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Math Help - find the general solution of the system...

  1. #1
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    find the general solution of the system...

    ok so i was working this problem and got stuck, im not sure what to do with the imaginary numbers that im getting?

    so here is the problem...



    i figured out that ...



    but im not sure where to go from here?

    any help would be greatly appreciated
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  2. #2
    Senior Member Sambit's Avatar
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    Divide the first equation by the second one: \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}=\frac{5x+2  y}{6x-y}

    that is \frac{dy}{dx}=\frac{5x+2y}{6x-y}

    y=vx gives \frac{dy}{dx}=v+x\frac{dv}{dx}

    Hence v+x\frac{dv}{dx}=\frac{5x+2vx}{6x-vx}=\frac{5+2v}{6-v}

    So x\frac{dv}{dx}=\frac{5+2v}{6-v}-v=\frac{5+2v-6v+v^2}{6-v}=\frac{v^2-4v+5}{6-v}

    Integrate this.


    \frac{v^2-4 v+5}{6-v}

    =\frac{-v^2+4 v-5}{v-6}

    =  -v-\frac{-17}{(v-6)}-2

    = -2\int dv-17\int \frac{1}{v-6} dv- \int v dv

    For the integrand \frac{1}{v-6}, substitute u = v-6
    = -1 7\int\frac{1}{u} du+ \int -2 dv- \int v dv

    = -17 log(u)+ \int-2 dv- \int v dv

    = -17 log(u)-\frac{v^2}{2}+ \int -2 dv

    = -17 log(u)-\frac{v^2}{2}-2v

    = -\frac{v^2}{2}-2v-17 log(v-6)


    Substitute back v=\frac{y}{x} to get the final result.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Sambit View Post
    Substitute back v=\frac{y}{x} to get the final result.

    All of this provides a prime integral F (x , y) of the autonomous system but not the solutions. That is, we determine the level sets F (x , y) = k where the orbits are contained. We can find the solutions for the system

    X'=AX \equiv\begin{bmatrix}{x'(t)}\\{y'(t)}\end{bmatrix}  =\begin{bmatrix}{6}&{-1}\\{5}&{\;\;2}\end{bmatrix} \begin{bmatrix}{x(t)}\\{y(t)}\end{bmatrix}

    using the standard method

    X=e^{tA}\begin{bmatrix}{C_1}\\{C_2}\end{bmatrix}
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  4. #4
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    Quote Originally Posted by Sambit View Post
    Divide the first equation by the second one: \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}=\frac{5x+2  y}{6x-y}

    that is \frac{dy}{dx}=\frac{5x+2y}{6x-y}

    y=vx gives \frac{dy}{dx}=v+x\frac{dv}{dx}

    Hence v+x\frac{dv}{dx}=\frac{5x+2vx}{6x-vx}=\frac{5+2v}{6-v}

    So x\frac{dv}{dx}=\frac{5+2v}{6-v}-v=\frac{5+2v-6v+v^2}{6-v}=\frac{v^2-4v+5}{6-v}

    Integrate this.


    \frac{v^2-4 v+5}{6-v}

    =\frac{-v^2+4 v-5}{v-6}

    =  -v-\frac{-17}{(v-6)}-2

    = -2\int dv-17\int \frac{1}{v-6} dv- \int v dv

    For the integrand \frac{1}{v-6}, substitute u = v-6
    = -1 7\int\frac{1}{u} du+ \int -2 dv- \int v dv

    = -17 log(u)+ \int-2 dv- \int v dv

    = -17 log(u)-\frac{v^2}{2}+ \int -2 dv

    = -17 log(u)-\frac{v^2}{2}-2v

    = -\frac{v^2}{2}-2v-17 log(v-6)


    Substitute back v=\frac{y}{x} to get the final result.
    That's sure is a lot of work and not very useful as Fernando points out, you want the solutions. One could also go the system route but what I would so is solve the first ODE for y, y=6x - \dot{x} and substitute into the second giving

    \ddot{x} - 8 \dot{x} + 17x = 0

    Characteristic equation

    m^2-8m + 17 = 0 so m = 4 \pm i

    soln: x =c_1e^{4t} \cos t + c_2 e^{4t} \sin t

    Then sub. this into y=6x - \dot{x}
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  5. #5
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    Quote Originally Posted by FernandoRevilla View Post
    All of this provides a prime integral F (x , y) of the autonomous system but not the solutions. That is, we determine the level sets F (x , y) = k where the orbits are contained. We can find the solutions for the system

    X'=AX \equiv\begin{bmatrix}{x'(t)}\\{y'(t)}\end{bmatrix}  =\begin{bmatrix}{6}&{-1}\\{5}&{\;\;2}\end{bmatrix} \begin{bmatrix}{x(t)}\\{y(t)}\end{bmatrix}

    using the standard method

    X=e^{tA}\begin{bmatrix}{C_1}\\{C_2}\end{bmatrix}
    ok so i will go with the standard method and have another go at the problem. thanks you guys!
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  6. #6
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    Quote Originally Posted by Danny View Post
    That's sure is a lot of work and not very useful as Fernando points out, you want the solutions. One could also go the system route but what I would so is solve the first ODE for y, y=6x - \dot{x} and substitute into the second giving

    \ddot{x} - 8 \dot{x} + 17x = 0

    Characteristic equation

    m^2-8m + 17 = 0 so m = 4 \pm i

    soln: x =c_1e^{4t} \cos t + c_2 e^{4t} \sin t

    Then sub. this into y=6x - \dot{x}
    thanks Danny! I was looking through another textbook and saw that similiar form that you gave... the author of the text was saying something to the effect of im going to read through it again and redo this problem, it doesnt seem bad after reading through the explanations posted! This forum is the reason why im passing Diffy Q!
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