Case 1 - two distinct real roots — Maths notes

Context: solving the homogeneous linear ODE with constant coefficients $ay'' + by' + cy = 0, \qquad a\neq 0.$ This note covers the case where the characteristic equation has two distinct real roots.

Derivation from scratch

WHAT we compute: $y = e^{rx}, \quad y' = r e^{rx}, \quad y'' = r^2 e^{rx}.$

HOW we substitute into $ay'' + by' + cy = 0$ : $a\,r^2 e^{rx} + b\,r\,e^{rx} + c\,e^{rx} = 0.$

Why this step? Each derivative just pulls down a power of $r$ , so $e^{rx}$ is a common factor.

Factor it out: $e^{rx}\big(ar^2 + br + c\big) = 0.$

Since $e^{rx} \neq 0$ for all real $x$ , the bracket must vanish:

Why we get TWO independent solutions

Each root gives a solution: $y_1 = e^{r_1 x}$ and $y_2 = e^{r_2 x}$ .

WHY combine them? The ODE is linear and second-order. Linear $\Rightarrow$ any sum/scaling of solutions is again a solution (superposition). Second-order $\Rightarrow$ the general solution needs exactly two arbitrary constants. So:

Figure — Case 1 - two distinct real roots

Worked examples

Characteristic equation: $r^2 + 5r + 6 = 0$ . Why? Replace $y''\to r^2,\ y'\to r,\ y\to 1$ .
Factor: $(r+2)(r+3)=0 \Rightarrow r_1=-2,\ r_2=-3$ . Why? These are the distinct real roots; discriminant $=25-24=1>0$ . ✔ Case 1.
General solution: $y = C_1 e^{-2x} + C_2 e^{-3x}$ . Why? Superpose the two independent exponentials.

$r^2 - r - 2 = 0 \Rightarrow (r-2)(r+1)=0 \Rightarrow r_1=2,\ r_2=-1$ .
$y = C_1 e^{2x} + C_2 e^{-x}$ .
Apply $y(0)=1$ : $C_1 + C_2 = 1$ . Why? At $x=0$ , both exponentials equal $1$ .
Differentiate: $y' = 2C_1 e^{2x} - C_2 e^{-x}$ . Apply $y'(0)=4$ : $2C_1 - C_2 = 4$ . Why? We need a second equation to pin both constants.
Add the two: $3C_1 = 5 \Rightarrow C_1 = \tfrac{5}{3}$ , then $C_2 = 1 - \tfrac{5}{3} = -\tfrac{2}{3}$ .
Answer: $y = \tfrac{5}{3}e^{2x} - \tfrac{2}{3}e^{-x}$ .

$r^2 - 9 = 0 \Rightarrow r = \pm 3$ . (Here $b=0$ , still two distinct reals.)
$y = C_1 e^{3x} + C_2 e^{-3x}$ . Why this matters: one term blows up, one decays — generic solutions grow unless $C_1=0$ .

Forecast-then-Verify

Recall Forecast: before solving

$y''+7y'+12y=0$ , predict the roots' signs and long-term behaviour. Forecast: $b,c>0$ with $b^2>4ac$ , so both roots are real and negative $\Rightarrow$ solution decays to 0. Verify: $r^2+7r+12=(r+3)(r+4)=0$ , roots $-3,-4$ (both negative). $y=C_1e^{-3x}+C_2e^{-4x}\to 0$ . ✔

Common mistakes

Recall Feynman: explain to a 12-year-old

Imagine a function that grows or shrinks at a rate proportional to its own size — like money in a bank with interest, $e^{rx}$ . The equation is asking: "find growth rates $r$ so that when I stack up the function and its speed and its acceleration with fixed weights, they cancel to zero." That cancelling is just a little quadratic puzzle. If the puzzle has two different answers $r_1$ and $r_2$ , then BOTH growth patterns work, and the full answer is just a mix ( $C_1$ of the first plus $C_2$ of the second), where you choose the mix to match where you start.

Connections

Characteristic equation — the algebra engine behind all constant-coefficient ODEs.
Case 2 repeated real root — what happens when $b^2-4ac=0$ .
Case 3 complex conjugate roots — when $b^2-4ac<0$ , gives oscillations via Euler's formula.
Wronskian and linear independence — proves $y_1,y_2$ span the solution space.
Superposition principle — why we add solutions.
Exponential function and its derivative — why $e^{rx}$ is the natural ansatz.

What ansatz do we try for

ay''+by'+cy=0

y=e^{rx}

, because its derivatives are scaled copies of itself.

What is the characteristic equation of

ay''+by'+cy=0

ar^2+br+c=0

Condition for Case 1 (two distinct real roots)?

Discriminant

b^2-4ac>0

General solution when roots are distinct reals

r_1\neq r_2

y=C_1e^{r_1x}+C_2e^{r_2x}

Why do we need two constants

C_1,C_2

The ODE is 2nd-order, so its solution family has two free parameters.

Why are

e^{r_1x}

and

e^{r_2x}

linearly independent when

r_1\neq r_2

Their ratio

e^{(r_1-r_2)x}

is non-constant; Wronskian

(r_2-r_1)e^{(r_1+r_2)x}\neq0

Roots of

y''+5y'+6y=0

r=-2,-3

, so

y=C_1e^{-2x}+C_2e^{-3x}

b^2-4ac=0

instead, which case and what changes?

Case 2 (repeated root); solution becomes

(C_1+C_2x)e^{r_1x}

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab humein $ay''+by'+cy=0$ type ka equation solve karna hota hai, toh trick simple hai: hum guess karte hain ki solution $y=e^{rx}$ hoga. Kyun? Kyunki exponential ka derivative apne hi jaisa hota hai — bas ek $r$ ka factor nikal aata hai. Isliye jab tum $e^{rx}$ ko equation mein daalte ho, $e^{rx}$ common ban ke bahar nikal jaata hai aur peeche reh jaata hai ek simple quadratic: $ar^2+br+c=0$ . Yahi characteristic equation hai.

Ab Case 1 ka matlab hai jab is quadratic ke do alag-alag real roots milein, yaani discriminant $b^2-4ac>0$ . Maan lo roots $r_1$ aur $r_2$ hain (different). Toh $e^{r_1x}$ bhi solution hai aur $e^{r_2x}$ bhi. Kyunki equation linear aur second-order hai, general solution dono ka mix hota hai: $y=C_1e^{r_1x}+C_2e^{r_2x}$ . Yeh do constants $C_1,C_2$ initial conditions se nikalte hain.

Ek important baat: dono constants mat bhoolna! Bahut students sirf $e^{r_1x}+e^{r_2x}$ likh dete hain — galat. Second-order ODE ko hamesha do free constants chahiye. Aur agar discriminant exactly zero ho jaye, toh roots same ho jaate hain — woh Case 2 hai, jahan ek extra $x$ multiply karna padta hai: $(C_1+C_2x)e^{rx}$ . Isliye Case 1 sirf tab jab roots strictly different hon.

Yaad rakhne ka mantra: Replace, Solve, Superpose — derivatives ko $r$ ki powers se replace karo, quadratic solve karo, aur dono solutions ko constants ke saath jod do. Bas ho gaya!

Case 1 - two distinct real roots

The big idea

Derivation from scratch

Why we get TWO independent solutions

Worked examples

Forecast-then-Verify

Common mistakes

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Ordinary Differential Equations

Connections