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

1. ## 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 $x'_2=3x'_1-4x_1+12t-2=0$

4. ## Re: Conversion of 2nd order differential equation into first order DE system Originally Posted by Vinod 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 $x_1,~x_2$

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

5. ## Re: Conversion of 2nd order differential equation into first order DE system Originally Posted by romsek You have the matrix wrong and there's no good reason to split up the terms that don't depend on $x_1,~x_2$

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

6. ## Re: Conversion of 2nd order differential equation into first order DE system Originally Posted by Vinod 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.

7. ## Re: Conversion of 2nd order differential equation into first order DE system Originally Posted by Vinod 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?