# Homework Help: Help with 2nd order differential equations needed

1. Jan 27, 2006

### 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.

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!

2. Jan 27, 2006

### 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.

3. Jan 27, 2006

### Brewer

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?

4. Jan 27, 2006

### 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

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

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

so...

5. Jan 27, 2006

### 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

6. Jan 27, 2006

### 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.