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Math Help - DE System won't check - can't find error

  1. #1
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    DE System won't check - can't find error

    Hi. I finished solving a system of DEs using Laplace transform some time ago. The values I figured do not solve the system when plugged back into the original. However, I cannot find my error. I am sure it is something small, but I have been laboring for over an hour and cannot find where I went wrong. I feel like I've double-checked everything, but that cannot be so because the answers do not check.

    The problem:

    Solve the following for u(t), v(t)

    u'(t) + u(t) - v(t) = 0 u(0)=1
    v'(t) - u(t) + v(t) = 2 v(0)=2

    My solution:

    u(t) = \frac{5 - 3e^{-2t}}{2}
    v(t) = \frac{3e^{-2t} + 5}{2}

    My work:

    Laplace transform the system to get:
    U(s) = L(u(t)), V(s) = L(v(t))

    sU(s) - 1 + U(s) - V(s) = 0
    sV(s) - 2 - U(s) + V(s) = 2

    simplifying,
    (s+1)U(s) - V(s) = 1
     - U(s) + (s+1)V(s) = 4

    solving for U(s), V(s)

    U(s) = \frac{s+5}{s(s+2)}
    V(s) = \frac{4s+5}{s(s+2)}

    partial fractions to inverse laplace later

    U(s) = \frac{5}{(2s)} - \frac{3}{2(s+2)}
    V(s) = \frac{3}{2(s+2)} + \frac{5}{(2s)}

    inverse laplace, obtaining u(t), v(t)

    u(t) = (5/2) - (3/2)e^{-2t}
    v(t) = (3/2)e^{-2t} + (5/2)

    simplify and you get what I wrote above.

    The problem is, when I plug them into the system, both equations work out to 0...the second one is not equal to 2.

    I have poured over this and cannot figure it out.

    Can someone spot my error?

    Thanks
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  2. #2
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    Quote Originally Posted by kpizle View Post
    Hi. I finished solving a system of DEs using Laplace transform some time ago. The values I figured do not solve the system when plugged back into the original. However, I cannot find my error. I am sure it is something small, but I have been laboring for over an hour and cannot find where I went wrong. I feel like I've double-checked everything, but that cannot be so because the answers do not check.

    The problem:

    Solve the following for u(t), v(t)

    u'(t) + u(t) - v(t) = 0 u(0)=1
    v'(t) - u(t) + v(t) = 2 [tex]v(0)=2[/math] (*)

    My solution:

    u(t) = \frac{5 - 3e^{-2t}}{2}
    v(t) = \frac{3e^{-2t} + 5}{2}

    My work:

    Laplace transform the system to get:
    U(s) = L(u(t)), V(s) = L(v(t))

    sU(s) - 1 + U(s) - V(s) = 0
    sV(s) - 2 - U(s) + V(s) = 2 (**)

    simplifying,
    (s+1)U(s) - V(s) = 1
     - U(s) + (s+1)V(s) = 4

    solving for U(s), V(s)

    U(s) = \frac{s+5}{s(s+2)}
    V(s) = \frac{4s+5}{s(s+2)}

    partial fractions to inverse laplace later

    U(s) = \frac{5}{(2s)} - \frac{3}{2(s+2)}
    V(s) = \frac{3}{2(s+2)} + \frac{5}{(2s)}

    inverse laplace, obtaining u(t), v(t)

    u(t) = (5/2) - (3/2)e^{-2t}
    v(t) = (3/2)e^{-2t} + (5/2)

    simplify and you get what I wrote above.

    The problem is, when I plug them into the system, both equations work out to 0...the second one is not equal to 2.

    I have poured over this and cannot figure it out.

    Can someone spot my error?

    Thanks
    My question is in (*) and (**). When taking the Laplace transform of the rhs of (*), why is L\{2\} = 2 and not L\{2\} = \frac{2}{s}
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  3. #3
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    Quote Originally Posted by Danny View Post
    My question is in (*) and (**). When taking the Laplace transform of the rhs of (*), why is L\{2\} = 2 and not L\{2\} = \frac{2}{s}
    Because I am an idiot and didn't take the laplace transform of that constant!

    ay yi yi...

    Thanks very much for looking over this. That's my problem exactly!
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