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Thread: Three variables?

  1. #1
    Super Member Showcase_22's Avatar
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    Three variables?

    Write $\displaystyle \ddot x -2 \dot x -3x=10 \sin t$ in the three variables $\displaystyle x, \ y= \dot x$ and $\displaystyle \dot s =1$ so that the equations are autonomous and first order.
    Clearly $\displaystyle \ddot x -2 \dot x -3x=10 \sin t$ is not a linear equation.

    My lecture notes state that "nonlinear equations cannot be written in the form $\displaystyle \dot x = Ax$ (where $\displaystyle x \in \mathbb{R}^n$ and $\displaystyle A$ is an $\displaystyle n \times n$ matrix)".

    So isn't this question literally asking me to do the impossible??
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    Clearly $\displaystyle \ddot x -2 \dot x -3x=10 \sin t$ is not a linear equation.

    My lecture notes state that "nonlinear equations cannot be written in the form $\displaystyle \dot x = Ax$ (where $\displaystyle x \in \mathbb{R}^n$ and $\displaystyle A$ is an $\displaystyle n \times n$ matrix)".

    So isn't this question literally asking me to do the impossible??
    It's not asing for 3 linear equations just 3 equations.

    Here
    $\displaystyle \frac{ds}{dt} = 1$ so $\displaystyle \frac{dx}{dt} = \frac{dx}{ds}\cdot \frac{ds}{dt} = \frac{dx}{ds}$. Simiarly $\displaystyle \frac{dy}{dt} = \frac{dy}{ds}. $

    Then $\displaystyle \frac{d^2x}{dt^2} = \frac{d}{dt} \left(\frac{dx}{dt}\right) = \frac{d}{ds} \left( \frac{dx}{ds}\right) \cdot \frac{ds}{dt} = \frac{d^2x}{ds^2} = \frac{dy}{ds}$ since $\displaystyle y = \frac{dx}{ds}$.

    So your three equations are

    $\displaystyle
    \frac{dt}{ds} = 1,
    $
    $\displaystyle
    \frac{dx}{ds} = y,
    $
    $\displaystyle
    \frac{dy}{ds} - 2 y - 3 x = 10 \sin t.
    $
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  3. #3
    Super Member Showcase_22's Avatar
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    Ohhh, I see!

    I have one more question though, how do I solve $\displaystyle \frac{dx}{ds}=y$ to get $\displaystyle x(t)$?

    $\displaystyle y=\frac{dx}{dt}$ but it's completely wrong to put $\displaystyle x(t)=yt+C$ since $\displaystyle y$ is a function of t.

    It's pretty easy to do the same for $\displaystyle \frac{dt}{ds}=1$ since this is equal to $\displaystyle \frac{ds}{dt}=1$ so $\displaystyle s(t)=t+C$.

    I'm just a little bit stuck on what $\displaystyle x$ and $\displaystyle y$ actually are.
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    Ohhh, I see!

    I have one more question though, how do I solve $\displaystyle \frac{dx}{ds}=y$ to get $\displaystyle x(t)$?

    $\displaystyle y=\frac{dx}{dt}$ but it's completely wrong to put $\displaystyle x(t)=yt+C$ since $\displaystyle y$ is a function of t.

    It's pretty easy to do the same for $\displaystyle \frac{dt}{ds}=1$ since this is equal to $\displaystyle \frac{ds}{dt}=1$ so $\displaystyle s(t)=t+C$.

    I'm just a little bit stuck on what $\displaystyle x$ and $\displaystyle y$ actually are.
    The middle step isn't right. $\displaystyle x$ and $\displaystyle y$ are unknown functions of $\displaystyle t$. You're treating y as constant. What you need to do is first solve the homogeneous system

    $\displaystyle
    \frac{dx}{ds} = y
    $
    $\displaystyle \frac{dy}{dx} = 3x + 2y$
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  5. #5
    Super Member Showcase_22's Avatar
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    $\displaystyle \frac{dx}{ds}=\frac{dx}{dt}=y$ and $\displaystyle \frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{1} {y}.\frac{dy}{dt} =3x+2y$.

    But the second equation is no longer a linear equation so it can't be put into matrix form.

    (sorry about this, i'm finding differential equations really hard!)
    Last edited by Showcase_22; Dec 13th 2009 at 12:41 AM.
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