Systems of first-order linear ODEs — matrix method: Question bank

4.6.21 · D5 · Maths › Ordinary Differential Equations › Systems of first-order linear ODEs — matrix method

Traps se pehle, notation ko ek baar pin down karte hain taaki kuch bhi undefined na rahe:

Baaki vocabulary ke reminders jo tumhe precisely use karne hain:

eigenpair $(λ, v)$ : ek number $λ$ aur ek nonzero vector $v$ jisme $A v = λ v$ — dekho Eigenvalues and Eigenvectors.
algebraic multiplicity: kitni baar $λ$ , $det (A - λ I) = 0$ ka root hai.
geometric multiplicity: $λ$ ke paas actually kitne independent eigenvectors hain.
defective: geometric $<$ algebraic multiplicity (bahut kam eigenvectors) — dekho Diagonalization.

Agla figure trajectory shapes ke liye tumhara visual dictionary hai jiska reference puri jagah hua hai — ise apne paas rakho.

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

True or false — justify karo

True or false: ek system

x^{'} = A x

mein hamesha exactly

n

eigenvalues hote hain, toh hamesha

n

independent solutions hoti hain.

False. Uske paas multiplicity ke saath

n

eigenvalues hote hain, lekin independent solutions independent eigenvectors se aati hain (plus generalized eigenvectors). Ek defective matrix mein kam eigenvectors hote hain, toh tumhe missing modes ko generalized eigenvectors se

(A - λ I) w = v

solve karke banana padta hai.

True or false: agar

A

ke har eigenvalue ka negative real part hai, toh har solution

t \to \infty

par

0

tak decay karti hai.

True. Har mode

e^{λ t}

carry karta hai, aur

∣ e^{λ t} ∣ = e^{(Re λ) t}

; agar har real part negative hai, toh har mode shrink karta hai, toh origin asymptotically stable hai — dekho Phase Portraits and Stability.

True or false: ek eigenvector ko

- 5

se multiply karne par system ka ek alag solution milta hai.

False, "new information" ke sense mein.

- 5 v

abhi bhi same

λ

ke liye ek eigenvector hai; solution

c e^{λ t} v

sirf

- 5

ko constant

c

mein absorb kar leta hai. Yeh same solution line span karta hai.

True or false: do distinct eigenvalues hamesha do linearly independent solutions dete hain.

True. Maano

c_{1} v_{1} + c_{2} v_{2} = 0

;

A

apply karo toh

c_{1} λ_{1} v_{1} + c_{2} λ_{2} v_{2} = 0

milta hai, pehle se

λ_{1}

times subtract karo toh

c_{2} (λ_{2} - λ_{1}) v_{2} = 0

milta hai. Kyunki

λ_{2} \neq = λ_{1}

aur

v_{2} \neq = 0

c_{2} = 0

, phir

c_{1} = 0

— toh eigenvectors, aur isliye dono modes, independent hain.

True or false: ek

2 \times 2

real matrix jiske complex eigenvalues hain, uske real eigenvectors hote hain.

False. Complex eigenvalues complex eigenvectors

v = p + i q

force karte hain; real solutions phir real part

p

aur imaginary part

q

se Euler's Formula ke zariye rebuild ki jaati hain.

True or false: general solution

x = c_{1} e^{λ_{1} t} + c_{2} e^{λ_{2} t}

correct hai jab tak tumne sahi

λ

's nikale.

False. Woh expression scalar hai — har term ko apna eigenvector carry karna chahiye:

x = c_{1} e^{λ_{1} t} v_{1} + c_{2} e^{λ_{2} t} v_{2}

v_{i}

ke bina koi direction nahi hai aur answer vector nahi hai.

True or false: agar

λ = 0

ek eigenvalue hai, toh woh mode ek aisa solution contribute karta hai jo ruka rehta hai.

True. Mode hai

e^{0 \cdot t} v = v

, ek constant vector. Toh

v

ke along equilibria ki ek poori line hai; uspe kuch nahi halta.

True or false:

det (A - λ I) = 0

ki jagah

det (λ I - A) = 0

use karne par eigenvalues badal jaate hain.

False. Dono polynomials sirf

(- 1)^{n}

factor se different hain, toh unke roots identical hain. Bas consistent raho taaki equation clearly padhe.

Error dhundho

Ek student likhta hai: "

A = (3013)

ka eigenvalue

3

do baar hai, toh solution hai

c_{1} e^{3 t} v_{1} + c_{2} e^{3 t} v_{2}

do eigenvectors ke saath."

Error: yeh matrix defective hai — iske paas sirf ek eigenvector

(0 1)

hai. Doosra mode ek generalized eigenvector

w

use karna padega jisme

(A - 3 I) w = v

, jo

e^{3 t} (t v + w)

deta hai ek genuine

t

factor ke saath.

Ek student complex

λ = α \pm i β

solve karta hai aur chaar solutions rakhta hai: do

α + i β

se aur do

α - i β

se.

Error: double-counting. Ek conjugate pair exactly do independent real solutions deta hai (ek single branch ke real aur imaginary parts). Conjugate branch kuch naya nahi add karta.

Ek student conclude karta hai: "

λ = 2 i

toh trajectories grow karti hain, kyunki

i

cheezein blow up karti hai."

Error: growth/decay real part

α = Re (λ)

se control hoti hai, yahan

0

hai. Imaginary part sirf rotation frequency set karta hai, toh trajectories closed circles hain (ek center), blow-up nahi.

Ek student

λ e^{λ t} v = e^{λ t} A v

e^{λ t}

cancel karta hai lekin worry karta hai ki kisi

t

ke liye yeh zero ho sakta hai.

Koi error nahi, aur worry bekar hai:

e^{λ t} \neq = 0

har real (aur complex)

t

ke liye, toh ise divide out karna hamesha legal hai. Yahi reason hai ki exponential ansatz algebraic eigenvalue equation par collapse karta hai.

Ek student generalized-eigenvector solution ko

e^{3 t} (t w + v)

likhta hai.

Error: roles swap ho gaye hain. Correct form hai

e^{λ t} (t v + w)

— eigenvector

v

t

multiply karta hai, kyunki

v

hi hai jise

(A - λ I)

annihilate karta hai.

Ek student kehta hai "origin hamesha ek solution hai, toh yeh stable hai."

Error:

x \equiv 0

hamesha ek equilibrium hai, lekin stability eigenvalues par depend karti hai. Positive real part nearby trajectories ko door bhejta hai — origin tab unstable hota hai.

Why questions

Hum

x = e^{λ t} v

guess kyun karte hain kisi aur function ki jagah?

Kyunki ek fixed direction

v

mein equation ko ek scalar equation

u^{'} = λ u

par reduce karna padta hai, jiska ek hi solution exponential hai. Exponential woh ek function hai jo differentiation ke under khud ko reproduce karta hai.

Repeated eigenvalue kabhi kabhi abhi bhi do eigenvectors kyun deta hai aur generalized ones ki zaroorat nahi padti?

Kyunki algebraic multiplicity aur geometric multiplicity equal ho sakti hain — jaise

A = 3 I

λ = 3

do baar hai lekin poora 2-dimensional eigenspace hai. Sirf jab eigenvectors kam padte hain (

A

defective hai) tabhi

t v + w

trick chahiye hoti hai.

Defective case mein

t

factor kyun appear karta hai?

e^{λ t} (t v + w)

substitute karo aur

(A - λ I) w = v

use karo toh

x^{'} = A x

ke dono sides balance ho jaate hain.

t

woh extra degree of freedom provide karta hai jo missing eigenvector nahi de paya tha.

Complex eigenvalue

α \pm i β

spirals kyun produce karta hai, straight-line motion kyun nahi?

Kyunki

e^{λ t} = e^{α t} (cos β t \pm i sin β t)

Euler's Formula se:

α

radius scale karta hai (grow/decay) jabki

β

rotate karta hai. Straight-line motion ke liye real eigenvector chahiye, jo complex

λ

ke paas nahi hota.

Koi bhi solution eigenvector solutions ke superposition ke roop mein kyun likhi ja sakti hai?

Kyunki system linear hai, toh solutions ka sum bhi solution hai, aur

n

independent eigenvector modes ek

n

-parameter family span karte hain — jo

n

free constants match karta hai jo ek first-order

n

-dimensional system ke paas hone chahiye.

Long-run behaviour decide karne mein akela real part ka sign kyun kaafi hai, eigenvectors ko ignore karke?

Eigenvectors sirf modes ki directions set karte hain; factor

e^{(Re λ) t}

decide karta hai ki har mode ka amplitude grow hoga ya shrink. Toh stability seedha

Re (λ)

se padhi jaati hai — dekho Phase Portraits and Stability.

Edge cases

Yahan naam ki har portrait shape ke liye upar wale figure ko refer karo.

Solution kya hoti hai jab

A

already diagonal ho,

A = diag (λ_{1}, λ_{2})

Variables already decoupled hain, toh

x_{1} = c_{1} e^{λ_{1} t}

x_{2} = c_{2} e^{λ_{2} t}

independently. Standard basis vectors eigenvectors hain — koi mixing undo karne ki zaroorat nahi.

Kya hota hai agar

A = 0

(zero matrix) ho?

Tab

x^{'} = 0

, toh har solution ek constant hai

x (t) = x_{0}

. Har direction

λ = 0

ke saath ek eigenvector hai; poora plane equilibria hai.

Kya hoga agar

A

ka ek eigenvalue

λ = 0

ho aur ek

λ < 0

ho?

λ = 0

mode ek fixed direction deta hai (equilibria ki ek line) jabki

λ < 0

mode us par decay karta hai. Trajectories us line ki taraf slide karti hain — ek degenerate / non-isolated stable case.

Phase portrait kaisa hoga jab

λ = α \pm i β

aur

α > 0

ho?

Ek outward spiral (unstable spiral / focus): har turn

e^{α t} > 1

se scale hota hai, toh radius barhta hai jabki

β

rotate karta rehta hai.

Phase portrait kaisa hoga jab

λ = α \pm i β

aur

α < 0

ho?

Ek inward spiral (stable spiral / focus): har turn

e^{α t} < 1

se scale hota hai, toh radius origin ki taraf shrink karta hai jabki

β

rotate karta rehta hai — trajectories spiral in karti hain.

Kya hoga agar do real eigenvalues ke opposite signs hon,

λ_{1} > 0 > λ_{2}

Ek saddle: trajectories

v_{1}

ke along push out hoti hain aur

v_{2}

ke along pull in hoti hain. Sirf woh points jo exactly

v_{2}

line par start karte hain origin tak pahunchte hain; baki sab escape kar jaate hain.

Kya hoga agar supposedly complex pair mein

β = 0

ho — yaani discriminant collapse ho jaaye?

Tab eigenvalues actually real aur equal hain (

λ = α

); koi rotation nahi hai. Tum repeated-eigenvalue case par pahunch gaye ho aur check karna chahiye ki

A

defective hai ya nahi.

Recall Traps ka one-line summary

Eigenvectors gino, eigenvalues nahi; har exponential term ke saath eigenvector attached rakho; growth $Re (λ)$ se aur rotation $Im (λ)$ se padho; aur degenerate boundaries hamesha check karo ( $λ = 0$ , repeated, defective, $α = 0$ ).