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Laplace Transform Solution to Second Order ODE IVP

  1. Nov 29, 2013 #1
    1. The problem statement, all variables and given/known data
    y''+6y=f(t), y(0)=0, y'(0)=-2

    f(t)= t for 0≤t<1 and 0 for t≥1

    2. Relevant equations


    3. The attempt at a solution

    L{y''}+6L{y}=L{t}-L{tμ(t-1)} where μ(t-1) is Unit Step

    Y(s)=L{y}
    sY(s)-y(0)=L{y'} and y(0)=0
    s2Y(s)-sy(0)-y'(0)+6Y(s) where y(0)=0 and y'(0)=-2

    LHS:
    s2Y(s)+2+6Y(s)
    Y(s)(s2+6)-2

    RHS:
    L{t}-L{tμ(t-1)}
    L{t}=1/s2
    L{tμ(t-1)}=L{(t-1)μ(t-1)+L{μ(t-1)}=e-s/s2+e-s/s
    L{t}-L{tμ(t-1)}=1/s2-e-s/s2-e-s/s

    Y(s)= -1/(s2+6) -e-s/(s2(s2+6))-e-s/s(s2+6)

    taking the inverse laplace of the first term
    -L-1{1/(s2+6)}=(-1/6)L-1{1/s}+(1/6)L-1{1/(s+6)}=(-1/6)+1/6e-6t)

    how could I take the laplace of the other 2 terms?
    I was thinking L-1{e-s/(s2(s2+6))}=y(t-1)μ(t-1)?

    or I could just break it up into (1/6)L-1{1/s2}-(1/6)L-1{1/(s2+6)} which I know the inverse laplace transforms for, however when I do so I get
    y(t)= (1/6)t-(1/6)+(1/6)e-6t
    which seems a bit bizarre to me, I would have assumed that the answer would have a unit step function

    Thank you for your time
     
  2. jcsd
  3. Nov 29, 2013 #2

    vela

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    I think you're fine up to here.

    You didn't do the algebra correctly. The first term is wrong.

    You didn't factor the denominator correctly. ##s^2 + 6 \ne s(s+6)##. Check your table for a Laplace transform with a quadratic denominator.

     
    Last edited: Nov 29, 2013
  4. Nov 29, 2013 #3
    definitely an algebra error present,

    Y(s)(s2+6)+2=1/s2-e-s/s2-e-s/s

    then bring the 2 over
    Y(s)(s2+6)=1/s2-e-s/s2-e-s/s-2

    divided out the (s2+6) term

    Y(s)=1/(s2(s2+6)) -e-s/(s2(s2+6)) - e-s/(s(s2+6)) - 2/(s2+6)
     
  5. Nov 29, 2013 #4

    So, with the algebra corrected I have
    Y(s)=1/(s2(s2+6)) -e-s/(s2(s2+6)) - e-s/(s(s2+6)) - 2/(s2+6)

    the inverse of the first term
    L-1{1/(s2(s2+6))} I can't find anything like this in my table, most of the entries have s2+k2 which I can't break 6 down into?
     
  6. Nov 29, 2013 #5

    vela

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    You have to use partial fractions on that term to break it up. If ##k^2=6##, then ##k = \sqrt{6}##, no?
     
  7. Nov 29, 2013 #6
    right! so I have 1/(s2(s2+6))=(1/6)(1/s2) - (1/6)(1/(s2+6)

    which the inverse laplace transform is

    (1/6)t-(√6/6)sin(√6t) ?
     
  8. Nov 29, 2013 #7

    vela

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    Almost. You're off by a factor of ##\sqrt{6}## in the second term:
    $$\frac{1}{6}\frac{1}{s^2+6} = \frac{1}{6\sqrt{6}} \frac{\sqrt{6}}{s^2+6} \rightarrow \frac{1}{6\sqrt{6}}\sin\sqrt6 t$$
     
  9. Nov 29, 2013 #8
    right, okay, and I can also apply this to the last term?

    L-1{2/(s2+6}=(2/√6)sin(6t)?

    how could I approach the exponential terms?
     
  10. Nov 29, 2013 #9

    vela

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    The exponential indicates a time shift. You found, for example, that
    $$\frac{1}{s^2+6} \rightarrow \frac{1}{6}t - \frac{1}{6\sqrt{6}} \sin \sqrt{6}t,$$ so
    $$e^{-s}\frac{1}{s^2+6} \rightarrow \frac{1}{6}(t-1)u(t-1) - \frac{1}{6\sqrt{6}} \sin \sqrt{6}(t-1)\,u(t-1).$$
     
  11. Nov 29, 2013 #10
    okay, so for

    $$e^{-s}\frac{1}{s(s^2+6)} \rightarrow \frac{1}{6}e^{-s}\frac{1}{s} -\frac{1}{6}e^{-s}\frac{s}{s^{2}+6} \rightarrow \frac{1}{6}u(t-1) - \frac{1}{6}cosh(\sqrt{6})u(t-1)$$?
     
  12. Nov 29, 2013 #11
    making the entire thing

    $$ \frac{1}{6}t - \frac{1}{6\sqrt{6}}sin(\sqrt{6}t) - \frac{1}{6}(t-1)u(t-1) + \frac{1}{6\sqrt{6}}sin(\sqrt{6}(t-1))u(t-1) - \frac{1}{6}u(t-1) + \frac{1}{6}cosh(\sqrt{6}t)u(t-1) - \frac{2}{\sqrt{6}}sin(\sqrt{6}t) $$
     
  13. Nov 29, 2013 #12

    vela

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    Where did you get cosh from? It should be cos. The argument of that function is slightly incorrect. The rest of the solution looks good.
     
    Last edited: Nov 29, 2013
  14. Nov 29, 2013 #13
    okay, fixed.

    $$ \frac{1}{6}t - \frac{1}{6\sqrt{6}}sin(\sqrt{6}t) - \frac{1}{6}(t-1)u(t-1) + \frac{1}{6\sqrt{6}}sin(\sqrt{6}(t-1))u(t-1) - \frac{1}{6}u(t-1) + \frac{1}{6}cos(\sqrt{6}t)u(t-1) - \frac{2}{\sqrt{6}}sin(\sqrt{6}t) $$

    I'm sorry but I'm not sure what you mean by the argument? should it be t-1 instead of t?
     
  15. Nov 29, 2013 #14

    vela

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    Yes, t-1 instead of t. Everywhere t appears in the unshifted function, you have to replace it by t-1.
     
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