Systems of first-order linear ODEs — matrix method: Exercises

4.6.21 · D4 · Maths › Ordinary Differential Equations › Systems of first-order linear ODEs — matrix method

Intuition Is page ko kaise use karein

Yeh problems ek ladder ki tarah chadhte hain. L1 sirf machine ko pehchanna kehta hai. L5 kehta hai kuch naya banana. Pehle har ek ko pen aur paper se try karo, PHIR collapsible [!recall]- solution kholo. Har worked step mein KIYA KYA aur KYUN dono explain kiya gaya hai. Saare tools parent note Systems of First-Order Linear ODEs — matrix method se aate hain; jab koi naya idea aata hai, hum use yahan rebuild karte hain.

Shuru karne se pehle, poore game ki ek reminder taaki koi bhi symbol unknown na rahe:

$x^{'} = A x$ ka matlab hai "vector $x$ ki rate of change, matrix $A$ ko $x$ se multiply karne ke barabar hai." Ek vector bas kuch numbers ki stack ki hui list hota hai (ek point, ya origin se ek arrow).
$A v = λ v$ kehta hai "arrow $v$ sirf stretch hota hai (number $λ$ se) jab $A$ usse act karta hai — uski direction unchanged rehti hai." Aisa arrow eigenvector kehlata hai; stretch factor $λ$ uski eigenvalue hai. Dekho Eigenvalues and Eigenvectors.
$det (A - λ I) = 0$ characteristic equation hai — woh filter jo un special $λ$ ko dhundta hai.

Level 1 — Recognition

Exercise 1.1

In vector functions mein se kaun sa $x^{'} = A x$ ka solution hai kisi matrix $A$ ke liye, sirf form se (yaani kya yeh eigenvector template se bana hai)? (a) $x (t) = e^{2 t} (3 1)$ (b) $x (t) = (e ^{2 t} + t e ^{2 t})$ (c) $x (t) = e^{- t} (- 5 2)$

Recall Solution

Template kya hai. Parent note ne derive kiya tha ki $x = e^{λ t} v$ system ko exactly solve karta hai tabhi jab $(λ, v)$ ek eigenpair ho. Toh ek "clean" single-mode solution mein ek fixed vector hona chahiye jo ek scalar exponential se multiply ho.

(a) $e^{2 t} (3 1)$ : fixed direction $(3 1)$ , $e^{2 t}$ se scale kiya. ✅ Template se match karta hai $λ = 2$ , $v = (3 1)$ ke saath.
(c) $e^{- t} (- 5 2)$ : same shape, $λ = - 1$ , $v = (- 5 2)$ . ✅
(b) mein ek component ke andar $+ t$ hai jo ek fixed eigenvector ko uniformly multiply nahi kar raha — dono components genuinely alag rates se grow karte hain, toh yeh ek single exponential mode nahi hai. Yeh ek defective solution $e^{λ t} (t v + w)$ ka piece ho sakta hai, lekin jaisa likha hai yeh plain eigenvector template nahi hai. ❌ "single-mode" ke liye.

Jawab: (a) aur (c).

Exercise 1.2

$A = (50 0 - 2)$ ke liye, eigenvalues aur eigenvectors by inspection likho, aur general solution bhi.

Recall Solution

By inspection kyun. $A$ diagonal hai: yeh $x_{1}$ aur $x_{2}$ ko bilkul mix nahi karta. Dono equations already decoupled hain: $x_{1}^{'} = 5 x_{1}$ , $x_{2}^{'} = - 2 x_{2}$ . Dekho Diagonalization.

Standard basis arrows $(0 1)$ aur $(1 0)$ direction mein unmoved hain — sirf stretch hote hain: $A (0 1) = 5 (0 1), A (1 0) = - 2 (1 0) .$ Toh eigenpairs hain $(5, (0 1))$ aur $(- 2, (1 0))$ . $x (t) = c_{1} e^{5 t} (0 1) + c_{2} e^{- 2 t} (1 0) .$

Level 2 — Application

Exercise 2.1

$x^{'} = A x$ , $A = (2112)$ , $x (0) = (0 4)$ solve karo.

Recall Solution

Step 1 — characteristic equation. Kyun: ek nonzero eigenvector tabhi exist karta hai jab $A - λ I$ singular ho, yaani $det = 0$ . $det (2 - λ 1 1 2 - λ) = (2 - λ)^{2} - 1 = 0.$ $(2 - λ)^{2} = 1 \Rightarrow 2 - λ = \pm 1 \Rightarrow λ = 1, 3.$

Step 2 — eigenvectors. Kyun: ab singular matrix ki null direction solve karo.

$λ = 3$ : $(- 1 1 1 - 1) v = 0 \Rightarrow - v_{1} + v_{2} = 0 \Rightarrow v = (1 1) .$
$λ = 1$ : $(1111) v = 0 \Rightarrow v_{1} + v_{2} = 0 \Rightarrow v = (- 1 1) .$

Step 3 — general solution. $x = c_{1} e^{3 t} (1 1) + c_{2} e^{t} (- 1 1) .$

Step 4 — IC apply karo $x (0) = (0 4)$ : $c_{1} (1 1) + c_{2} (- 1 1) = (0 4)$ . $c_{1} + c_{2} = 4$ , $c_{1} - c_{2} = 0 \Rightarrow c_{1} = c_{2} = 2.$ $x (t) = 2 e^{3 t} (1 1) + 2 e^{t} (- 1 1) .$

Exercise 2.2

$x^{'} = A x$ , $A = (- 3 0 4 - 1)$ solve karo (sirf general solution).

Recall Solution

Step 1. $A$ upper-triangular hai, toh uski eigenvalues diagonal entries hain: $det (A - λ I) = (- 3 - λ) (- 1 - λ) = 0 \Rightarrow λ = - 3, - 1$ padh lo.

Step 2.

$λ = - 3$ : $(0042) v = 0 \Rightarrow 4 v_{2} = 0 \Rightarrow v_{2} = 0 \Rightarrow v = (0 1) .$
$λ = - 1$ : $(- 2 0 40) v = 0 \Rightarrow - 2 v_{1} + 4 v_{2} = 0 \Rightarrow v_{1} = 2 v_{2} \Rightarrow v = (1 2) .$

Step 3. $x = c_{1} e^{- 3 t} (0 1) + c_{2} e^{- t} (1 2) .$ Dono exponents negative hain → har trajectory origin par decay karti hai (ek stable node), dekho Phase Portraits and Stability.

Level 3 — Analysis

Exercise 3.1

$A = (12 - 2 1)$ . Real solutions dhundho aur trajectory geometry describe karo.

Recall Solution

Step 1. $det (A - λ I) = (1 - λ)^{2} + 4 = 0 \Rightarrow (1 - λ)^{2} = - 4 \Rightarrow 1 - λ = \pm 2 i \Rightarrow λ = 1 \pm 2 i .$ Toh $α = 1$ (real part), $β = 2$ (imaginary part). $α, β$ ki parwah kyun? Euler's Formula ke through, $e^{(α + i β) t} = e^{α t} (cos β t + i sin β t)$ : $α$ growth/decay set karta hai, $β$ rotation speed set karta hai.

Step 2 — $λ = 1 + 2 i$ ke liye eigenvector. $(A - λ I) = (- 2 i 2 - 2 - 2 i)$ . Top row: $- 2 i v_{1} - 2 v_{2} = 0 \Rightarrow v_{2} = - i v_{1} .$ $v_{1} = 1$ lo: $v = (- i 1)$ . Real aur imaginary parts mein split karo: $v = (0 1) + i (- 1 0)$ , toh $p = (0 1)$ , $q = (- 1 0)$ .

Step 3 — parent formula use karke do real solutions assemble karo $α = 1, β = 2$ ke saath: $x_{1} = e^{t} (p cos 2 t - q sin 2 t) = e^{t} (s i n 2 t c o s 2 t),$ $x_{2} = e^{t} (p sin 2 t + q cos 2 t) = e^{t} (- c o s 2 t s i n 2 t) .$ $x = e^{t} [c_{1} (sin 2 t cos 2 t) + c_{2} (- cos 2 t sin 2 t)] .$

Geometry. $α = 1 > 0$ ka matlab hai radius badhta hai: ek outward spiral (unstable spiral / focus). Rotation period $2 π / β = π$ hai. Neeche figure dekho.

Figure — Systems of first-order linear ODEs — matrix method

Exercise 3.2

$A = (0 - 4 10)$ ke liye, origin par equilibrium ka type bina poora solution likhe decide karo, phir $λ$ dhundh kar confirm karo.

Recall Solution

Pehle reasoning. $A$ ka trace (diagonal ka sum) $0 + 0 = 0$ hai; determinant $(0) (0) - (1) (- 4) = 4 > 0$ hai. Ek $2 \times 2$ system ke liye, eigenvalues $λ^{2} - (tr) λ + det = 0$ satisfy karte hain, yaani $λ^{2} + 4 = 0$ . Zero trace + positive determinant ⇒ purely imaginary eigenvalues ⇒ ek center (closed orbits, na growing na decaying).

Confirm karo. $λ^{2} + 4 = 0 \Rightarrow λ = \pm 2 i$ . Yahan $α = 0$ : koi growth nahi, toh trajectories origin ke around closed ellipses hain. ✅

Level 4 — Synthesis

Exercise 4.1

Defective system $x^{'} = A x$ , $A = (4 - 1 12)$ , $x (0) = (0 1)$ solve karo.

Recall Solution

Step 1. $det (A - λ I) = (4 - λ) (2 - λ) - (1) (- 1) = (4 - λ) (2 - λ) + 1.$ Expand karo: $λ^{2} - 6 λ + 8 + 1 = λ^{2} - 6 λ + 9 = (λ - 3)^{2} = 0 \Rightarrow λ = 3$ (double).

Step 2 — eigenvector. $(A - 3 I) = (1 - 1 1 - 1)$ . Row: $v_{1} + v_{2} = 0 \Rightarrow v = (- 1 1)$ . Sirf ek independent eigenvector → defective (parent note dekho).

Step 3 — generalized eigenvector $w$ with $(A - 3 I) w = v$ : $(1 - 1 1 - 1) (w _{2} w _{1}) = (- 1 1) \Rightarrow w_{1} + w_{2} = 1.$ $w_{1} = 1, w_{2} = 0$ lo: $w = (0 1)$ . Yeh freedom kyun? $v$ ka koi bhi multiple jod kar milne wala koi bhi $w$ kaam karta hai; yeh sirf $c_{1}$ ko re-label karta hai.

Step 4 — do independent solutions. $x_{1} = e^{3 t} (- 1 1), x_{2} = e^{3 t} (t (- 1 1) + (0 1)) .$ $x = c_{1} e^{3 t} (- 1 1) + c_{2} e^{3 t} (t (- 1 1) + (0 1)) .$

Step 5 — IC apply karo $x (0) = (0 1)$ . $t = 0$ par $t$ -term vanish ho jaata hai: $c_{1} (- 1 1) + c_{2} (0 1) = (0 1) .$ Bottom component: $- c_{1} = 0 \Rightarrow c_{1} = 0$ . Top: $c_{1} + c_{2} = 1 \Rightarrow c_{2} = 1.$ $x (t) = e^{3 t} (t (- 1 1) + (0 1)) = e^{3 t} (- t t + 1) .$

Exercise 4.2

Ek $2 \times 2$ real matrix $A$ banao jiska system $x^{'} = A x$ ka general solution ho $x = c_{1} e^{- t} (1 1) + c_{2} e^{2 t} (- 1 1) .$

Recall Solution

KIYA HUM reverse-engineer karte hain. Humein eigenpairs bataye gaye hain: $(- 1, (1 1))$ aur $(2, (- 1 1))$ . Koi bhi $A$ jo $A v_{i} = λ_{i} v_{i}$ ke taur par act kare kaam karega, toh hum Diagonalization use karte hain: $A = P D P^{- 1}$ jahan $P$ mein columns eigenvectors hain aur $D$ eigenvalues ki diagonal matrix hai. $P = (11 1 - 1), D = (- 1 0 02) .$ $det P = - 2$ , toh $P^{- 1} = \frac{1}{- 2} (- 1 - 1 - 1 1) = (\frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2}) .$

$A = P D P^{- 1}$ compute karo: $D P^{- 1} = (- 1 0 02) (\frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2}) = (- \frac{1}{2} 1 - \frac{1}{2} - 1) .$ $A = P (D P^{- 1}) = (11 1 - 1) (- \frac{1}{2} 1 - \frac{1}{2} - 1) = (\frac{1}{2} - \frac{3}{2} - \frac{3}{2} \frac{1}{2}) .$ $A = (\frac{1}{2} - \frac{3}{2} - \frac{3}{2} \frac{1}{2}) .$ Check karo: $tr A = 1 = λ_{1} + λ_{2} = (- 1) + 2$ ✅ aur $det A = \frac{1}{4} - \frac{9}{4} = - 2 = λ_{1} λ_{2}$ ✅.

Level 5 — Mastery

Exercise 5.1

Ek system mein $A = (0 - 1 1 - 2)$ hai. Equilibrium ko fully classify karo (type + stability), ise solve karo, aur answer ko second-order ODE $y^{''} + 2 y^{'} + y = 0$ se connect karo.

Recall Solution

Step 1 — connection. Maano $x_{1} = y$ , $x_{2} = y^{'}$ . Tab $x_{1}^{'} = x_{2}$ aur $x_{2}^{'} = y^{''} = - 2 y^{'} - y = - x_{1} - 2 x_{2}$ , jo exactly $x^{'} = A x$ deta hai is $A$ ke saath. Yeh standard trick hai jo kisi bhi second-order ODE ko first-order system mein convert karti hai.

Step 2 — eigenvalues. $tr A = - 2$ , $det A = (0) (- 2) - (1) (- 1) = 1$ . Characteristic: $λ^{2} + 2 λ + 1 = (λ + 1)^{2} = 0 \Rightarrow λ = - 1$ (double). Note karo yeh $y^{''} + 2 y^{'} + y = 0$ ka wahi characteristic polynomial hai.

Step 3 — defective? $(A + I) = (1 - 1 1 - 1)$ , rank 1, toh sirf ek eigenvector $v$ : $v_{1} + v_{2} = 0 \Rightarrow v = (- 1 1)$ . Defective.

Step 4 — generalized eigenvector. $(A + I) w = v$ : $w_{1} + w_{2} = 1$ . $w = (0 1)$ lo. $x = c_{1} e^{- t} (- 1 1) + c_{2} e^{- t} (t (- 1 1) + (0 1)) .$

Step 5 — classification. Dono modes $e^{- t}$ carry karte hain (ek polynomial $t$ factor ke saath): $λ < 0$ , repeated, ek eigenvector ⇒ improper (degenerate) stable node. Sab kuch origin par decay karta hai. Yeh $y = (c_{1} + c_{2} t) e^{- t}$ se match karta hai, second-order ODE ka critically-damped solution — same physics, do languages.

Exercise 5.2

Matrix exponential use karke dikhao ki $x (t) = e^{A t} x_{0}$ , $x^{'} = A x$ , $x (0) = x_{0}$ ko solve karta hai, aur $A = (20 0 - 1)$ ke liye $e^{A t}$ evaluate karo.

Recall Solution

Step 1 — $e^{A t}$ ka matlab. Scalar series $e^{a t} = \sum_{k \geq 0} \frac{( a t ) ^{k}}{k !}$ ke analogy se, hum Matrix Exponential define karte hain $e^{A t} = \sum_{k = 0}^{\infty} \frac{( A t ) ^{k}}{k !} = I + A t + \frac{A ^{2} t ^{2}}{2 !} + \dots .$ Yeh har square $A$ ke liye converge karta hai.

Step 2 — yeh system kyun solve karta hai. Term by term differentiate karo (allowed kyunki series nicely converge karti hai): $\frac{d}{d t} e^{A t} = A + A^{2} t + \frac{A ^{3} t ^{2}}{2 !} + \dots = A (I + A t + \frac{A ^{2} t ^{2}}{2 !} + \dots) = A e^{A t} .$ Toh $x = e^{A t} x_{0}$ se $x^{'} = A e^{A t} x_{0} = A x$ milta hai. ✅ Aur $t = 0$ par, $e^{A \cdot 0} = I$ , toh $x (0) = x_{0}$ . ✅

Step 3 — diagonal $A$ ke liye evaluate karo. Ek diagonal matrix ke liye, powers entrywise act karte hain: $A^{k} = (2^{k} 0 0 (- 1)^{k})$ . Har entry ko apni scalar exponential series ki tarah sum karo: $e^{A t} = (\sum_{k} \frac{( 2 t ) ^{k}}{k !} 0 0 \sum_{k} \frac{( - t ) ^{k}}{k !}) = (e^{2 t} 0 0 e^{- t}) .$ Tab $x (t) = e^{A t} x_{0} = (x _{0, 2} e ^{- t} x _{0, 1} e ^{2 t})$ — exactly wahi jo decoupled equations $x_{1}^{'} = 2 x_{1}, x_{2}^{'} = - x_{2}$ predict karti hain.

Recall Ladder ka ek-line summary

Template pehchano ::: $x = e^{λ t} v$ ko $A v = λ v$ chahiye Apply karo ::: real distinct $λ$ → exponentials superpose karo Analyze karo ::: complex $λ = α \pm i β$ → spiral, stability $α$ ke sign se Synthesize karo ::: defective → generalized eigenvector $e^{λ t} (t v + w)$ ; $A = P D P^{- 1}$ se reverse-engineer karo Master karo ::: 2nd-order ODEs systems hain; $e^{A t}$ saare solutions ek saath package karta hai