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Math Help - Initial value problem

  1. #1
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    Initial value problem

    Hi.

    Solve the Initial value problem:

    \dot{u}(t) = \begin{pmatrix} a & 1 \\ 0 & b \end{pmatrix}\cdot u(t) \ , \ t\in [0,T] \ , \ u(0) = (1,1)^t

    (theres nothing said about 'a' and 'b')

    does anyone know how to solve it, because i dont

    Thanks so much!

    Rapha
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  2. #2
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    Quote Originally Posted by Rapha View Post
    Hi.

    Solve the Initial value problem:

    \dot{u}(t) = \begin{pmatrix} a & 1 \\ 0 & b \end{pmatrix}\cdot u(t) \ , \ t\in [0,T] \ , \ u(0) = (1,1)^t

    (there's nothing said about 'a' and 'b')
    Let u(t) = \begin{bmatrix}x(t)\\y(t)\end{bmatrix}. Then \begin{bmatrix}\dot{x}(t)\\\dot{y}(t)\end{bmatrix} = \begin{bmatrix}a & 1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x(t)\\y(t)\end{bmatri  x}. This says that \dot{x} = ax+y,\ \dot{y} = by, with initial conditions x(0)=y(0)=1. The equation for y should be easy to solve. Then substitute the formula for y into the equation for x. Again, it will be a reasonably easy ordinary differential equation (but there will be two different forms of the solution, depending on whether a=b or not).
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  3. #3
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    Hey Opalg,

    I really appreciate your answer, thank you

    Quote Originally Posted by Opalg View Post
    Let u(t) = \begin{bmatrix}x(t)\\y(t)\end{bmatrix}. Then \begin{bmatrix}\dot{x}(t)\\\dot{y}(t)\end{bmatrix} = \begin{bmatrix}a & 1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x(t)\\y(t)\end{bmatri  x}. This says that \dot{x} = ax+y,\ \dot{y} = by, with initial conditions x(0)=y(0)=1. The equation for y should be easy to solve.
    Indeed,  y(t) = e^{bt} \Rightarrow \dot{y} = y'(t) = b*e^{bt}

    Quote Originally Posted by Opalg View Post
    Then substitute the formula for y into the equation for x. Again, it will be a reasonably easy ordinary differential equation (but there will be two different forms of the solution, depending on whether a=b or not).
     \dot{x} = ax(t)+e^{bt}

    I dont get it
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  4. #4
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    Quote Originally Posted by Rapha View Post
     \dot{x} = ax(t)+e^{bt}

    I don't get it
    If a ≠ b then you should be able to find a solution of the form x(t) = Ae^{at} + Be^{bt}, for suitable constants A and B.

    If a = b then the solution will be of the form x(t) = (At+B)e^{at}.
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