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2nd order DE w/ various solutions

  1. Oct 6, 2014 #1

    I'm currently finding a solution (i.e. y) for y′′ +t(y′)2 = 0.

    The answer is:

    y=(1/k)ln|(k−t)/(k+t)|+c2 if c1 =k^2 >0; y=(2/k)arctan(t/k)+c2 if c1 =−k^2 <0; y=−2t−1 +c2 if c1 =0; also y=c

    (NOTE: c1 or c2 is just c_1 or c_2)

    I've gotten the solutions as y = 0 or y=(2/k)arctan(t/k)+c2, depending on the initial conditions, but don't seem to understand how the first solution (i.e. y=(1/k)ln|(k−t)/(k+t)|+c2) provided in the answer was derived. Also, I don't quite understand why the restrictions are being stated (i.e. if c1 > k^2). If anyone could please explain why there are multiple restrictions and different solutions for each, and also how the first solution was derived, that would be extremely helpful.

    Thank you!
  2. jcsd
  3. Oct 7, 2014 #2


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    Homework Helper

    I think you mean [itex]y = -2t^{-1} + c_2[/itex] in the case [itex]c_1 = 0[/itex]; [itex]-2t + C[/itex], which is what you have, is not actually a solution.

    For non-constant solutions, the ODE may be rearranged as [tex]\frac{y''}{y'^2} = -t,[/tex] and integrating once with respect to [itex]t[/itex] will yield [tex]
    y' = \frac{2}{t^2 - c_1}.[/tex] How would you proceed from here? It depends on whether [itex]c_1[/itex] is strictly positive (in which case it may be convenient to set [itex]c_1 = k^2[/itex] for some [itex]k > 0[/itex]), strictly negative (in which case it may be convenient to set [itex]c_1 = -k^2[/itex] for some [itex]k > 0[/itex]), or zero.
  4. Oct 7, 2014 #3
    Hmmm...do you mind possibly explaining how you integrated and went from [tex]\frac{y''}{y'^2} = -2t,[/tex]to[tex]y' = \frac{2}{t^2 - c_1}[/tex]? I may just not be seeing something, but any help would be much appreciated!
    Last edited: Oct 7, 2014
  5. Oct 9, 2014 #4


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    Let u= y'. Then y''= u' so the differential equation becomes [tex]u^{-2}du/dt= -2t[/tex] so [tex]u^{-2}du= -2t dt[/tex]. Integrating both sides [tex]-u^{-1}= -t^2+ C[/tex] or [tex]u= dy/dt= 1/(t- C)[/tex].
  6. Oct 9, 2014 #5
    Was the [tex] t^2 [/tex] in the denominator [tex] t-C [/tex] in the final step intentionally left out for any particular reason? If not, thank you so much for your help!
  7. Oct 12, 2014 #6


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    Science Advisor

    That was a typo. The last equation should have been
    [tex]u= \frac{dy}{dt}= \frac{1}{t^2- C}[/tex]
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