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Initial value problems with matrices

  1. Nov 1, 2008 #1
    Given y'=[matrix]y + [matrix of constants]. y(0)=[initial conditions] Would I treat the matrix with variation of parameters? Or should I ignore the constants on right hand side and find the general solution. Then when it comes to solving the initial conditions, then do the calculations with it?

    Thanks in advance.
     
  2. jcsd
  3. Nov 1, 2008 #2

    gabbagabbahey

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    Well, if you had y' -Ay=B, and y(0)=C ,where A,B,C are known constants, what would you do?....Why not do the same thing with matrices?
     
  4. Nov 1, 2008 #3
    Yeah, I supposed I would've just done a separable equation with +B. Then initial condition will also be involved with B. So a constant wouldn't require me to use other methods of solving.

    Thanks for explaining in an apprehensive way.
     
  5. Nov 2, 2008 #4
    Wait sorry, after trying the question out. How should I treat the matrix of constants from the beginning? I just ignored it and continued calculating for a general solution, then when it was time to calculate for initial conditions, I couldn't reason to where those constants would've come into play.
     
  6. Nov 2, 2008 #5

    gabbagabbahey

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    Perhaps you should show me what your general solution for the question I posted earlier (the one without matrices) is.
     
  7. Nov 2, 2008 #6
    I ended up y= (eAt+CA+B)/A where C is the constant to be solved by initial condition.
     
  8. Nov 2, 2008 #7

    gabbagabbahey

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    Hmm... that doesn't look quite right: using that solution, y(0)=(eAC+B)/A....not C...I think you'd better show me your steps.
     
  9. Nov 2, 2008 #8
    y'-Ay=B
    dy/dt=B+Ay
    [tex]\int dy/(B+Ay)[/tex]=[tex]\int dt[/tex]
    1/A(ln|B+Ay|)=t+C
    B+Ay=eA(t+C)
    y=(eA(t+C)-B)/A

    Oops, it was -B sorry. I didn't solve for C because you said to solve for general solution? So that would mean only leave it with unidentified constants?

    um.. but if I solved for C, then the solution is
    y=(eA(t+(ln(B+AC)/A))-B)/A
     
  10. Nov 2, 2008 #9

    HallsofIvy

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    For any equation of the form Y'= Ay+ C, you can first solve Y'= AY to get Y(t)= DeAt, with D an unknown constant matrix, and then look for a "specific solution" which will be a constant: if Y is a constant then Y'= 0 so AY+ C= 0 and Y= A-1C (assuming A has an inverse).
    The general solution to the entire equation is Y= DeAt+ A-1C and you can find D by using the initial condition.
     
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