Homogeneous with constant coefficients — characteristic equation: Worked examples

4.6.10 · D3 · Maths › Ordinary Differential Equations › Homogeneous with constant coefficients — characteristic equa

Yeh page ek drill hai har case ke liye jo ek constant-coefficient linear ODE ki characteristic equation mein aa sakta hai. Hum pehle saari possible root shapes list karte hain (the "scenario matrix"), phir har cell ke liye ek example karte hain. Agar aapne teen cases abhi tak nahi padhe, toh pehle parent padhein: Homogeneous with constant coefficients — characteristic equation.

The scenario matrix

Discriminant hai $== Δ = b^{2} - 4 a c ==$ — woh cheez jo $r = \frac{- b \pm Δ}{2 a}$ mein square root ke andar hai. Iska sign, aur kuch special coefficient values, har possible problem ko in cells mein divide karte hain. Poori discussion mein, $n$ matlab hai ODE ka order (sabse uunchi derivative), jo characteristic polynomial ki degree ke barabar hai:

Cell	Trigger	Root type	Answer shape
A Distinct real, dono negative	$Δ > 0$ , $b > 0$ , $c > 0$	$r_{1}, r_{2} < 0$	decaying $c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}$
A2 Distinct real, dono positive	$Δ > 0$ , $b < 0$ , $c > 0$	$r_{1}, r_{2} > 0$	pure growth $c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}$
B Distinct real, opposite signs	$Δ > 0$ , $c < 0$	ek $+$ , ek $-$	ek term blow up karti hai
C Ek root exactly zero	$c = 0$	$r = 0$ aur $r = - b / a$	constant $+$ exponential
D Repeated real root	$Δ = 0$	$r$ double	$(c_{1} + c_{2} x) e^{r x}$
E Pure imaginary ( $α = 0$ )	$b = 0$ , $c > 0$	$\pm i β$	undamped $cos, sin$
F Complex with decay	$Δ < 0$ , $b > 0$	$α \pm i β$ , $α < 0$	damped oscillation
G Complex with growth	$Δ < 0$ , $b < 0$	$α \pm i β$ , $α > 0$	growing oscillation
H Higher order ( $n \geq 3$ )	degree $\geq 3$	upar waalon ka mix	factors ka product
I Word problem (physics)	—	koi bhi	real-world reading
J Exam twist (given a solution)	reverse-engineer	—	ODE banao

Har example neeche apni cell(s) ke saath tagged hai. (Yahan $α$ real part hai aur $β$ imaginary part hai ek complex root $r = α \pm i β$ ka.)

Cell A — Distinct real roots, dono negative (pure decay)

Worked example Cell A:

$y^{''} + 5 y^{'} + 6 y = 0$ Pehle forecast karo: saare coefficients positive hain — growth expect karte ho ya decay? Padhne se pehle guess karo.

Char. eq.: $r^{2} + 5 r + 6 = 0$ . Yeh step kyun? $y^{''} \to r^{2}$ , $y^{'} \to r$ , $y \to 1$ replace karo; ODE collapse hokar algebra ban jaata hai.
Discriminant: $Δ = 5^{2} - 4 (1) (6) = 25 - 24 = 1 > 0$ . Yeh step kyun? $Δ$ ka sign case decide karta hai. $Δ > 0$ → do distinct real roots (Cell A/A2/B/C).
Roots: $(r + 2) (r + 3) = 0 \Rightarrow r = - 2, - 3$ . Yeh step kyun? Factoring se pata chalta hai kahan polynomial zero hai — woh exponents hamare do solutions hain.
Dono roots negative → dono $e^{- 2 x}, e^{- 3 x}$ shrink karte hain. Toh: $y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} .$ Yeh step kyun? Distinct real roots se independent exponentials milte hain (Case 1); linearly combine karo.

Verify: $y = e^{- 2 x}$ wapas daalo: $y^{''} = 4 e^{- 2 x}$ , $y^{'} = - 2 e^{- 2 x}$ , toh $4 - 10 + 6 = 0$ . ✓ Tumhara forecast decay hona chahiye tha — negative exponents se match karta hai.

Cell A2 — Distinct real roots, dono positive (pure growth)

Worked example Cell A2:

$y^{''} - 5 y^{'} + 6 y = 0$ Forecast: $y^{'}$ term ab negative hai lekin $c = 6 > 0$ . Kya dono roots same sign ke honge ya opposite? Growth ya decay?

Char. eq.: $r^{2} - 5 r + 6 = 0$ . Yeh step kyun? $y = e^{r x}$ substitute karo: har derivative $y^{(k)}$ become karta hai $r^{k}$ , toh ODE is polynomial mein $r$ mein convert ho jaata hai.
Discriminant: $Δ = (- 5)^{2} - 4 (1) (6) = 25 - 24 = 1 > 0$ . Yeh step kyun? $Δ > 0$ → do distinct real roots.
Roots: $(r - 2) (r - 3) = 0 \Rightarrow r = 2, 3$ . Yeh step kyun? Sum of roots $= - b / a = 5 > 0$ aur product $= c / a = 6 > 0$ force karte hain dono roots positive ko — yeh $b < 0, c > 0$ ka fingerprint hai.
$y = c_{1} e^{2 x} + c_{2} e^{3 x} .$ Yeh step kyun? Dono exponents positive → har term grows; yeh system koi oscillation ke bina bhaag jaata hai.

Verify: $y = e^{2 x}$ : $4 - 10 + 6 = 0$ ✓; $y = e^{3 x}$ : $9 - 15 + 6 = 0$ ✓. Forecast: pure growth — dono positive roots se match karta hai.

Cell B — Distinct real, opposite signs (ek term explode karti hai)

Cell C — Ek root exactly zero ( $c = 0$ )

Cell D — Repeated real root ( $Δ = 0$ )

Worked example Cell D:

$y^{''} + 4 y^{'} + 4 y = 0$ Forecast: $Δ = 16 - 16 = 0$ . Kitne exponentials milenge, aur "missing" doosre solution ke baare mein kya karna padega?

Char. eq.: $r^{2} + 4 r + 4 = (r + 2)^{2} = 0 \Rightarrow r = - 2$ (double). Yeh step kyun? $Δ = 0$ matlab dono roots merge hokar ek ho gayi hain.
$e^{- 2 x}$ sirf ek solution hai — $n = 2$ equation ko do independent solutions chahiye. Yeh step kyun? Second-order ODE ko hamesha do independent basic solutions chahiye hote hain (dekho Wronskian and Linear Independence of Solutions).
Doosra solution hai $x e^{- 2 x}$ . Yeh step kyun? (extra $x$ ka summary): Jab polynomial factor hoti hai $a (D - r_{0})^{2}$ ke roop mein, $D = \frac{d}{d x}$ ke saath, toh $(D - r_{0}) y = e^{r_{0} x}$ solve karne par tumhe ek constant integrate karna padta hai (exponential integrating factor se cancel ho jaata hai), aur ek constant integrate karne se $x$ mein linear term aata hai — isliye $x e^{r_{0} x}$ . Kyunki $x$ , $1$ ka constant multiple nahi hai, yeh doosra solution genuinely independent hai $e^{r_{0} x}$ se. (Full derivation: Second Order Linear ODE — Reduction of Order.) $y = (c_{1} + c_{2} x) e^{- 2 x} .$

Verify: $y = x e^{- 2 x}$ daalo. Phir $y^{'} = e^{- 2 x} - 2 x e^{- 2 x}$ , $y^{''} = - 4 e^{- 2 x} + 4 x e^{- 2 x}$ . Sum: $(- 4 + 4 x) + 4 (1 - 2 x) + 4 x = - 4 + 4 x + 4 - 8 x + 4 x = 0$ ✓.

Cell E — Pure imaginary roots ( $b = 0$ , undamped)

Worked example Cell E:

$y^{''} + 9 y = 0$ Forecast: koi $y^{'}$ term nahi matlab koi friction nahi. Growth, decay, ya forever-oscillation?

Char. eq.: $r^{2} + 9 = 0 \Rightarrow r^{2} = - 9 \Rightarrow r = \pm 3 i$ . Yeh step kyun? Koi bhi real number $- 9$ tak square nahi hota; roots purely imaginary hain ( $α = 0, β = 3$ ).
Euler $e^{i β x} = cos β x + i sin β x$ use karte hue (dekho Euler's Formula and Complex Exponentials), real solutions hain $cos 3 x, sin 3 x$ . Yeh step kyun? ODE real hai, toh hum $e^{3 i x}$ ke real aur imaginary parts lete hain real solutions paane ke liye.
$α = 0$ ke saath envelope $e^{0 \cdot x} = 1$ kabhi shrink nahi karti: $y = c_{1} cos 3 x + c_{2} sin 3 x .$

Neeche wali figure $c_{1} = 1, c_{2} = 0$ solution $y = cos 3 x$ plot karti hai. Do coral dashed lines ko dekho $y = \pm 1$ par: woh envelope hain, aur kyunki $α = 0$ hai, woh bilkul flat rehti hain — wave na grow karti hai na shrink. Lavender wiggles gino: kyunki $β = 3$ hai, exactly teen full oscillations fit hote hain $2 π$ lambai ke har interval mein. Yeh Cell E ka visual signature hai — steady, undying oscillation.

Verify: $y = cos 3 x$ : $y^{''} = - 9 cos 3 x$ , toh $- 9 cos 3 x + 9 cos 3 x = 0$ ✓. Pure oscillation — tumhara forecast forever-wiggling hona chahiye tha.

Cell F — Complex with decay (damped oscillation)

Worked example Cell F:

$y^{''} + 2 y^{'} + 10 y = 0$ , $y (0) = 2, y^{'} (0) = 0$ Forecast: thoda friction ( $b = 2$ ), stiff spring ( $c = 10$ ). Ek wiggle jo fade ho, ya bina wiggle ke sirf fade?

Char. eq.: $r^{2} + 2 r + 10 = 0$ . $Δ = 4 - 40 = - 36 < 0$ → complex roots. Yeh step kyun? $y = e^{r x}$ substitute karo taaki ODE is polynomial mein convert ho; phir $Δ < 0$ padhkar pata chalo ki roots complex hain.
Roots: $r = \frac{- 2 \pm - 36}{2} = - 1 \pm 3 i$ . Toh $α = - 1, β = 3$ . Yeh step kyun? $α < 0$ → decaying envelope; $β = 3$ → oscillation frequency.
General: $y = e^{- x} (c_{1} cos 3 x + c_{2} sin 3 x)$ .
$y (0) = 2$ apply karo: $x = 0$ par, $e^{0} = 1, cos 0 = 1, sin 0 = 0$ , toh $c_{1} = 2$ . Yeh step kyun? $x = 0$ par evaluate karne se $sin$ term kill ho jaata hai aur $c_{1}$ pin ho jaata hai.
$y^{'} (0) = 0$ apply karo: differentiate karo (product rule!): $y^{'} = e^{- x} [(- c_{1} + 3 c_{2}) cos 3 x + (- c_{2} - 3 c_{1}) sin 3 x] .$ $x = 0$ par: $- c_{1} + 3 c_{2} = 0 \Rightarrow c_{2} = \frac{c _{1}}{3} = \frac{2}{3}$ . Yeh step kyun? Substitute karne se pehle poore product ko differentiate karna zaroori hai (parent mein mistakes list dekho).
$y = e^{- x} (2 cos 3 x + \frac{2}{3} sin 3 x) .$

Verify: $x = 0$ par, $y = 2$ ✓ aur $y^{'} = (- 2 + 3 \cdot \frac{2}{3}) = 0$ ✓. Yeh ek Damped Harmonic Oscillator hai — ek fading wiggle.

Cell G — Complex with growth (negative damping)

Cell H — Higher order (ek cubic jo cases mix karta hai)

Worked example Cell H:

$y^{'''} - y^{''} + y^{'} - y = 0$ Forecast: degree-3 polynomial → teen roots. Kya ek equation ek decay aur ek wiggle dono combine kar sakti hai?

Char. eq.: $r^{3} - r^{2} + r - 1 = 0$ . Yeh step kyun? Same substitution $y = e^{r x}$ ; yahan $n = 3$ toh teesri derivative $r^{3}$ deti hai, ek cubic produce karke.
Grouping se factor karo: $r^{2} (r - 1) + (r - 1) = (r - 1) (r^{2} + 1) = 0$ . Yeh step kyun? Grouping se real root $r = 1$ aur complex pair expose hoti hai.
Roots: $r = 1$ (real), aur $r^{2} + 1 = 0 \Rightarrow r = \pm i$ ( $α = 0, β = 1$ ). Yeh step kyun? Ek real root → ek exponential; ek imaginary pair → $cos, sin$ .
$y = c_{1} e^{x} + c_{2} cos x + c_{3} sin x .$ Yeh step kyun? Superposition: $n$ -th order equation ko $n = 3$ independent pieces chahiye (dekho Linear Differential Operators and Superposition).

Verify: $y = cos x$ : $y^{'''} = sin x, y^{''} = - cos x, y^{'} = - sin x$ . Phir $sin x - (- cos x) + (- sin x) - cos x = sin x + cos x - sin x - cos x = 0$ ✓. Aur $y = e^{x}$ : $1 - 1 + 1 - 1 = 0$ ✓.

Cell I — Word problem (real-world reading)

Worked example Cell I: ek shock absorber

Ek car suspension satisfy karta hai $m y^{''} + c y^{'} + k y = 0$ mass $m = 1 kg$ , damping $c = 6 N\cdots/m$ , stiffness $k = 9 N/m$ ke saath. Yahan independent variable time $t$ (seconds mein) hai $x$ ki jagah — sirf variable ka naam badlata hai; derivatives $y^{''} = \frac{d ^{2} y}{d t ^{2}}$ , $y^{'} = \frac{d y}{d t}$ aur poora method identical hai. $y$ displacement (metres) hai. Motion describe karo. Forecast: kya yeh bounce-and-fade hai, ya slide-back-with-no-bounce?

Char. eq.: $r^{2} + 6 r + 9 = 0$ . Yeh step kyun? $m = 1$ se divide karo, phir $y = e^{r t}$ substitute karo: har time-derivative $y^{(k)}$ $r^{k}$ ban jaata hai, yeh polynomial $r$ mein deta hai.
Discriminant: $Δ = 36 - 36 = 0$ → repeated root (Cell D territory). Yeh step kyun? $Δ = 0$ critically damped boundary hai — koi oscillation ke bina sabse fast return.
Root: $r = - 3$ (double). $y = (c_{1} + c_{2} t) e^{- 3 t} .$
Reading: koi $cos / sin$ nahi → koi bouncing nahi. $e^{- 3 t}$ displacement ko zero ki taraf pull karta hai; $t$ se zyada se zyada ek baar overshoot ho sakta hai. Yeh exactly critical damping hai — engineers $c$ ko $Δ = 0$ hit karne ke liye tune karte hain.

Verify (units + math): $Δ = c^{2} - 4 mk = 36 - 4 (1) (9) = 0$ ✓ (units $N^{2} s^{2} / m^{2}$ dono terms par consistent). Root check $y = t e^{- 3 t}$ : Cell-D verification pattern se ODE satisfy karta hai.

Cell J — Exam twist (solution se ODE banao)

Worked example Cell J: reverse-engineer

Ek second-order constant-coefficient ODE ka general solution hai $y = e^{- 2 x} (c_{1} cos x + c_{2} sin x)$ . ODE find karo. Forecast: solution ki shape roots batati hai. $α$ aur $β$ kya hain?

Roots padhkar nikalo: envelope $e^{- 2 x}$ → $α = - 2$ ; $cos x, sin x$ ki frequency → $β = 1$ . Toh $r = - 2 \pm i$ . Yeh step kyun? Complex-case form $e^{α x} (c_{1} cos β x + c_{2} sin β x)$ directly $α, β$ encode karta hai.
Polynomial rebuild karo: $r = - 2 \pm i$ matlab factors $(r + 2 - i) (r + 2 + i)$ . $(r + 2)^{2} - i^{2} = (r + 2)^{2} + 1 = r^{2} + 4 r + 5.$ Yeh step kyun? Conjugate factors multiply karne se imaginary part clear ho jaata hai, real coefficients milte hain.
Char. eq. → ODE: $r^{2} + 4 r + 5 = 0$ correspond karta hai $y^{''} + 4 y^{'} + 5 y = 0.$ Yeh step kyun? Substitution $r^{k} \leftrightarrow y^{(k)}$ reverse karo.

Verify: aage solve karo: $Δ = 16 - 20 = - 4$ , $r = \frac{- 4 \pm 2 i}{2} = - 2 \pm i$ ✓ — diye gaye solution se match karta hai.

Recall Matrix ke across quick self-test

Root pair $- 1 \pm 4 i$ → kaun sa cell aur kya solution? ::: Cell F, $y = e^{- x} (c_{1} cos 4 x + c_{2} sin 4 x)$ (decaying oscillation). $a y^{''} + b y^{'} + cy = 0$ mein $c = 0$ kaun sa root force karta hai? ::: $r = 0$ , constant solution $c_{1}$ deta hai (Cell C). Damped system mein $Δ = 0$ ka physical matlab? ::: Critical damping — koi oscillation ke bina sabse fast return (Cell D/I). Solution $(c_{1} + c_{2} x) e^{5 x}$ kaun se discriminant sign se aaya? ::: $Δ = 0$ , repeated root $r = 5$ . Dono roots real aur positive ( $b < 0, c > 0$ ) → behaviour? ::: Pure growth, koi oscillation nahi (Cell A2).

Connections

Homogeneous with constant coefficients — characteristic equation
Linear Differential Operators and Superposition
Euler's Formula and Complex Exponentials
Wronskian and Linear Independence of Solutions
Method of Undetermined Coefficients
Damped Harmonic Oscillator
Second Order Linear ODE — Reduction of Order