Case 1 - two distinct real roots: Question bank

Before we start, the objects in play — so no symbol appears un-earned:

$y$ is the unknown function we are solving for; $y^{'}$ is its slope (first derivative), $y^{''}$ its curvature (second derivative).
$a y^{''} + b y^{'} + cy = 0$ is our ODE, with fixed number weights $a, b, c$ and $a \neq = 0$ .
$e^{r x}$ is the exponential ansatz: the one shape whose derivative is just itself times a number, $\frac{d}{d x} e^{r x} = r e^{r x}$ (see Exponential function and its derivative).
$a r^{2} + b r + c = 0$ is the Characteristic equation; its roots $r_{1}, r_{2}$ come from the quadratic formula, and $b^{2} - 4 a c$ is the discriminant that decides which case we are in.
Case 1 means $b^{2} - 4 a c > 0$ : two different real roots $r_{1} \neq = r_{2}$ .

True or false — justify

TF1. If

y_{1}

and

y_{2}

both solve the ODE, then

C_{1} y_{1} + C_{2} y_{2}

also solves it.

True — the equation is linear, so scaling and adding solutions keeps you in the solution set; this is exactly the Superposition principle.

TF2.

y = e^{r_{1} x} + e^{r_{2} x}

(no constants) is the general solution in Case 1.

False — it is one particular solution. The general solution needs the two free constants

C_{1}, C_{2}

because a 2nd-order ODE has a two-parameter family.

TF3. Two distinct real roots always mean the solution decays to zero as

x \to \infty

False — decay needs both roots negative. If either root is positive (e.g.

r = \pm 3

), a generic solution grows without bound.

TF4.

e^{r_{1} x}

and

e^{r_{2} x}

are linearly independent whenever

r_{1} \neq = r_{2}

True — their ratio

e^{(r_{1} - r_{2}) x}

is non-constant, so neither is a fixed multiple of the other (confirmed by a nonzero Wronskian).

TF5. Since

e^{r x} \neq = 0

for every real

x

, we are allowed to divide it out of the substituted equation.

True — the exponential never hits zero, so cancelling it introduces no lost solutions and leaves the pure algebra

a r^{2} + b r + c = 0

TF6. If the roots are

r_{1} = 0

and

r_{2} = - 4

, one of the two basis solutions is the constant function

1

True —

e^{0 \cdot x} = 1

, so

y = C_{1} + C_{2} e^{- 4 x}

; the constant term is a perfectly valid, non-trivial solution.

TF7. A discriminant of exactly

0

still counts as Case 1 because "the quadratic formula gives a root."

False —

b^{2} - 4 a c = 0

gives one repeated root; the two exponentials coincide and lose independence. That is Case 2 repeated real root, not Case 1.

TF8. Swapping the labels

r_{1} \leftrightarrow r_{2}

changes the general solution.

False — it only renames the constants;

C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x}

describes the very same family of functions.

TF9. The characteristic equation depends on the initial conditions

y (0), y^{'} (0)

False — the roots come only from

a, b, c

. Initial conditions enter later, fixing

C_{1}, C_{2}

from the general solution.

Spot the error

SE1. "

r^{2} - r - 2 = 0

, so by factoring

(r - 2) (r + 1)

the roots are

- 2

and

1

The roots are where each factor is zero:

r - 2 = 0 \Rightarrow r = 2

and

r + 1 = 0 \Rightarrow r = - 1

. The roots are

+ 2

and

- 1

, not the numbers inside the brackets.

SE2. "Discriminant is

b^{2} - 4 a c = 25 - 24 = 1

, which is not positive, so this isn't Case 1."

1

is positive (

1 > 0

), so it is Case 1. The trap is misreading a small positive number as failing the strict inequality.

SE3. "For

y^{''} - 9 y = 0

we get

r^{2} - 9 = 0

, and since

b = 0

there's only one root."

r^{2} - 9 = 0

gives

r = \pm 3

— two distinct real roots. A zero middle coefficient does not reduce the root count; it just makes them symmetric.

SE4. "

y = C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x}

, so

y^{'} = C_{1} r_{1} e^{r_{1} x} + C_{2} e^{r_{2} x}

The chain rule pulls down

r_{2}

from the second term too:

y^{'} = C_{1} r_{1} e^{r_{1} x} + C_{2} r_{2} e^{r_{2} x}

. Forgetting one factor of

r

corrupts every later step.

SE5. "Both roots solve the ODE, so their product

e^{r_{1} x} \cdot e^{r_{2} x} = e^{(r_{1} + r_{2}) x}

is also a solution."

Superposition permits sums and scalings, not products.

e^{(r_{1} + r_{2}) x}

generally has a growth rate

r_{1} + r_{2}

that is not a root, so it need not satisfy the ODE.

SE6. "Since

e^{r x} (a r^{2} + b r + c) = 0

, either

e^{r x} = 0

or the bracket is zero — let's try

e^{r x} = 0

e^{r x}

is never zero for real

x

, so that branch is empty; only the bracket can vanish, which is precisely why we get the characteristic equation.

Why questions

WQ1. Why do we guess

e^{r x}

and not, say,

sin (r x)

or a polynomial?

Because

e^{r x}

is the unique shape whose derivatives are just scaled copies of itself, so

e^{r x}

factors out and the ODE collapses into simple algebra in

r

WQ2. Why does the algebra come out as a quadratic rather than some other equation?

The highest derivative is second order (

y^{''}

), which contributes

r^{2}

; matching each derivative to a power of

r

produces

a r^{2} + b r + c

, a degree-2 polynomial.

WQ3. Why must a second-order ODE have exactly two arbitrary constants?

Recovering

y

from

y^{''}

requires two integrations, each adding one constant; equivalently you need two initial data (

y

and

y^{'}

) to pin a unique solution.

WQ4. Why does

r_{1} \neq = r_{2}

guarantee independence, while

r_{1} = r_{2}

destroys it?

When the roots differ, the ratio

e^{(r_{1} - r_{2}) x}

changes with

x

(independent). When they're equal the ratio is

1

— the same function twice — so you've only found one direction of the solution space.

WQ5. Why do we even check the Wronskian if we can already see the ratio is non-constant?

The Wronskian gives a single, rigorous nonzero test valid at every point, and it generalises to cases where the "eyeball the ratio" trick isn't obvious.

WQ6. Why does the sign of the roots, not their size, decide long-term behaviour?

e^{r x}

, a positive

r

grows and a negative

r

decays as

x \to \infty

; the exponent's sign sets the direction, while its magnitude only sets the speed.

Edge cases

EC1. What happens to the general solution if exactly one root is

0

That term becomes

C_{1} e^{0 \cdot x} = C_{1}

, a constant, so

y = C_{1} + C_{2} e^{r_{2} x}

— the solution levels off toward

C_{1}

rather than toward zero.

EC2. What if both distinct roots are negative but very close, like

r_{1} = - 3, r_{2} = - 3.0001

Still Case 1 (they differ), so

y = C_{1} e^{- 3 x} + C_{2} e^{- 3.0001 x}

; both decay. The formulas don't break, though numerically the two exponentials look nearly identical.

EC3. What if the roots are equal in magnitude but opposite in sign,

r = \pm k

You get

y = C_{1} e^{k x} + C_{2} e^{- k x}

— one growing, one decaying "saddle" behaviour; the solution blows up unless the initial data forces

C_{1} = 0

EC4. As

b^{2} - 4 a c \to 0^{+}

, what happens to the two roots and to Case 1?

The roots slide toward each other and merge; in the limit they coincide, Case 1 breaks down, and you must switch to Case 2 repeated real root with the extra

x

factor.

EC5. If

b^{2} - 4 a c < 0

instead, why can't we stay in Case 1?

The square root of a negative number gives complex roots, so no real distinct roots exist; this is Case 3 complex conjugate roots, where solutions oscillate.

EC6. Could a genuine Case-1 solution ever be the zero function

y \equiv 0

Yes — choosing

C_{1} = C_{2} = 0

gives the trivial solution, which always satisfies a homogeneous ODE but carries no information; it's the one member of the family everyone forgets.

EC7. If someone hands you

r_{1} = 2, r_{2} = - 1

but writes

y = C_{1} e^{- x} + C_{2} e^{2 x}

, is that wrong?

No — the order and naming of terms is free, so this is the identical general solution; only the set of exponentials matters, not their sequence.

Connections

Case 1 - two distinct real roots (index 4.6.11) — the parent note these traps stress-test.
Characteristic equation — the algebra whose roots every card hinges on.
Case 2 repeated real root / Case 3 complex conjugate roots — the two "next door" cases many traps disguise themselves as.
Wronskian and linear independence — the rigorous independence test behind TF4/WQ5.
Superposition principle — why sums (not products) of solutions are allowed (TF1, SE5).
Exponential function and its derivative — why $e^{r x}$ is the natural guess (WQ1).