Results 1 to 8 of 8

Math Help - solving a simple DE using the Laplace Transform - am i doing it right?

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    136

    solving a simple DE using the Laplace Transform - am i doing it right?

    Hi all. I have been trying to relearn the laplace transform. I want to solve this second order ode

    y^{''}+y^{'}+y=1

    I know this can be easier done using the characteristic equation but I want to use the laplace tansform for learning purposes

    here is my solution - is this right? have i done something stupid?


    step 1 - L(y^{''})+L(y^{'})+L(y)=L(1)

    =>  S^{2}L(y) - Sy(0) - y^{'}(0) + SL(y)-y(0)+L(y)=L(1)

    *here i am confused. With no initial conditions does the equation become*

    =>  S^{2}L(y) + SL(y) +L(y)=1/S

    then L(y) = \frac{1}{S\cdot(S^{2}+S+1)}

    step 2 - I factorise the quadratic and that becomes

    L(y) = \frac{1}{S\cdot(S+(\frac{1}{2}+\frac{\sqrt {3}}{2}))(S+(\frac{1}{2}-\frac{\sqrt{3}}{2}))}

    *i am sure what i did here with partial fractions is wrong*

    L(y) = \frac{1}{S}+\frac{A}{(S+(\frac{1}{2}+\frac{\sqrt {3}}{2}))}+\frac{B}{(S+(\frac{1}{2}-\frac{\sqrt{3}}{2}))}

    step 3 - take the inverse laplace transform

    if i take the inverse laplace transform i get the result from the characteristic equation

    Ae^{\frac{1}{2}+\frac{\sqrt {3}}{2}} + Be^{\frac{1}{2}-\frac{\sqrt {3}}{2}} + 1

    are my steps correct?

    thanks for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    May 2011
    Posts
    22
    One thing is obviously wrong. e^(.5+/-sqrt(3)/2) is a constant. y is supposed to be a function of something, be it x or t. Moreover, this equation has a negative discriminant, I can see that right off the bat since b^2-4ac is -3 (the coefficients a b and c are all 1), so you should be getting complex roots.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    First it looks like you assumed that all of your initial conditions are zero.

    When you have

    \mathcal{L}\{y\}=\frac{1}{s(s^2+s+1)}=\frac{1}{s} \cdot \frac{1}{(s+\frac{1}{2})^2+\left(\frac{\sqrt{3}}{2  } \right)^2}

    This is an irreducible quadratic so the inverse transform will be sines and/or cosines multiplied by an expoential. This comes from the shifting theorem.

    \mathcal{L}\{e^{at}f(t) \}=F(s-a)

    we are just using it in reverse.

    Also the factor of \frac{1}{s}
    says that you can just integrate the inverse transform of the 2nd fraction (The convolution theorem) so you get

    y(t)=\int_{0}^{t}\frac{2}{\sqrt{3}}e^{-\frac{\tau}{2}}\sin\left( \frac{\sqrt{3}}{2}\tau\right)d\tau
    Last edited by TheEmptySet; May 31st 2011 at 03:24 PM. Reason: missing square
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2006
    Posts
    136
    thanks TheEmptySet

    I kind of see what you mean, but unfortunately i am more confused

    how did you get \mathcal{L}\{y\}=\frac{1}{s(s^2+s+1)}=\frac{1}{s} \cdot \frac{1}{(s+\frac{1}{2})+\left(\frac{\sqrt{3}}{2} \right)^2}

    to avoid the irreducible quadratic could i not just search for y^{''}+y^{'}+y=0 and then tag on a solution. Would this be easier


    also apologies, the quadratic has complex roots
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by chogo View Post
    thanks TheEmptySet

    I kind of see what you mean, but unfortunately i am more confused

    how did you get \mathcal{L}\{y\}=\frac{1}{s(s^2+s+1)}=\frac{1}{s} \cdot \frac{1}{(s+\frac{1}{2})+\left(\frac{\sqrt{3}}{2} \right)^2}

    to avoid the irreducible quadratic could i not just search for y^{''}+y^{'}+y=0 and then tag on a solution. Would this be easier


    also apologies, the quadratic has complex roots
    Sorry I can see why you are confused I left a square off one of my terms.

    s^2+s+1=s^2+s+\frac{1}{4}-\frac{1}{4}+1

    Now if we factor the first 3 terms we get

    s^2+s+\frac{1}{4}-\frac{1}{4}+1=\left(s+\frac{1}{2} \right)^2+ \left(\frac{\sqrt{3}}{2} \right)^2

    I hope this clears itup.
    Last edited by TheEmptySet; May 31st 2011 at 04:31 PM. Reason: haha I did it again
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2006
    Posts
    136
    wow i see, thats really clever!

    i will let it digest further

    One final thing, i assume the initial conditions are zero. Should you avoid solving a DE like this using the Laplace transform if you do not have the initial conditions

    thanks for your help and time
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by chogo View Post
    wow i see, thats really clever!

    i will let it digest further

    One final thing, i assume the initial conditions are zero. Should you avoid solving a DE like this using the Laplace transform if you do not have the initial conditions

    thanks for your help and time
    Well The Laplace trasform works really well when you do have initial conditions but if you don't know them you could just suppose that

    y(0)=c_1 \quad y'(0)=c_2

    Now if you take the Laplace transform you get

    s^2Y-sc_1-c_2+sY-c_1+Y=\frac{1}{s}

    Now solving for Y gives

    (s^2+s+1)Y-sc_1-c_1-c_2=\frac{1}{s}

    (s^2+s+1)Y= sc_1+(c_1+c_2)+\frac{1}{s} \iff Y =\frac{sc_1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2}  )^2}+\frac{c_1+c_2}{(s+\frac{1}{2})^2+(\frac{\sqrt  {3}}{2})^2}+\frac{1}{s}\frac{1}{(s+\frac{1}{2})^2+  (\frac{\sqrt{3}}{2})^2}

    Now we just have to fix the first one (add one half c_1 and subtract to balance) so we can use the shifting theorem

     Y =\frac{c_1(s+\frac{1}{2})}{(s+\frac{1}{2})^2+( \frac{\sqrt{3}}{2} )^2}+\frac{\frac{1}{2}c_1+c_2}{(s+ \frac{1}{2} )^2+( \frac{\sqrt{3}}{2} )^2}+\frac{1}{s}\frac{1}{(s+ \frac{1}{2} )^2+( \frac{\sqrt{3}}{2} )^2}

    So the inverse transform is

    y =c_1 e^{-\frac{t}{2}}\cos\left( \frac{\sqrt{3}}{2}t \right) + (\frac{1}{2}c_1+c_2)\frac{2}{\sqrt{3}}e^{-\frac{t}{2}}\sin\left( \frac{\sqrt{3}}{2}t \right) +\frac{2}{\sqrt{3}}\int_{0}^{t} e^{-\frac{\tau}{2}}\sin\left( \frac{\sqrt{3}}{2}\tau \right) d\tau

    So as you may have noticed this may be a bit more work.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2006
    Posts
    136
    thank you very much for your time and patience. I really appreciate the help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Simple Inverse Laplace Transform
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: March 16th 2011, 02:20 PM
  2. Simple Laplace transform problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 4th 2009, 04:15 AM
  3. Solving a nonhomogeneous second order ODE using Laplace transform
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 3rd 2009, 01:05 AM
  4. Replies: 1
    Last Post: January 25th 2009, 09:50 PM
  5. Replies: 7
    Last Post: December 11th 2008, 09:16 PM

Search Tags


/mathhelpforum @mathhelpforum