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Math Help - 2nd order DE constant coefficients and boundary conditions.

  1. #1
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    2nd order DE constant coefficients and boundary conditions.

    Some help with this problem.Find the solution of the equation
    v''- 4v'+5v=0,such that v=-1 and v'=1 when x=pi=3.14159
    Thank you for your time.
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  2. #2
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    What ideas have you had so far?
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  3. #3
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    Quote Originally Posted by AkilMAI View Post
    Some help with this problem.Find the solution of the equation
    v''- 4v'+5v=0,such that v=-1 and v'=1 when x=pi=3.14159
    Thank you for your time.
    You can treat this as a polynomial.

    m^2-4m+5=0

    Solve for m
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  4. #4
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    and then when I get m I can use the initial conditions to find the constant c1 and c2(for example) from the general solution form?
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  5. #5
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    Quote Originally Posted by AkilMAI View Post
    and then when I get m I can use the initial conditions to find the constant c1 and c2(for example) from the general solution form?
    When you get m, post it.

    Your solution will be of the form

    m=\alpha\pm i\beta

    v=e^{x\alpha}(C_1\cos{\beta}+C_2i\sin{\beta})
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  6. #6
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    now I get this system
    e^[pi+2](A*cos[2x]+B*i*sin[2x])=-1
    2e^[pi+2](B*i*cos[2x]-A*sin[2x])=1...how can I solve it?
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  7. #7
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    sorry about that...
    e^[pi+2](A*cos[2pi]+B*i*sin[2x])=-1=>e^[pi+2]*B*i=-1
    2e^[pi+2](B*i*cos[2pi]-A*sin[2x])=1.=>2e^[pi+2]*A=1
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  8. #8
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    Quote Originally Posted by AkilMAI View Post
    sorry about that...
    e^[pi+2](A*cos[2pi]+B*i*sin[2x])=-1=>e^[pi+2]*B*i=-1
    2e^[pi+2](B*i*cos[2pi]-A*sin[2x])=1.=>2e^[pi+2]*A=1
    \displaystyle m^2-4m+5=0\Rightarrow m=\frac{4\pm\sqrt{16-4(5)}}{2}=\frac{4\pm 2i}{2}=2\pm i

    \alpha=2, \ \ \ \beta=1

    v=e^{2x}\left(C_1\cos(x)+C_2\sin(x)\right)

    v(\pi)=e^{2\pi}\left(C_1\cos(\pi)+C_2\sin(\pi)\rig  ht)=-1\Rightarrow e^{2\pi}\left(-1\cdot C_1)=-1

    \displaystyle C_1e^{2\pi}=1\Rightarrow C_1=e^{-2\pi}

    v'=e^{2x-2\pi}\left((C_2e^{-2\pi}+2)\cos(x)+(2e^{2\pi}C_2-1)\sin(x)\right)

    \displaystyle v'(\pi)=(C_2e^{-2\pi}+2)\cos(\pi)+(2e^{2\pi}C_2-1)\sin(\pi)=1\Rightarrow -(C_2e^{-2\pi}+2)=1\Rightarrow C_2=-3e^{2\pi}

    v(x)=e^{2x-2\pi}(\cos(x)-3e^{4\pi}\sin(x))
    Last edited by dwsmith; February 5th 2011 at 12:12 PM.
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