Homogeneous with constant coefficients — characteristic equation: Exercises

4.6.10 · D4 · Maths › Ordinary Differential Equations › Homogeneous with constant coefficients — characteristic equa

Har case ke liye quick reference (hum yeh labels baar baar use karenge):

$Δ$ kyun sab kuch decide karta hai — picture. Characteristic polynomial $a r^{2} + b r + c$ variable $r$ mein ek parabola hai. Roots wahan hain jahan woh parabola horizontal axis ko cross karti hai. Agla figure teeno cases ek saath dikhata hai: parabola axis ko kitni baar touch karti hai, wahi $Δ$ disguise mein hai.

Figure — Homogeneous with constant coefficients — characteristic equation

Do crossings (left, red): parabola axis ke neeche jaati hai, toh woh use do jagah kaatti hai → do distinct real roots → do exponentials.
Ek touch (middle, red): parabola apne vertex par axis ko bas kiss karti hai → do roots merge ho gaye hain ek mein → $Δ = 0$ , aur yahi merging wajah hai ki hum ek solution kum pad rahe hain aur $x$ -factor $x e^{r x}$ dalna padta hai.
Koi crossing nahi (right, red): parabola axis ke bilkul upar floating hai → koi real root nahi. Roots real line chhod kar complex pair $α \pm i β$ ban gaye hain.

Aur kyun complex roots ka matlab waves hota hai? Agla figure complex root $α \pm i β$ ko ek point ke roop mein plane mein dikhata hai. $e^{i β x}$ se multiply karne par woh point ek circle ke around rotate karta hai jaise $x$ badhta hai — aur ek point ka shadow jo circle ke around jaata hai woh exactly $cos$ / $sin$ wave hoti hai. Vertical axis se doori, jo $α$ se control hoti hai, wahi circle ko spiral in ( $α < 0$ ) ya out ( $α > 0$ ) karaati hai.

Level 1 — Recognition

Yahan characteristic equation pehle se factored di gayi hai, ya trivially factor karna easy hai. Skill: roots padho aur sahi shape chuno.

Problem 1.1

Kisi ODE ki characteristic equation $(r - 4) (r + 1) = 0$ hai. General solution $y (x)$ likho.

Recall Solution 1.1

WHAT: roots nikalo. Har factor ko zero set karo: $r - 4 = 0 \Rightarrow r = 4$ aur $r + 1 = 0 \Rightarrow r = - 1$ . WHY shape: do alag real numbers → $Δ > 0$ → Case 1 (parabola figure ke left panel mein do crossings). $y = c_{1} e^{4 x} + c_{2} e^{- x} .$

Problem 1.2

Characteristic equation $(r - 7)^{2} = 0$ hai. General solution likho.

Recall Solution 1.2

WHAT: $(r - 7)^{2} = 0$ ka matlab $r = 7$ do baar counted — ek repeated root ( $Δ = 0$ , "just touches" middle panel). WHY extra $x$ : ek root se sirf ek independent solution $e^{7 x}$ milta hai; doosra independent solution $x e^{7 x}$ hai (prove kiya gaya hai Second Order Linear ODE — Reduction of Order mein). $y = (c_{1} + c_{2} x) e^{7 x} .$

Problem 1.3

Characteristic equation $r^{2} + 9 = 0$ hai. General solution likho.

Recall Solution 1.3

WHAT: $r^{2} = - 9$ , toh $r = \pm 3 i$ . $α \pm i β$ form mein $α = 0$ , $β = 3$ hai ("no crossing" right panel — parabola $r^{2} + 9$ axis tak kabhi nahi pahunchti). WHY sin aur cos: $α = 0$ ka matlab exponential envelope $e^{0 \cdot x} = 1$ (koi growth/decay nahi), aur $β = 3$ oscillation frequency hai. Euler $e^{\pm 3 i x}$ ko real waves mein convert karta hai. $y = c_{1} cos 3 x + c_{2} sin 3 x .$

Problem 1.4 — zero-root edge case

Characteristic equation $r (r - 5) = 0$ hai. General solution likho.

Recall Solution 1.4

WHAT: do roots hain $r = 0$ aur $r = 5$ . Root $r = 0$ easily miss ho jaata hai kyunki yeh $e^{0 \cdot x} = e^{0} = 1$ — ek plain constant banata hai. WHY ek constant count karta hai: $y = c_{1} e^{0 \cdot x} = c_{1}$ sach mein ODE solve karta hai (yeh ek ODE jaise $y^{''} - 5 y^{'} = 0$ se aata hai, jisme $y$ khud kabhi nahi hota, toh koi bhi constant survive karta hai). Yeh ek real, independent solution hai. $y = c_{1} e^{0 \cdot x} + c_{2} e^{5 x} = c_{1} + c_{2} e^{5 x} .$ Agar equation $r^{2} = 0$ hoti (repeated zero root), toh wahi $x$ -factor rule lagti hai: $y = (c_{1} + c_{2} x) e^{0 \cdot x} = c_{1} + c_{2} x$ — ek straight line.

Level 2 — Application

Ab tum khud characteristic equation form karo aur solve karo. Leading coefficient $a$ par dhyan rakho.

Problem 2.1

$y^{''} + 7 y^{'} + 10 y = 0$ solve karo.

Recall Solution 2.1

Char. eq. ( $y^{''} \to r^{2}$ , $y^{'} \to r$ , $y \to 1$ replace karo): $r^{2} + 7 r + 10 = 0$ . Factor karo: $(r + 2) (r + 5) = 0 \Rightarrow r = - 2, - 5$ . Dono real aur distinct hain. $y = c_{1} e^{- 2 x} + c_{2} e^{- 5 x} .$

Problem 2.2

$y^{''} - 4 y^{'} + 4 y = 0$ solve karo.

Recall Solution 2.2

Char. eq.: $r^{2} - 4 r + 4 = 0$ . Discriminant check karo: $Δ = (- 4)^{2} - 4 (1) (4) = 16 - 16 = 0$ → repeated root. $(r - 2)^{2} = 0 \Rightarrow r = 2$ (double). $y = (c_{1} + c_{2} x) e^{2 x} .$

Problem 2.3

$y^{''} + 6 y^{'} + 13 y = 0$ solve karo.

Recall Solution 2.3

Char. eq.: $r^{2} + 6 r + 13 = 0$ . Discriminant $Δ = 36 - 52 = - 16 < 0$ → complex. Quadratic formula: $r = \frac{- 6 \pm - 16}{2} = \frac{- 6 \pm 4 i}{2} = - 3 \pm 2 i$ . Toh $α = - 3$ (decay), $β = 2$ (frequency). $y = e^{- 3 x} (c_{1} cos 2 x + c_{2} sin 2 x) .$

Problem 2.4

$2 y^{''} - 3 y^{'} - 2 y = 0$ solve karo — note karo $a = 2$ , not $1$ .

Recall Solution 2.4

Char. eq.: $2 r^{2} - 3 r - 2 = 0$ . $2$ ko drop mat karo. Quadratic formula: $r = \frac{3 \pm 9 + 16}{4} = \frac{3 \pm 5}{4}$ , jo $r = 2$ aur $r = - \frac{1}{2}$ deta hai. $y = c_{1} e^{2 x} + c_{2} e^{- x /2} .$

Level 3 — Analysis (initial conditions + interpretation)

Problem 3.1

$y^{''} - y^{'} - 6 y = 0$ ko $y (0) = 1$ , $y^{'} (0) = 8$ ke saath solve karo.

Recall Solution 3.1

Roots: $r^{2} - r - 6 = (r - 3) (r + 2) = 0 \Rightarrow r = 3, - 2$ . General: $y = c_{1} e^{3 x} + c_{2} e^{- 2 x}$ . $y (0) = 1$ apply karo: $x = 0$ par, $e^{0} = 1$ , toh $c_{1} + c_{2} = 1$ . Differentiate karo: $y^{'} = 3 c_{1} e^{3 x} - 2 c_{2} e^{- 2 x}$ . $x = 0$ par: $3 c_{1} - 2 c_{2} = 8$ . Pair solve karo: pehle se, $c_{2} = 1 - c_{1}$ ; substitute karo: $3 c_{1} - 2 (1 - c_{1}) = 8 \Rightarrow 5 c_{1} = 10 \Rightarrow c_{1} = 2$ , isliye $c_{2} = - 1$ . $y = 2 e^{3 x} - e^{- 2 x} .$

Problem 3.2

$y^{''} + 2 y^{'} + y = 0$ ko $y (0) = 3$ , $y^{'} (0) = - 5$ ke saath solve karo.

Recall Solution 3.2

Roots: $r^{2} + 2 r + 1 = (r + 1)^{2} = 0 \Rightarrow r = - 1$ (double). General: $y = (c_{1} + c_{2} x) e^{- x}$ . $y (0) = 3$ apply karo: $x = 0$ deta hai $(c_{1} + 0) \cdot 1 = c_{1} = 3$ . Differentiate karo (product rule!): $y^{'} = c_{2} e^{- x} - (c_{1} + c_{2} x) e^{- x} = (c_{2} - c_{1} - c_{2} x) e^{- x}$ . $x = 0$ par: $c_{2} - c_{1} = - 5 \Rightarrow c_{2} = - 5 + 3 = - 2$ . $y = (3 - 2 x) e^{- x} .$

Problem 3.3

$y^{''} + 4 y^{'} + 5 y = 0$ ko $y (0) = 2$ , $y^{'} (0) = - 1$ ke saath solve karo, aur motion describe karo.

Recall Solution 3.3

Roots: $r^{2} + 4 r + 5 = 0$ , $Δ = 16 - 20 = - 4$ , toh $r = \frac{- 4 \pm 2 i}{2} = - 2 \pm i$ . Isliye $α = - 2, β = 1$ . General: $y = e^{- 2 x} (c_{1} cos x + c_{2} sin x)$ . $y (0) = 2$ apply karo: $e^{0} (c_{1} \cdot 1 + c_{2} \cdot 0) = c_{1} = 2$ . Differentiate karo (product rule): $y^{'} = - 2 e^{- 2 x} (c_{1} cos x + c_{2} sin x) + e^{- 2 x} (- c_{1} sin x + c_{2} cos x) .$ $x = 0$ par: $- 2 c_{1} + c_{2} = - 1 \Rightarrow c_{2} = - 1 + 2 c_{1} = - 1 + 4 = 3$ . $y = e^{- 2 x} (2 cos x + 3 sin x) .$ Interpretation: $α = - 2 < 0$ → amplitude envelope $e^{- 2 x}$ shrink hoti hai, toh yeh ek damped oscillation hai — ek wiggle jo zero tak fade ho jaati hai (dekho Damped Harmonic Oscillator). Neeche figure is exact solution ko (red mein) dikhata hai apne shrinking envelope $\pm 13 e^{- 2 x}$ (dashed black) ke andar trapped: labelled axes $x$ (horizontal) aur $y (x)$ (vertical) hain, aur arrows oscillation aur decaying envelope ko point out karte hain.

Level 4 — Synthesis (higher order, mixed roots)

Sab kuch scale up hota hai: ek $n$ -th order equation degree- $n$ characteristic polynomial deta hai aur $== n ==$ independent solutions chahiye hote hain (from Wronskian and Linear Independence of Solutions).

Problem 4.1

Third-order ODE $y^{'''} - 6 y^{''} + 11 y^{'} - 6 y = 0$ solve karo.

Recall Solution 4.1

Char. eq.: $r^{3} - 6 r^{2} + 11 r - 6 = 0$ . Chhote integer roots try karo (constant $6$ ke factors): $r = 1$ deta hai $1 - 6 + 11 - 6 = 0$ ✓. Toh factor theorem se $(r - 1)$ cubic ko exactly divide karta hai. Ise divide kaise karein — synthetic division. Neeche figure $(r - 1)$ se divide karne ke liye synthetic-division table dikhata hai: leading $1$ neeche laao, root $1$ se multiply karo, agle coefficient mein add karo, repeat karo. Bottom row quotient coefficients $1, - 5, 6$ aur final remainder $0$ deta hai (jo confirm karta hai ki $r = 1$ sach mein ek root hai):

Toh $r^{3} - 6 r^{2} + 11 r - 6 = (r - 1) (r^{2} - 5 r + 6) = (r - 1) (r - 2) (r - 3)$ . Roots $r = 1, 2, 3$ — teen distinct real roots, teen exponentials. $y = c_{1} e^{x} + c_{2} e^{2 x} + c_{3} e^{3 x} .$

Problem 4.2

$y^{'''} - 3 y^{''} + 3 y^{'} - y = 0$ solve karo.

Recall Solution 4.2

Char. eq.: $r^{3} - 3 r^{2} + 3 r - 1 = 0$ . Binomial pehchano: yeh $(r - 1)^{3} = 0$ hai, toh $r = 1$ multiplicity three ke saath. WHY teen terms $1, x, x^{2}$ : ek root jo $m$ baar repeated ho woh $e^{r x}$ ko $1, x, \dots, x^{m - 1}$ se multiply karke contribute karta hai — wahi "sneak in an $x$ " idea, extended. $y = (c_{1} + c_{2} x + c_{3} x^{2}) e^{x} .$

Problem 4.3

Fourth-order ODE $y^{(4)} + 8 y^{''} + 16 y = 0$ solve karo.

Recall Solution 4.3

Char. eq.: $r^{4} + 8 r^{2} + 16 = 0$ . Maano $u = r^{2}$ : $u^{2} + 8 u + 16 = (u + 4)^{2} = 0 \Rightarrow u = - 4$ (double). Toh $r^{2} = - 4 \Rightarrow r = \pm 2 i$ , aur kyunki $u = - 4$ ek double root tha, pair $\pm 2 i$ repeated hai. Yahan $α = 0, β = 2$ . WHY trig par bhi extra $x$ : ek repeated complex pair $cos, sin$ mein se har ek ko $1$ aur $x$ se multiply karta hai — bilkul wahi repeated-root rule oscillations par apply kiya gaya. $y = (c_{1} + c_{2} x) cos 2 x + (c_{3} + c_{4} x) sin 2 x .$

Problem 4.4 — doosron ke beech repeated zero root

$y^{(4)} - y^{'''} = 0$ solve karo.

Recall Solution 4.4

Char. eq.: $r^{4} - r^{3} = 0 \Rightarrow r^{3} (r - 1) = 0$ . Roots: $r = 0$ multiplicity three ke saath aur $r = 1$ ek baar. WHY $x$ ke powers: triple root $r = 0$ deta hai $e^{0 \cdot x}$ times $1, x, x^{2}$ — yaani bas $1, x, x^{2}$ — aur $r = 1$ deta hai $e^{x}$ . $y = c_{1} + c_{2} x + c_{3} x^{2} + c_{4} e^{x} .$

Level 5 — Mastery (banao / prove karo)

Problem 5.1 (reverse engineering)

Ek second-order homogeneous ODE with constant coefficients dhundho jiska general solution $y = e^{- x} (c_{1} cos 3 x + c_{2} sin 3 x)$ ho.

Recall Solution 5.1

Solution hamare ko kya batata hai: $α = - 1, β = 3$ , toh roots $r = - 1 \pm 3 i$ hain. Apne roots se polynomial banao: $(r - (- 1 + 3 i)) (r - (- 1 - 3 i)) = (r + 1 - 3 i) (r + 1 + 3 i) = (r + 1)^{2} - (3 i)^{2} = (r + 1)^{2} + 9.$ Expand karo: $(r + 1)^{2} + 9 = r^{2} + 2 r + 1 + 9 = r^{2} + 2 r + 10$ . ODE padho ( $r^{2}, r, 1$ ke coefficients $y^{''}, y^{'}, y$ ban jaate hain): $y^{''} + 2 y^{'} + 10 y = 0.$ Check: $Δ = 4 - 40 = - 36 < 0$ ✓ complex, aur $r = \frac{- 2 \pm 6 i}{2} = - 1 \pm 3 i$ ✓.

Problem 5.2 (mixed spectrum ke saath banao)

Woh lowest-order constant-coefficient ODE construct karo jiske solutions mein $x e^{2 x}$ aur $cos x$ shamil hon.

Recall Solution 5.2

Har solution ko required roots mein decode karo:

$x e^{2 x}$ appear karna matlab $r = 2$ ek double root hai → factor $(r - 2)^{2}$ .
$cos x$ (yaani $e^{0 \cdot x} cos 1 x$ ) matlab pair $r = \pm i$ → factor $(r^{2} + 1)$ . Lowest degree = exactly inhe multiply karo: $(r - 2)^{2} (r^{2} + 1)$ , degree-4 polynomial → 4th order ODE. Expand karo: $(r - 2)^{2} = r^{2} - 4 r + 4$ , phir $(r^{2} - 4 r + 4) (r^{2} + 1) = r^{4} - 4 r^{3} + 4 r^{2} + r^{2} - 4 r + 4 = r^{4} - 4 r^{3} + 5 r^{2} - 4 r + 4.$ ODE: $y^{(4)} - 4 y^{'''} + 5 y^{''} - 4 y^{'} + 4 y = 0.$

Problem 5.3 (reduction fact prove karo)

$y^{''} - 2 r y^{'} + r^{2} y = 0$ ke liye (deliberate double root $r$ ), directly verify karo ki $y = x e^{r x}$ ek solution hai.

Recall Solution 5.3

Derivatives compute karo (product rule): $y = x e^{r x}$ , $y^{'} = e^{r x} + r x e^{r x} = (1 + r x) e^{r x}$ , $y^{''} = r e^{r x} + (1 + r x) r e^{r x} = (2 r + r^{2} x) e^{r x}$ . $y^{''} - 2 r y^{'} + r^{2} y$ mein substitute karo: $(2 r + r^{2} x) e^{r x} - 2 r (1 + r x) e^{r x} + r^{2} x e^{r x} .$ Bracket collect karo: $2 r + r^{2} x - 2 r - 2 r^{2} x + r^{2} x = (2 r - 2 r) + (r^{2} x - 2 r^{2} x + r^{2} x) = 0$ . Poora $0 \cdot e^{r x} = 0$ ✓. Toh $x e^{r x}$ sach mein double-root ODE solve karta hai — yahi wajah hai ki $x$ -factor koi trick nahi balki ek theorem hai.

Connections

Homogeneous with constant coefficients — characteristic equation (parent)
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