Results 1 to 7 of 7

Math Help - Undetermined Coefficients

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    35

    Undetermined Coefficients

     y''+2y'+5y=\left\{\begin{array}{cc}1,&\mbox{ if }<br />
t\leq t\leq\frac{\pi}{2}\\0, & \mbox{ if } t>\frac{\pi}{2}\end{array}\right

    where  y(0)=0, y'(0)=0 assume that y and y' are continuous at  t=\frac{\pi}{2}


    I don't know where to start this problem. any suggestions?
    Last edited by questionboy; October 25th 2010 at 04:29 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Solve the DE separately in each region, without applying the conditions. Then apply the conditions to find the constants of integration.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,535
    Thanks
    1389
    Each separate solution will involve two unknown constants, for a total of four. The boundary conditions will give two equations for those. Use the fact that the solution function must be differentiable at t= \pi/2 to find the other two- that is, the two functions and the derivatives of the two functions must be the same at t= \pi/2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2009
    Posts
    35
    what is the final answer should like?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    35
    Y(t)=ce^x+ce^(-3x)

    Yp(t)=1

    y(t)=ce^x+ce^(-3x)-3
    what should i do
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,535
    Thanks
    1389
    Go back and start all over again! Your characteristic equation is r^2+ 2r+ 5= 0 which does NOT have roots 1 and -3!

    What are the roots of r^2+2r+ 5= 0? What are two independent solutions to y"+ 2y'+ 5y= 0?

    Now, with right hand side "1", try a solution of the form y= A, a constant. What "A" satisifies y"+ 2y'+ 5y= 1?

    Now, you should have two functions, Y1 and Y2, say, so that Y1 is correct between 0 and \pi/2 (I am assuming you meant " 0\le t\le \pi/2" and not t\le t\le \pi/2 as you had at first) and Y2 is correct for x> \pi/2. Y1 will, of course, involve two unknown constants (and don't write them both as "c"!). You can determine those from the initial conditions y(0)= 0 and y'(0)= 0. Once you know those, you know Y1(t) completely and can evaluate it at t= \pi/2. Y2 will also involve two unknown constants. You can determine those from Y2(\pi/2)= Y1(\pi/2) and Y2'(\pi/2)= Y1'(\pi/2).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2009
    Posts
    35
    yh=c1e^-t\cos(2t)+c2e^-t\sin(2t)


    y1=-1/5e^-t\cos(2t)+-1/10e^-t\sin(2t)+1/5

    y2=Ae^-t\cos(2t)+Be^-t\sin(2t)

    what should i do now?

    set Y1=Y2?
    Last edited by questionboy; October 25th 2010 at 07:36 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Undetermined Coefficients
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 10th 2011, 09:52 PM
  2. undetermined coefficients
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 7th 2010, 02:19 AM
  3. Undetermined coefficients
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: July 10th 2009, 01:48 AM
  4. IVP Undetermined Coefficients
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 14th 2007, 06:16 PM
  5. Undetermined Coefficients
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 5th 2007, 08:23 AM

Search Tags


/mathhelpforum @mathhelpforum