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Thread: Conversion of 2nd order differential equation into first order DE system

  1. #1
    Senior Member Vinod's Avatar
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    Conversion of 2nd order differential equation into first order DE system

    I want to convert this 2nd order differential equation into first order differential equation system.(matrix form).

    $y''-3y'-4y+12t-2=0$.

    Answer: My attempt is putting $x_1=y, x'_1=y'=x_2,x'_2=y''$ so i get $x'_2=3x'_1-4x_1+12t-2=0$

    Answer provided to me is $\vec{x'}=\small\begin{pmatrix}1&2\\3&2\end {pmatrix}$$\vec{x}+t \small\begin{pmatrix}2\\-4\end{pmatrix}$

    How to arrive at the answer?
    Last edited by Vinod; Nov 12th 2018 at 06:58 AM.
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    Senior Member Vinod's Avatar
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    Re: Conversion of 2nd order differential equation into first order DE system

    $x'_1=x_2$ $x'_2=3x'_1-4x_1+12t-2$
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    Senior Member Vinod's Avatar
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    Re: Conversion of 2nd order differential equation into first order DE system

    Quote Originally Posted by Vinod View Post
    $x'_1=x_2$ $x'_2=3x'_1-4x_1+12t-2$
    Hello,

    So, I converted the ordinary differential equation into a matrix form as follows:-$\vec{x'}$=$\small\begin {pmatrix}0&1\\3&4\end {pmatrix}$$\vec{x}$+t$\small\begin {pmatrix}0\\-12\end {pmatrix}$+$\small\begin {pmatrix}0\\2\end {pmatrix}$ Is this correct?
    Last edited by Vinod; Nov 25th 2018 at 11:32 PM.
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    Re: Conversion of 2nd order differential equation into first order DE system

    Quote Originally Posted by Vinod View Post
    Hello,

    So, I converted the ordinary differential equation into a matrix form as follows:-$\vec{x'}$=$\small\begin {pmatrix}0&1\\3&4\end {pmatrix}$$\vec{x}$+t$\small\begin {pmatrix}0\\-12\end {pmatrix}$+$\small\begin {pmatrix}0\\2\end {pmatrix}$ Is this correct?
    You have the matrix wrong and there's no good reason to split up the terms that don't depend on $x_1,~x_2$

    I'd say the answer is

    $\begin{pmatrix}x_1^\prime \\x_2^\prime\end{pmatrix} = \begin{pmatrix}0 &1 \\ -4 &3\end{pmatrix}\begin{pmatrix}x_1 \\x_2 \end{pmatrix} + \begin{pmatrix}0 \\ 12t-2\end{pmatrix}$
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    Senior Member Vinod's Avatar
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    Re: Conversion of 2nd order differential equation into first order DE system

    Quote Originally Posted by romsek View Post
    You have the matrix wrong and there's no good reason to split up the terms that don't depend on $x_1,~x_2$

    I'd say the answer is

    $\begin{pmatrix}x_1^\prime \\x_2^\prime\end{pmatrix} = \begin{pmatrix}0 &1 \\ -4 &3\end{pmatrix}\begin{pmatrix}x_1 \\x_2 \end{pmatrix} + \begin{pmatrix}0 \\ 12t-2\end{pmatrix}$
    Hello,

    Original DE is $y"-3y'-4y+12t-2=0$. But I made mistake whlie calculating $x'_2$ in my post#1 and #2.$x'_2=4x_1+3x_2-12t+2$ So i think $\vec{x'_2}=\small\begin {pmatrix}0&1\\4&3\end {pmatrix}\vec {x}+ \small\begin {pmatrix} 0\\-12t+2\end {pmatrix}$ would be correct.
    Last edited by Vinod; Nov 26th 2018 at 12:56 AM.
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    Senior Member Vinod's Avatar
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    Re: Conversion of 2nd order differential equation into first order DE system

    Quote Originally Posted by Vinod View Post
    Hello,

    Original DE is $y"-3y'-4y+12t-2=0$. But I made mistake whlie calculating $x'_2$ in my post#1 and #2.$x'_2=4x_1+3x_2-12t+2$ So i think $\vec{x'_2}=\small\begin {pmatrix}0&1\\4&3\end {pmatrix}\vec {x}+ \small\begin {pmatrix} 0\\-12t+2\end {pmatrix}$ would be correct.
    Hello,
    After solving $\vec{x'}=\small\begin{pmatrix}0&1\\4&3\end {pmatrix}\vec{x}+\small\begin{pmatrix}0\\-12t+2\end {pmatrix}$

    I get the general solution as $ C_1e^{4t}\small\begin{pmatrix}1\\4\end {pmatrix}+C_2e^{-t}\small\begin{pmatrix}-1\\1\end {pmatrix}$. Is this correct? What will be a particular solution to this DE system.
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    Senior Member Vinod's Avatar
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    Re: Conversion of 2nd order differential equation into first order DE system

    Quote Originally Posted by Vinod View Post
    Hello,
    After solving $\vec{x'}=\small\begin{pmatrix}0&1\\4&3\end {pmatrix}\vec{x}+\small\begin{pmatrix}0\\-12t+2\end {pmatrix}$

    I get the general solution as $ C_1e^{4t}\small\begin{pmatrix}1\\4\end {pmatrix}+C_2e^{-t}\small\begin{pmatrix}-1\\1\end {pmatrix}$. Is this correct? What will be a particular solution to this DE system.
    Can this differential equation system in matrix form $ \vec{x'}=\small\begin{pmatrix}0&1\\4&3\end {pmatrix}\vec{x}+\small\begin{pmatrix}0\\-12t+2\end {pmatrix}$ be solved? Or i have to use laplace transform?
    Last edited by Vinod; Dec 1st 2018 at 05:19 AM.
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