mg}{k} - \left(\frac{mg}{k} - v_0 \right) e^{-kt/m}$

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

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

--Chris

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