Integrating absolute value function

Question

I'm working on a problem drawing phase plane diagrams in my applied mathematics course. I'm supposed to draw the phase line diagram of $x''+\vert x\vert=0.$ In the process, I get to the differential equation $$\frac{dy}{dx}=\frac{-\vert x\vert}{y}$$ and need to integrate but am apparently blanking on what this ends up being with the absolute values. I haven't taken an ODE course for 4 years so wondered if someone could go step by step with this integration. It'd be greatly appreciated so I can remember how and then draw my diagram.

Answer 1

Consider solving the pair of odes

$$ \frac{dy}{dx}= \begin{cases} \frac{-x}{y},\quad x\geq 0 \\ \frac{x}{y},\quad x < 0 \end{cases}.$$

Answer 2

This doesn't really make any sense unless it is an initial value problem. Remember that any continous function is integrable and for $a\leq0$ and $b\geq0:$

$$\int_{a}^{b}|x| = \int_{a}^{0} -x + \int_{0}^{b}x.$$

Solve the problem accordingly.

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \int\verts{x}\,\dd x = x\verts{x} - \int x\ \sgn\pars{x}\,\dd x =x\verts{x} - \int\verts{x}\,\dd x $$

$$ \int\verts{x}\,\dd x = \half\,x\verts{x} +\ \mbox{a constant} $$