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Write the following linear differential equations with constant coefficients in the form of the linear system $\dot{x}=Ax$ and solve:

(a) $\ddot{x}+\dot{x}-2x=0$

(b) $\ddot{x}+x=0$

(c)$\dddot{x}-2\ddot{x}-\dot{x}+2x=0$

Hint: Let $x_1 = x, x_2 =\dot{x} $, etc.

I have tried to do this in the following way but I do not know if I am doing well:

Let $x_1 = x, x_2 =\dot{x}, x_3=\ddot{x}, x_4=\dddot{x} $ , then the previous system becomes (a) $x_3+x_2-2x_1=0$, (b) $x_3+x_1=0$, (c) $x_4-x_3-x_2+2x_1=0$ and thus $\begin{bmatrix}x_1\\x_2 \\x_3 \\ x_4\end{bmatrix}=\begin{bmatrix}\frac{x_2}{3}\\x_2 \\\frac{-x_2}{3} \\ 0\end{bmatrix}$. Is this what it should be or is it different? Thank you very much.

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  • $\begingroup$ should be different as the equations in all 3 parts should not be combined. $\endgroup$
    – cineel
    Commented Oct 23, 2021 at 2:25

2 Answers 2

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Let $x_1 =\dot{x}, x_2=\ddot{x}, x_3=\dddot{x}$. No need to substitute $x$ as we want the final equations to be in the form of $Ax$. So $$ \Biggl\{ \begin{array}{c} x_2+x_1=2x \\ x_2=-x \\ x_3-2x_2-x_1=-2x \end{array}$$ Substituting $x_2$ in the first equation, we get $x_1=3x$. Finally, we substitute $x_1, x_2$ in the last equation and we get $x_3=-x$. Now solve the final system $$ \Biggl\{ \begin{array}{c} \dot{x}=3x \\ \ddot{x}=-x \\ \dddot{x}=-x \end{array}$$

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  • $\begingroup$ this answer doesnt make sense as it seems like parts (a) , (b) , (c) were all combined. when they are separate questions. $\endgroup$
    – cineel
    Commented Oct 23, 2021 at 2:16
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For case (a) we can write higher order differential equations as a system with a very simple change of variable. We’ll start by defining the following two new functions. $$x_1(t) = x(t)$$ and $$x_2(t) = \dot{x}(t).$$ Now notice that if we differentiate both sides of these we get, $$\dot{x_1} = \dot{x} = x_2$$ and $$\dot{x_2} = - \dot{x} + 2x = -x_2 + 2x_1$$ and we have the system $$\begin{cases} \dot{x_1} = x_2\\ \dot{x_2} = 2x_1 - x_2 \end{cases}$$ and $$A = \begin{bmatrix} 0 & 1\\ 2 & -1\\ \end{bmatrix} $$ and then you can solve $\dot{x}=Ax$ system.

(b) $$A = \begin{bmatrix} 0 & 1\\ -1 & 0\\ \end{bmatrix} $$

(c) take $x_1 = x$, $x_2 = \dot{x}$ and $x_3= \ddot{x}$ $\Rightarrow$ $\dot{x_3}=\dddot{x}$, $\dot{x_2}=\ddot{x}$ and $\dot{x_1}=\dot{x}=x_2$.

$$\begin{cases} \dot{x_1} = x_2\\ \dot{x_2} = x_3\\ \dot{x_3} = -2x_1 + x_2 + 2x_3 \end{cases}$$ and $$A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ -2 & 1 & 2 \end{bmatrix} $$

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