Case 1 - two distinct real roots: Visual walkthrough

This page draws the derivation that the parent note wrote in symbols. We start from a picture of what the equation is even asking, and end with the general solution — one figure per step. No symbol appears before you can see what it means.

Step 1 — What does the equation even ask?

WHAT. Our whole task is this one line:

weight a acceleration of y y^{''} + weight b speed of y y^{'} + weight c height of y y = 0.

Read every symbol out loud:

$y$ is a function — a curve whose height at position $x$ we can plot.
$y^{'}$ (say "y-prime") is its slope at each point: how fast the curve is rising.
$y^{''}$ ("y-double-prime") is the slope's slope: how fast the rising itself is changing — its bend.
$a, b, c$ are just fixed numbers (weights), and $a \neq = 0$ so there really is a $y^{''}$ term.

WHY read it this way. The equation says: take the curve, its slope, and its bend, multiply each by its weight, add — and the result must be flat zero at every single $x$ . That is a very demanding balance. Most curves fail it. We are hunting the special curves that pass.

PICTURE. Below, one candidate curve $y$ is drawn in red. At the marked point, the black arrows show its height, its slope, and its bend. The balance $a y^{''} + b y^{'} + cy$ must cancel to $0$ everywhere, not just here.

Figure — Case 1 - two distinct real roots

Step 2 — Which curve is its own slope? (Why $e^{r x}$ )

WHAT. We need a shape where $y^{'}$ and $y^{''}$ are just scaled copies of $y$ . The exponential does exactly this:

y = e^{r x} ⟹ y^{'} = scale r e^{r x}, y^{''} = scale r^{2} e^{r x} .

Here $e \approx 2.718$ is a fixed number, and $r$ is an unknown growth rate we will solve for. When $r > 0$ the curve grows; when $r < 0$ it decays; the bigger $∣ r ∣$ , the steeper.

WHY this tool and no other. We could try polynomials, sines, anything — but taking a derivative usually changes a function's shape. The exponential is the unique shape whose derivative is the same shape, just rescaled (see Exponential function and its derivative). That is the one property Step 1 demanded. So $e^{r x}$ is not a lucky guess; it is forced.

PICTURE. The red curve $e^{r x}$ and its slope curve are laid on top of each other — they are the same shape, one just stretched vertically by the factor $r$ . That overlap is the whole reason the exponential works.

Step 3 — Substitute and watch $e^{r x}$ factor out

WHAT. Put the three copies from Step 2 into the equation:

a y^{''} r^{2} e^{r x} + b y^{'} r e^{r x} + c y e^{r x} = 0.

Every term carries the same block $e^{r x}$ . Pull it out front like a common factor:

e^{r x} (pure algebra in r a r^{2} + b r + c) = 0.

WHY. The exponential does its job and then steps aside: it converts a calculus problem (derivatives) into an algebra problem (a quadratic in $r$ ). That is the payoff of choosing $e^{r x}$ .

PICTURE. Think of it as peeling a label. The $e^{r x}$ is a factor riding on every term; peeling it off leaves a bare quadratic. The figure shows the three terms lined up, each wearing the same $e^{r x}$ sticker, then the sticker lifted off.

Step 4 — Kill the exponential: the characteristic equation

WHAT. We have $e^{r x} (a r^{2} + b r + c) = 0$ . A product is zero only if a factor is zero. But $e^{r x}$ is never zero — the exponential curve never touches the $x$ -axis, for any real $x$ . So the other factor must vanish:

a r^{2} + b r + c = 0 the == characteristic equation == .

WHY. This is the exact step where "which curves solve the ODE?" becomes "which numbers $r$ solve a quadratic?" Solving quadratics we already know how (see Characteristic equation).

PICTURE. The red exponential is drawn hugging the axis but never crossing it — visual proof that $e^{r x} \neq = 0$ , so it cannot be the factor that is zero. The bracket is the only candidate.

Step 5 — Solve the quadratic; when are the roots two distinct reals?

WHAT. The quadratic formula gives

r_{1, 2} = \frac{- b \pm b ^{2} - 4 a c discriminant D}{2 a} .

The part under the root, $D = b^{2} - 4 a c$ , is the discriminant. It decides everything:

$D > 0$ : the $\pm$ splits into two different real numbers $r_{1} \neq = r_{2}$ — this is Case 1.
$D = 0$ : the $\pm$ collapses, one repeated root — that is Case 2 repeated real root.
$D < 0$ : the square root of a negative — complex roots, Case 3 complex conjugate roots.

WHY the discriminant. It is precisely the quantity that measures the gap $r_{1} - r_{2} = D / a$ between the roots. A positive gap means two genuinely separate growth rates.

PICTURE. The parabola $y = a r^{2} + b r + c$ is drawn. Its two crossings of the $r$ -axis (in red) are $r_{1}$ and $r_{2}$ . When $D > 0$ the parabola cuts the axis at two points — that is what "two distinct real roots" looks like.

Step 6 — Two roots → two solution curves

WHAT. Each root feeds back through the ansatz to give a solution:

y_{1} = e^{r_{1} x}, y_{2} = e^{r_{2} x} .

Because $r_{1} \neq = r_{2}$ , these two curves have different steepness — they are not the same shape rescaled.

WHY they are independent. Form their ratio:

\frac{y _{1}}{y _{2}} = \frac{e ^{r_{1} x}}{e ^{r_{2} x}} = e^{(r_{1} - r_{2}) x} .

Since $r_{1} - r_{2} \neq = 0$ , this ratio genuinely changes with $x$ — it is not a fixed number. Two curves that aren't constant multiples of each other carry independent information (see Wronskian and linear independence).

PICTURE. Both curves plotted from the same start. One climbs (or decays) faster than the other; the red shaded gap between them keeps changing, showing the ratio is not constant.

Recall Wronskian check (why "not constant ratio" is enough)

The Wronskian $W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2} = (r_{2} - r_{1}) e^{(r_{1} + r_{2}) x}$ . Since $r_{1} \neq = r_{2}$ and $e^{(\dots) x} \neq = 0$ , we get $W \neq = 0$ everywhere — the formal certificate of independence.

Step 7 — Superpose: the general solution

WHAT. The equation is linear: if $y_{1}$ and $y_{2}$ each make the weighted sum zero, so does any blend

y (x) = how much of y_{1} C_{1} e^{r_{1} x} + how much of y_{2} C_{2} e^{r_{2} x}

where $C_{1}, C_{2}$ are free constants you tune to match a starting height and starting slope.

WHY exactly two constants. The equation is second-order — its highest term is $y^{''}$ . Such equations always have a 2-parameter family of solutions. Two independent curves + two dials $C_{1}, C_{2}$ = every solution, none missing (see Superposition principle).

PICTURE. Several blends of the two red base curves are drawn faintly, sweeping out the whole family; two of them are highlighted as picked by particular $C_{1}, C_{2}$ .

Step 8 — The three flavours of Case 1 (all sign combinations)

WHAT. "Two distinct real roots" still hides three qualitatively different pictures, depending on the signs of $r_{1}, r_{2}$ :

Both negative	Opposite signs	Both positive
$r_{1}, r_{2} < 0$	$r_{1} > 0 > r_{2}$	$r_{1}, r_{2} > 0$
every term decays	one grows, one decays	every term grows
$y \to 0$	saddle: generic $y$ blows up	$y \to \pm \infty$
e.g. $y^{''} + 5 y^{'} + 6 y = 0$	e.g. $y^{''} - 9 y = 0$	e.g. $y^{''} - 5 y^{'} + 6 y = 0$

WHY show all three. The formula is identical, but the behaviour is not. You must never be surprised by a growing solution just because you once saw a decaying one.

PICTURE. Three side-by-side panels: decay-to-zero, saddle (one term up one down), and blow-up — the key growing/decaying term in red in each.

The one-picture summary

The single figure above compresses the whole journey: equation → exponential guess → factor out → quadratic → two roots → two curves → blended family.

Recall Feynman retelling — the walkthrough in plain words

We were handed a curve-hunting puzzle: find a curve whose height, tilt, and bend, mixed with fixed weights, cancel to flat zero everywhere. Only one kind of curve has a tilt and bend that are just rescaled copies of itself — the exponential, like money growing at a steady interest rate. So we guessed $e^{r x}$ with a mystery growth-rate $r$ . Plugging in, the exponential appears in every term and politely factors out, leaving a plain quadratic in $r$ . Since the exponential never actually hits zero, the quadratic itself must be zero — that's the characteristic equation. Solve it. If the number under the square root is positive, we get two different growth rates: two curves, one for each. They aren't the same shape (their ratio keeps drifting), so together they cover everything. Blend them with two dials $C_{1}, C_{2}$ , twist the dials to match where you started, and you're done. And always peek at the signs of the two rates — both down means it fades to nothing, mixed means it usually runs away, both up means it explodes.

Connections

Case 1 - two distinct real roots (index 4.6.11) — the parent this page illustrates.
Exponential function and its derivative — Step 2's whole justification.
Characteristic equation — the quadratic born in Steps 3–4.
Wronskian and linear independence — Step 6's independence certificate.
Superposition principle — Step 7's blending rule.
Case 2 repeated real root — the $D = 0$ edge in Step 8.
Case 3 complex conjugate roots — the $D < 0$ sibling.