Help with 2nd order differential equations needed

[Brewer]

I have come across the following question when revising for my upcoming exam, and wondered if anyone wouldn't mind giving me a hand and some hints as how to solve it.

A car of mass m is moving on a road. The car engive provides a moving force F_{m}. The frictionalforce is proportional to the speed, v, of the car: F_{f} = kv. Write down Newtons Second Law as a second order differential of the motion of the car. Solve this equation to find the distance, s, from an initial point as a function of time, t, assuming initial conditions s(0)=0 and v(0) = 0

So far I have:

F_{m} - k\frac{ds}{dt} = m\frac{d^2s}{dt^2}

And now I'm stuck as to the solution of the equation, as its a 2nd order non-homogenous differential equation, but doesn't have a term in s. Is this a problem when solving this, or do I just put it =0?

Any hints would be appreciated!

Apologies if the symbols don't come out as anticipated. Its my first time using them!

 

[benorin]

Make the substitution,

v(t)=\frac{ds}{dt}\mbox{ so that }\frac{dv}{dt}=\frac{d^2s}{dt^2}

and the DE becomes a first order, namely

F_{m} - kv = m\frac{dv}{dt},

that you can solve. Once you have a solution for v, plug it into v(t)=\frac{ds}{dt}, and that'll give you the solution to the given DE. Note that we didn't turn a second order DE into a first order DE, but rather a system of 2 first order DEs.

 

[Brewer]

That thought had occurred to me, however the answer I got for v made no sense.

I ended up with the following processes:

f_{m} - kv = m\frac{dv}{dt}
\int dt = m\int\frac{dv}{F_{m} - kv}
t=\frac{-m}{k} \ln(F_{m} - kv) + c

which I then rearranged to make v the subject. However I don't think that this is correct way to do this is it? I think I probably used the wrong technique when integrating?

 

[benorin]

Here it goes...

Rearrange the DE to be

\frac{dv}{dt}+\frac{k}{m}v =\frac{f_{m}}{m}

Then the associated homogeneous equation is

\frac{dv}{dt}+\frac{k}{m}v =0,

Put v(t)=Ae^{\lambda t}\mbox{ so that }\frac{dv}{dt}=A\lambda e^{\lambda t} and this, when substituted into the homo. eqn., gives

A\lambda e^{\lambda t}+\frac{k}{m}Ae^{\lambda t} =0

or Ae^{\lambda t}\left( \lambda +\frac{k}{m}\right) =0

and since e^{\lambda t}\neq 0 for any t, we have

\lambda +\frac{k}{m} =0\mbox{ and hence }\lambda =-\frac{k}{m}

which gives v_H(t)=Ae^{-\frac{k}{m}t} as the general solution to the associated homogeneous equation.

Now, since the non-homogeneous equation, namely

\frac{dv}{dt}+\frac{k}{m}v =\frac{f_{m}}{m},

differs only by a constant term, let us solve for an arbitrary constant as a particular solution, say v_P(t)=B, whose derivative is 0, and it must satisfy

0+\frac{k}{m}B =\frac{f_{m}}{m},

so B =\frac{f_{m}}{k}=v_P(t), and hence the most general solution to the given DE (the one in v) is v(t)=v_H(t)+v_P(t), that is

v(t)=Ae^{-\frac{k}{m}t} + \frac{f_{m}}{k},

The initial condition is v(0) = 0, from which we can determine A:

v(0)=A\cdot 1 + \frac{f_{m}}{k}\Rightarrow A=-\frac{f_{m}}{k}.

Thus the solution for v is

v(t)=-\frac{f_{m}}{k}e^{-\frac{k}{m}t} + \frac{f_{m}}{k}=-\frac{f_{m}}{k}\left( e^{-\frac{k}{m}t}-1\right)

but recall we had made the substitution

v(t)=\frac{ds}{dt}

so...

 

[Brewer]

Thank you. That helped a lot. I think I actually was on the right lines, I'd just confused myself, but your way seems to make a lot more sense to me.

Thanks again

 

[benorin]

The above way is only so lengthy as I explained the means by which the technique is derived...

Note that from

t=\frac{-m}{k} \ln(F_{m} - kv) + c
-\frac{k}{m}t-c= \ln(F_{m} - kv)
v(t)=\frac{F_{m}}{k}-\frac{1}{k}e^{-\frac{k}{m}t-c} =\frac{F_{m}}{k}-\frac{1}{k}e^{-\frac{k}{m}t}e^{-c}

and then reason that e^{-c}=A for some positive A... same answer as mine, if you follow it out.