Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - Differential Equations (Laplace Transforms)

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    81

    Differential Equations (Laplace Transforms)

    Use Laplace Transforms to solve these IVP (partial fractions):

    1. x''+4'+8x=e^-t x(0)=x'(0)=0
    I don't know how to work with the imaginary roots.

    2. x''''-x=0 x(0)=1 x'(0)=x''(0)=x'''(0)=0

    3. x''''+13x''+36x=0 x(0)=x''(0)=0 x'(0)=2 x'''(0)=-13



    Apply the convolution theorem to find the inverse Laplace transforms for:
    4. F(s) = 1/(s(s^2+4))

    5. F(s) = 1/(s(s^2+4s+5))


    Apply the convolution theorem to derive the indicated solution x(t) of the given DE w/ initial conditions x(0)=x'(0) = 0

    6.x''+4x'+13x=f(t);
    x(t)=1/3 int[0,t] f(t-tau)e^-2(tau)*sin(3*tau) dtau

    Solve the IVP

    7. mx''+cx'+kx=f(t);
    x(0) = x'(0)=0

    m=1
    k=4
    c=5
    f(t)=1 if 0=<t=<2
    f(t)=0 if t>=2






    Help on any is useful.
    Last edited by thedoge; April 15th 2008 at 09:00 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by thedoge View Post
    Use Laplace Transforms to solve these IVP (partial fractions):

    apply the convolution theorem to find the inverse Laplace transforms for:
    F(s) = 1/(s(s^2+4))

    F(s) = 1/(s(s^2+4s+5))


    any help on any is useful.
    for the first one

    \frac{1}{s(s^2+4)}=\underbrace{\frac{1}{s}}_{f(t)=  1} \cdot \underbrace{\frac{1}{s^2+4}}_{g(t)=\frac{1}{2}\sin  (2t)}

    so the inverse transform is

    \int_{0}^{t}f(t-\tau)g(\tau)d \tau=\int_{0}^{t}\frac{1}{2}\sin(2\tau)d\tau=
     -\frac{1}{4}\cos(2\tau)|_{0}^{t}=-\frac{1}{4}\cos(2t)+\frac{1}{4}

    for the next one

    \frac{1}{s(s^2+4s+5)}=\frac{1}{s[(s+2)^2+1}

    \underbrace{\frac{1}{s}}_{1} \cdot \underbrace{\frac{f(t)=1}{(s+2)^2+1}}_{g(t)=e^{-2t}\sin(t)}

    so evaluate this integral

    \int_0^{t}e^{-2\tau}\sin(\tau) d\tau
    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
    Apply the convolution theorem to derive the indicated solution x(t) of the given DE w/ initial conditions x(0)=x'(0) = 0

    x''+4x'+13x=f(t);
    x(t)=1/3 int[0,t] f(t-tau)e^-2(tau)*sin(3*tau) dtau

    taking the Laplace transform we get

    s^2X+4sX+13X=F(s) \iff (s^2+4s+13)X=F(s)

    X=\frac{F(s)}{s^2+4s+13}=\underbrace{F(s)}_{f(t)} \cdot \underbrace{\frac{1}{(s+2)^2+9}}_{g(t)=\frac{1}{3}  e^{-2t}\sin(3t)}

    by the convolution theorem we get

    x(t)=\frac{1}{3}\int_{0}^{t}f(t-\tau)e^{-2\tau}\sin(3\tau)d\tau
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2006
    Posts
    81
    Can you clarify how you got g(t)=1/2sin(2t)?

    It's probably something simple, but I'm missing it.

    I think it has to do with the imaginary roots. As I said, I'm not sure how to work with those.
    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 thedoge View Post
    Can you clarify how you got g(t)=1/2sin(2t)?

    It's probably something simple, but I'm missing it.
    There are few ways the most common is to memorize some common formulas and use them back wards. If you have taken a course in complex variables I will show you how to evaluate the contour integral

    \mathcal{L}(sin(at))=\frac{a}{s^2+a^2}

    so we want to use this in reverse so

    \frac{1}{s^2+a^2}=\frac{1}{2} \cdot \frac{2}{s^2+a^2}

    by using the above formula backwards we get

    \frac{1}{2}\sin(2t)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2006
    Posts
    81
    Thanks again, that makes perfect sense.

    I'm trying to decide between PDEs and Complex Variables. I think PDEs would be more useful in my field (atmospheric science), but I don't really know what the Complex Variables course is about.

    So, for the first problem, I have it here:

    <br />
X \left( s \right) = \left( s+1 \left( {s}^{2}+4\,s+8 \right) <br />
 \right) ^{-1}<br />

    With Roots: 2+-2i, what happens now?
    I can tell there will be an e^-2t and a cos(2t) sin(2t), but im hazy for the steps to get to the final answer
    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 thedoge View Post
    Thanks again, that makes perfect sense.

    I'm trying to decide between PDEs and Complex Variables. I think PDEs would be more useful in my field (atmospheric science), but I don't really know what the Complex Variables course is about.

    So, for the first problem, I have it here:

    <br />
X \left( s \right) = \left( s+1 \left( {s}^{2}+4\,s+8 \right) <br />
\right) ^{-1}<br />

    With Roots: 2+-2i, what happens now?
    I can tell there will be an e^-2t and a cos(2t) sin(2t), but im hazy for the steps to get to the final answer
    The transform would be

     <br />
s^2X+4sX+8X=\frac{1}{s+1}<br />

    solving for X gives

    X=\frac{1}{(s+1)(s^2+4s+8)}

    using partial fractions we get

    X=\frac{1}{(s+1)(s^2+4s+8)}=\frac{1}{5}\frac{1}{(s-1)}-\frac{1}{5}\frac{(s+3)}{s^2+4s+8}

    Lets look at 2nd term complete the square of the denominator

    -\frac{1}{5}\frac{(s+3)}{s^2+4s+8}= -\frac{1}{5} \left( \frac{(s+2)+1}{(s+2)^2+4} \right) =-\frac{1}{5}\left( \frac{s+2}{(s+2)^2+4}+\frac{1}{(s+2)^2+4}\right)

    now using the s-axis theorem backwards we get

    \mathcal{L}^{-1}\left( -\frac{1}{5}\left( \frac{s+2}{(s+2)^2+4}+\frac{1}{(s+2)^2+4}\right) \right) =-\frac{1}{5} \left( e^{-2t}\cos(2t)+e^{-2t}\frac{1}{2}\sin(2t)\right)

    You should be able to finish from here
    Last edited by TheEmptySet; April 15th 2008 at 07:23 PM. Reason: missing e
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Oct 2006
    Posts
    81
    I'd still appreciate seeing the procedure for 2, 3 and 7 if someone feels like it.

    I'm really rusty at this stuff.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    number 2

    Quote Originally Posted by thedoge View Post
    I'd still appreciate seeing the procedure for 2, 3 and 7 if someone feels like it.

    I'm really rusty at this stuff.
     s^4X-s^3-X=\frac{1}{s} \iff X= \frac{s^4+1}{s(s^4-1)}

    by partial fractions we get

    X=\frac{1}{s}+\frac{1}{2}\frac{1}{(s-1)}+\frac{1}{2}\frac{1}{(s+1)}+\frac{s}{s^2+1}

    taking the inverse Transform we get

    x(t)=1+\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}+\cos(t)

    note that the following can also be written as

    \frac{1}{2}e^{t}+\frac{1}{2}e^{-t}=\cosh(t)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    #7

    starting with

    x''+5x'+4x=f(t)

    taking the LaPlace transform we get

    s^2X+5sX+4X=F(s) \iff (s^2+5s+4)X=F(s)

    X=\frac{F(s)}{(x+4)(x+1)}

    by partial fractions we get

    X=F(s)\frac{1}{3}\frac{1}{(s+1)}-F(s)\frac{1}{3}\frac{1}{(s+4)}

    using the convolution theorem we get

    x(t)=\frac{1}{3}\int_{0}^{t}f(t-\tau)e^{-\tau}d\tau-\frac{1}{3}\int_{0}^{t}f(t-\tau)e^{-4\tau}d\tau
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Oct 2006
    Posts
    81
     s^4X-s^3-X=\frac{1}{s} \iff X= \frac{s^4+1}{s(s^4-1)}


    I'm really confused how you have \frac{1}{s} and the s^3.

    It's =0 and the equation should be s^4X-X, right?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by thedoge View Post
     s^4X-s^3-X=\frac{1}{s} \iff X= \frac{s^4+1}{s(s^4-1)}


    I'm really confused how you have \frac{1}{s} and the s^3.

    It's =0 and the equation should be s^4X-X, right?
     <br />
x''''-x=0 ;x(0)=1 x'(0)=x''(0)=x'''(0)=0<br /> <br />

    You need to remember the intial conditions

    \mathcal{L} (x^{(4)})=s^4X-s^3x(0)-s^2x'(0)-sx''(0)-x'''(0)=s^4X-s^3(1)-s^2(0)-s(0)-0=

    s^4X-s^3

    in general we gave

    \mathcal{L}(x^{(n)})=s^nx(0)-s^{n-1}x'-...-sx^{(n-1)}-x^{n}<br /> <br />

    where x^{(n)} is the nth derivative

    You are right the right hand side should be zero, I wrote down a 1 by mistake.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    I should have gotten

    s^4X-s^3-X=0 \iff X=\frac{s^3}{(s^4-1)}

    by partial fractions I get...

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

    Taking the inverse Laplace transform I get

    x(t)=\frac{1}{4}e^{t}+\frac{1}{4}e^{-t}+\frac{1}{2}\cos(t)

    or as before

    x(t)=\frac{1}{2}\cosh(t)+\frac{1}{2}\cos(t)

    sorry about solving the wrong problem.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Oct 2006
    Posts
    81
    Thank you very much! I think I finally understand these now You were able to clarify what I was missing. I did plenty on my own, but I wasn't sure how to deal with these few I posted.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Oct 2006
    Posts
    81
    on 3...

    I have it to
    <br />
X(s) = \frac{-13s^3+4s-36}{s^2(s^2+13)}<br />

    Now do I do partial fractions like this?
    <br />
X(s) = \frac{Ax+B}{s^2} + \frac{Cx+D}{s^2+13}<br />
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 9th 2012, 07:03 PM
  2. Using Laplace Transform to solve Differential equations
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 10th 2011, 02:44 AM
  3. Laplace Transforms and differential equations
    Posted in the Differential Equations Forum
    Replies: 10
    Last Post: August 15th 2010, 01:33 PM
  4. Solution of Differential equations by Laplace Transforms
    Posted in the Differential Equations Forum
    Replies: 12
    Last Post: August 14th 2010, 03:44 PM
  5. Differential Equation Using Laplace Transforms
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 6th 2009, 07:40 AM

Search Tags


/mathhelpforum @mathhelpforum