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Math Help - Three variables?

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

    Write \ddot x -2 \dot x -3x=10 \sin t in the three variables x, \ y= \dot x and \dot s =1 so that the equations are autonomous and first order.
    Clearly \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 \dot x = Ax (where x \in \mathbb{R}^n and A is an 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 \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 \dot x = Ax (where x \in \mathbb{R}^n and A is an 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
    \frac{ds}{dt} = 1 so \frac{dx}{dt} = \frac{dx}{ds}\cdot \frac{ds}{dt} = \frac{dx}{ds}. Simiarly \frac{dy}{dt} = \frac{dy}{ds}.

    Then \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 y = \frac{dx}{ds}.

    So your three equations are

     <br />
\frac{dt}{ds} = 1,<br />
     <br />
\frac{dx}{ds} = y,<br />
     <br />
\frac{dy}{ds} - 2 y - 3 x = 10 \sin t.<br />
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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 \frac{dx}{ds}=y to get x(t)?

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

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

    I'm just a little bit stuck on what x and 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 \frac{dx}{ds}=y to get x(t)?

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

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

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

     <br />
\frac{dx}{ds} = y<br />
    \frac{dy}{dx} = 3x + 2y
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  5. #5
    Super Member Showcase_22's Avatar
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    \frac{dx}{ds}=\frac{dx}{dt}=y and \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; December 13th 2009 at 01:41 AM.
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