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Help with 2nd order differential equations needed

  1. Jan 27, 2006 #1
    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:

    [tex]F_{m} - k\frac{ds}{dt} = m\frac{d^2s}{dt^2}[/tex]

    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. jcsd
  3. Jan 27, 2006 #2


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    Make the substitution,

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

    and the DE becomes a first order, namely

    [tex]F_{m} - kv = m\frac{dv}{dt},[/tex]

    that you can solve. Once you have a solution for v, plug it into [tex]v(t)=\frac{ds}{dt},[/tex] 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.
  4. Jan 27, 2006 #3
    That thought had occured to me, however the answer I got for v made no sense.

    I ended up with the following processes:

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

    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?
  5. Jan 27, 2006 #4


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    Here it goes...

    Rearrange the DE to be

    [tex]\frac{dv}{dt}+\frac{k}{m}v =\frac{f_{m}}{m} [/tex]

    Then the associated homogeneous equation is

    [tex]\frac{dv}{dt}+\frac{k}{m}v =0,[/tex]

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

    [tex]A\lambda e^{\lambda t}+\frac{k}{m}Ae^{\lambda t} =0[/tex]

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

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

    [tex]\lambda +\frac{k}{m} =0\mbox{ and hence }\lambda =-\frac{k}{m}[/tex]

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

    Now, since the non-homogeneous equation, namely

    [tex]\frac{dv}{dt}+\frac{k}{m}v =\frac{f_{m}}{m},[/tex]

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

    [tex]0+\frac{k}{m}B =\frac{f_{m}}{m},[/tex]

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

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

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

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

    Thus the solution for v is

    [tex]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)[/tex]

    but recall we had made the substitution


  6. Jan 27, 2006 #5
    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
  7. Jan 27, 2006 #6


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

    Note that from

    [tex]t=\frac{-m}{k} \ln(F_{m} - kv) + c[/tex]
    [tex]-\frac{k}{m}t-c= \ln(F_{m} - kv) [/tex]
    [tex]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}[/tex]

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