SymPy — symbolic algebra, calculus, ODE solving: Visual walkthrough

Prerequisite trails you can wander back to: Limits and the definition of the derivative, Ordinary Differential Equations — characteristic equation method, Taylor and Maclaurin Series, and the home base the parent SymPy note.

Step 1 — What a derivative is, drawn

WHAT. Before any equation, we need one idea: the derivative. Pick a curve — a height that changes as you move right. The derivative at a point is the steepness of the curve there — how fast the height climbs or falls.

WHY this tool and not another. We are about to meet the symbol $y^{'}$ and $y^{''}$ . If we don't nail down what they mean geometrically, the whole page is symbol-pushing. So: earn the symbol first. We use the slope-of-the-tangent-line picture because it is the one thing $y^{'}$ measures, nothing more.

PICTURE. Look at the figure. The white curve is some function $y (x)$ . The amber line just kisses the curve at one point — that's the tangent line. Its tilt is $y^{'}$ at that point.

Figure — SymPy — symbolic algebra, calculus, ODE solving

y^{''} = "rate of change of" \frac{d}{d x} the slope y^{'}

Read it aloud: " $y^{''}$ is the rate of change of the slope." Nothing more mysterious than that.

Step 2 — Reading the equation $y^{''} + y = 0$ as a sentence

WHAT. We rewrite the equation so its meaning is loud:

y^{''} = - y

WHY. An equation with symbols is a sentence in disguise. Rearranging to isolate $y^{''}$ turns it into a physical rule we can picture. We move $y$ to the right (subtract it from both sides) purely so the "curving" sits alone on the left.

PICTURE. Each symbol is doing a job:

how the curve bends y^{''} = opposite direction - current height y

The sentence: "however high you are, bend back toward zero by exactly that much." High up → bend down hard. Below zero → bend up hard. At zero → don't bend. Look at the figure: the amber arrows are the bending force $y^{''}$ , always pointing back at the centre line, always as long as the height itself.

Step 3 — The educated guess $y = e^{r x}$

WHAT. We try a candidate solution:

y = e^{r x}

Here $e \approx 2.718$ is a fixed number, and $r$ is an unknown number we get to choose.

WHY this function and not another. We need a function whose derivative looks like itself, because the equation compares $y$ with $y^{''}$ — apples must meet apples. The exponential is the unique function with the magic property

\frac{d}{d x} e^{r x} = r e^{r x}

— differentiating just multiplies by $r$ . That means $y$ , $y^{'}$ , $y^{''}$ are all copies of $e^{r x}$ scaled by powers of $r$ , so the equation will collapse into simple algebra in $r$ . No other elementary function keeps its shape under differentiation, so this is the right tool.

PICTURE. The figure shows $e^{r x}$ and its derivative sitting on top of each other, differing only by a vertical stretch factor $r$ — same shape, rescaled.

take slope \frac{d}{d x} our guess e^{r x} = the stretch r same shape back e^{r x}

Step 4 — Plug in and cancel: the characteristic equation

WHAT. Differentiate the guess twice and substitute into $y^{''} + y = 0$ .

y = e^{r x}, y^{'} = r e^{r x}, y^{''} = r^{2} e^{r x}

Substituting:

y^{''} r^{2} e^{r x} + y e^{r x} = 0

WHY. This is the whole point of the guess: the equation now contains only $e^{r x}$ as a common factor. We factor it out:

e^{r x} (r^{2} + 1) = 0

Now here's the key move. $e^{r x}$ is never zero — an exponential is always positive, it can't vanish. So the only way the product is zero is if the other factor is zero:

r^{2} + 1 = 0

This tiny equation is called the characteristic equation. We just traded a calculus problem for an algebra problem — that's the trick.

PICTURE. The figure shows the exponential curve staying strictly above zero everywhere (so it can never be the "zero" factor), forcing $r^{2} + 1 = 0$ .

never 0 e^{r x} \cdot must be 0 (r^{2} + 1) = 0

Step 5 — Solving $r^{2} + 1 = 0$ : meeting $i$

WHAT. Solve for $r$ :

r^{2} = - 1 ⟹ r = \pm i

WHY the symbol $i$ . We need a number whose square is $- 1$ . No ordinary number works — squaring any real number gives something $\geq 0$ . So mathematicians defined a new number, called $i$ , by the single rule

i^{2} = - 1.

That's all $i$ means: "the thing that squares to $- 1$ ." We introduce it only because the equation demanded it — the algebra hit a wall, and $i$ is the doorway through the wall.

PICTURE. The figure is the number line bent into a plane: real numbers run left–right, and $i$ points up, off the line. The two solutions $+ i$ and $- i$ sit above and below the centre — mirror images.

Step 6 — From $e^{i x}$ to real waves (Euler's bridge)

WHAT. Our two solutions are $y = e^{i x}$ and $y = e^{- i x}$ . But we wanted a real height — a curve you can draw. Euler's identity converts the imaginary exponential into ordinary waves:

e^{i x} = cos x + i sin x

WHY. We are stuck with $e^{i x}$ , which looks unplottable. Euler's identity is the translator from "exponential of imaginary" to "sine and cosine." It's not a trick pulled from air — it falls straight out of the Taylor and Maclaurin Series of $e$ , $cos$ and $sin$ (the parent note's series tool): line up the three power series and the pattern matches exactly.

PICTURE. The figure shows the point $e^{i x}$ travelling around a unit circle as $x$ grows. Its horizontal shadow traces $cos x$ ; its vertical shadow traces $sin x$ . Circular motion is two oscillations at right angles — exactly the wobble we forecast in Step 2.

e^{i x} = horizontal shadow cos x + i vertical shadow sin x

Step 7 — Assembling the general solution

WHAT. The two imaginary solutions combine into two real building blocks, $cos x$ and $sin x$ . Because the equation is linear (no $y^{2}$ , no $y \cdot y^{'}$ ), any weighted mix of solutions is again a solution. So:

y (x) = how much sine C_{1} sin x + how much cosine C_{2} cos x

WHY two constants. A second-derivative equation ( $y^{''}$ ) needs two facts to pin down a unique curve — where you start ( $y (0)$ ) and how fast you start ( $y^{'} (0)$ ). Each fact eats one constant. That is exactly why dsolve hands back $C_{1}$ and $C_{2}$ : it doesn't yet know your starting conditions.

PICTURE. The figure overlays several members of the family — different $C_{1}, C_{2}$ give waves of different height and phase, but every one wobbles, confirming Step 2's spring forecast.

x = symbols('x')
y = Function('y')
dsolve(Eq(y(x).diff(x, 2) + y(x), 0), y(x))
# -> Eq(y(x), C1*sin(x) + C2*cos(x))

Step 8 — The degenerate & edge cases (never leave a gap)

WHAT. What if the equation were slightly different? We check the neighbours so no scenario surprises you.

WHY. The judge of understanding is coverage: you must know why this equation gives waves and a cousin gives something else.

Equation	Characteristic eqn	Roots $r$	Solution shape	Picture
$y^{''} + y = 0$	$r^{2} + 1 = 0$	$\pm i$ (imaginary)	$cos x, sin x$ — pure waves	Step 7
$y^{''} - y = 0$	$r^{2} - 1 = 0$	$\pm 1$ (real)	$e^{x}, e^{- x}$ — run away / decay	below
$y^{''} = 0$	$r^{2} = 0$	$0, 0$ (repeated)	$1, x$ — straight line	below
$y^{'} - 2 y = 0$	$r - 2 = 0$	$2$ (single)	$e^{2 x}$ — growth, one constant	parent §4

The sign inside the characteristic equation decides everything: $+$ gives imaginary roots and oscillation; $-$ gives real roots and exponential growth/decay; a repeated root at $0$ gives a straight line. The figure draws all three side by side.

The one-picture summary

This single figure compresses all eight steps: the equation's meaning (spring), the exponential guess, the algebra collapse, the imaginary roots on the vertical axis, Euler's circle, and the two real waves that come out.

Recall Feynman retelling — the whole walkthrough in plain words

We started with a curve and learned that $y^{'}$ is its steepness and $y^{''}$ is how the steepness bends. The equation $y^{''} + y = 0$ literally says "bend back toward the middle as hard as you're far from it" — that's a spring, so we guessed the answer wobbles. To find it we tried $y = e^{r x}$ , the one function that stays itself when you take its slope, so the calculus turned into simple algebra: $r^{2} + 1 = 0$ . That needs a number squaring to $- 1$ , so $i$ walked in, giving $r = \pm i$ . Euler's identity then said $e^{i x}$ is just a point spinning around a circle, whose two shadows are $cos x$ and $sin x$ . Mix those two waves with any amounts $C_{1}, C_{2}$ and you have every possible solution — exactly what dsolve prints. Change one sign and the roots turn real and the wobble becomes runaway growth. That's the whole story: a spring, a clever guess, a doorway called $i$ , and two waves.

Recall Why does

dsolve return two constants for $y^{''} + y = 0$ but only one for $y^{'} - 2 y = 0$ ? The order of the highest derivative equals the number of independent starting facts you must supply, hence the number of arbitrary constants: second order → $C_{1}, C_{2}$ ; first order → $C_{1}$ alone.

Recall Why is factoring out

$e^{r x}$ legal and useful? $e^{r x}$ is never zero, so dividing (or factoring) by it can't lose or invent solutions — and it strips the calculus away, leaving the pure algebra $r^{2} + 1 = 0$ .

Connections

Limits and the definition of the derivative — Step 1's slope idea, made rigorous.
Ordinary Differential Equations — characteristic equation method — the hand method dsolve automates.
Taylor and Maclaurin Series — where Euler's $e^{i x} = cos x + i sin x$ (Step 6) comes from.
Lambdify — bridging SymPy to NumPy for plotting — turn $C_{1} sin x + C_{2} cos x$ into numbers to plot.
NumPy — numerical arrays — the numeric counterpart that would only ever see the decimals.