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need help with my math problem thanks for the help

Consider the second order homogeneous equation $y''(t) + by'(t) + cy(t) = 0$ where $b,c$ are real constants. This equation is equivalent to a 2x2 system of differential $\overrightarrow x = \begin{pmatrix} y(t) \\ y'(t) \\ \end{pmatrix}$. Find that system.

Attempt:

I have found the roots of the general characteristic equation and put them in the general solution using exponentials, but I am unsure how to proceed from here. How do I find the 2x2 matrix from here ?

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    You are making a mountain out of a molehill. Simply note that $$x'=\begin{pmatrix}y'\\y''\end{pmatrix}=\begin{pmatrix}y'\\-by'-cy\end{pmatrix}=\begin{pmatrix}\cdot&\cdot\\\cdot&\cdot\end{pmatrix}\begin{pmatrix}y\\ y'\end{pmatrix}$$ and identify the missing entries in the $2\times2$ matrix on the RHS. – 2017-02-03

3 Answers 3

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You have $x_1 = y$ and $x_2 = y'$. Therefore, $x_1' = x_2$ and $x_2' = -c x_1 -b x_2$. Can you put this in matrix form?

(Going from second and higher order to first order is a standard, mechanical procedure, which is required to use most packages that integrate differential equations.)

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Hint: Let $y(t)=y_1(t)$

$y'(t)=y_2(t)$

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$$\begin{cases} x_1(t) :=y(t)\\x_2(t):=y'(t)\end{cases}\Rightarrow \begin{cases} x'_1(t) =y'(t)=x_2(t)\\x'_2(t)=y''(t)=-cx_1(t)-bx_2(t)\end{cases}$$ $$\Rightarrow \begin{bmatrix}{x'_1(t)}\\{x'_2(t)}\end{bmatrix}=\begin{bmatrix}{0}&{1}\\{-c}&{-b}\end{bmatrix}\begin{bmatrix}{x_1(t)}\\{x_2(t)}\end{bmatrix}.$$