Homogeneous with constant coefficients — characteristic equation

1. Setup and definitions

WHY does homogeneous + linear let us add solutions? Because the operator $L[y]=a_n y^{(n)}+\dots+a_0 y$ is linear: $L[c_1y_1+c_2y_2]=c_1L[y_1]+c_2L[y_2]$ . If $L[y_1]=0$ and $L[y_2]=0$ , then any combination is also $0$ . So the general solution is a linear combination of independent basic solutions — and an $n$ -th order equation needs $==n=="$ independent ones.

2. Derivation from scratch (second order)

Take $ay'' + by' + cy = 0$ , $a\neq 0$ .

Step 1 — Guess $y=e^{rx}$ . Why this step? Only exponentials reproduce themselves under differentiation, so they're the natural candidate to make the combination cancel.

Step 2 — Differentiate: $y'=re^{rx}$ , $y''=r^2e^{rx}$ .

Step 3 — Substitute: $a r^2 e^{rx} + b r e^{rx} + c e^{rx} = 0 \;\Rightarrow\; e^{rx}(ar^2+br+c)=0.$ Why this step? Factor out the common $e^{rx}$ to isolate the algebra.

Step 4 — Divide by $e^{rx}$ (never zero): $\boxed{ar^2 + br + c = 0}$ the characteristic equation. Solve it with the quadratic formula $r=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ .

The discriminant $\Delta=b^2-4ac$ splits everything into 3 cases.

3. The three cases (and WHY each form appears)

Why does the repeated root need $xe^{rx}$ ?

If $r$ is a double root then $ar^2+br+c=a(r-r_0)^2$ , so the operator factors as $a(D-r_0)^2 y=0$ where $D=\frac{d}{dx}$ . Solving $(D-r_0)u=0$ gives $u=c_2e^{r_0x}$ , then $(D-r_0)y=u$ is a first-order linear ODE whose solution is $y=(c_1+c_2x)e^{r_0x}$ . The factor $x$ appears because the integrating factor cancels the exponential and leaves a constant to integrate, producing a linear-in- $x$ term. That's why — not a magic rule.

Why do complex roots give sin and cos?

With $r=\alpha\pm i\beta$ , solutions are $e^{(\alpha+i\beta)x}$ and $e^{(\alpha-i\beta)x}$ . These are complex; but the ODE has real coefficients, so we want real solutions. Take real and imaginary parts: $e^{\alpha x}\cos\beta x$ and $e^{\alpha x}\sin\beta x$ — both are solutions and they are independent. (This is Euler's formula doing all the work.)

Figure — Homogeneous with constant coefficients — characteristic equation

4. Worked examples

Worked example Example 4 — Complex with decay + initial conditions

Solve $y''+2y'+5y=0$ , $y(0)=1,\ y'(0)=0$ .

Char. eq.: $r^2+2r+5=0\Rightarrow r=\frac{-2\pm\sqrt{4-20}}{2}=-1\pm 2i$ . So $\alpha=-1,\beta=2$ .
General: $y=e^{-x}(c_1\cos2x+c_2\sin2x)$ .
Apply $y(0)=1$ : $c_1=1$ . Why? At $x=0$ , $e^0=1$ , $\cos0=1,\sin0=0$ .
$y'=e^{-x}(-c_1\cos2x-c_2\sin2x) + e^{-x}(-2c_1\sin2x+2c_2\cos2x)$ . At $0$ : $-c_1+2c_2=0\Rightarrow c_2=\tfrac12$ .
Answer: $y=e^{-x}\big(\cos2x+\tfrac12\sin2x\big)$ . A damped oscillation.

5. Common mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine a swing. The rule for how it moves only involves the swing's position, its speed, and how fast its speed changes — all multiplied by fixed numbers. We bet the answer looks like a special number $e$ raised to a power, because that's the one shape that stays the same when you measure how fast it changes. Plugging the bet in turns the hard "rate of change" question into an easy "solve a number puzzle" ( $ar^2+br+c=0$ ). If the puzzle gives two normal numbers → two stretchy exponential curves. If it gives one repeated number → you sneak in an extra $x$ . If it gives "imaginary" numbers → the swing actually wiggles, so you get waves (sin and cos), maybe fading away.

Connections

Linear Differential Operators and Superposition
Euler's Formula and Complex Exponentials
Wronskian and Linear Independence of Solutions
Method of Undetermined Coefficients (non-homogeneous next step)
Damped Harmonic Oscillator (physics application of Case 3)
Second Order Linear ODE — Reduction of Order (explains $xe^{rx}$ )

Flashcards

Why guess

y=e^{rx}

for constant-coefficient linear ODEs?

Exponentials are the only functions whose derivatives are scaled copies of themselves, so they turn the ODE into a polynomial equation in

r

What is the characteristic equation of

ay''+by'+cy=0

ar^2+br+c=0

, obtained by substituting

y=e^{rx}

and dividing by

e^{rx}

Case

\Delta>0

(distinct real roots) general solution?

y=c_1e^{r_1x}+c_2e^{r_2x}

Case

\Delta=0

(repeated root

r

) general solution?

y=(c_1+c_2x)e^{rx}

Case

\Delta<0

(roots

\alpha\pm i\beta

) general solution?

y=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)

Why does the repeated root need an extra factor of

x

The operator factors as

(D-r_0)^2

; solving the chained first-order equations introduces a linear-in-

x

term, giving a second independent solution

xe^{rx}

Why do complex roots produce sin and cos?

Euler's formula:

e^{(\alpha\pm i\beta)x}=e^{\alpha x}(\cos\beta x\pm i\sin\beta x)

; taking real and imaginary parts gives two real independent solutions.

r=\alpha\pm i\beta

, what does

\alpha

control vs

\beta

\alpha

controls exponential growth/decay of the envelope;

\beta

controls the oscillation frequency.

Solve

y''+4y=0

r^2+4=0\Rightarrow r=\pm2i

, so

y=c_1\cos2x+c_2\sin2x

How many independent solutions does an order-

n

homogeneous linear ODE have?

Exactly

n

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab humein aisi ODE milti hai jaise $ay''+by'+cy=0$ jisme saare coefficients constant hain, to seedha trick hai: hum guess karte hain ki solution $y=e^{rx}$ hoga. Kyun? Kyunki sirf exponential function aisa hai jiska derivative khud ka hi scaled version hota hai. Jaise hi yeh plug karte ho, $e^{rx}$ common nikal kar cancel ho jaata hai, aur bachta hai ek simple polynomial $ar^2+br+c=0$ — isko characteristic equation kehte hain. Matlab calculus ka tough problem ek easy algebra problem ban gaya.

Ab roots ke teen cases hain. Agar do alag real roots mile ( $\Delta>0$ ), to answer hai $c_1e^{r_1x}+c_2e^{r_2x}$ . Agar ek hi root repeat ho ( $\Delta=0$ ), to sirf $e^{rx}$ likhne se ek solution kam pad jaata hai, isliye ek extra $x$ multiply karke $xe^{rx}$ lete hain — $(c_1+c_2x)e^{rx}$ . Aur agar roots complex ho ( $\alpha\pm i\beta$ ), to Euler formula use karke $e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)$ milta hai, jisme $\alpha$ batata hai growth ya decay aur $\beta$ batata hai oscillation kitni tez hai.

Yeh matter kyun karta hai? Physics mein spring, pendulum, electrical circuits — sab isi tarah ke equations dete hain. $\alpha<0$ ho to oscillation dheere dheere mar jaata hai (damping), $\alpha=0$ ho to pure wave chalti rehti hai. Toh ek hi formula trick se tum predict kar sakte ho ki system spread karega, shaant hoga, ya hilta rahega.

Sabse common galti: repeated root pe $c_1e^{rx}+c_2e^{rx}$ likh dena — yeh galat hai kyunki yeh ek hi constant ban jaata hai. Hamesha yaad rakho: "Different, Double, Dizzy" — alag roots, double root mein $x$ daalo, aur complex mein sin-cos aata hai.

Homogeneous with constant coefficients — characteristic equation

1. Setup and definitions

2. Derivation from scratch (second order)

3. The three cases (and WHY each form appears)

Why does the repeated root need $xe^{rx}$ ?

Why do complex roots give sin and cos?

4. Worked examples

5. Common mistakes (Steel-manned)

Connections

Flashcards

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Ordinary Differential Equations

Connections