Case 2 - repeated real root — reduction of order: Question bank

4.6.12 · D5 · Maths › Ordinary Differential Equations › Case 2 - repeated real root — reduction of order

Shuru karne se pehle anchor facts yaad karo:

Ek repeated root $r$ tab aata hai jab discriminant $b^{2} - 4 a c = 0$ ho, jisse ek root milta hai $r = - \frac{b}{2 a}$ .
Do independent solutions hain $e^{r x}$ aur $x e^{r x}$ , isliye $y = (C_{1} + C_{2} x) e^{r x}$ .
Independence Wronskian $W = e^{2 r x} \neq = 0$ se certified hoti hai.

Error dhundo

r^{2} - 6 r + 9 = 0

ke roots

3

aur

- 3

hain, isliye

y = C_{1} e^{3 x} + C_{2} e^{- 3 x}

Error:

r^{2} - 6 r + 9 = (r - 3)^{2}

hai, ek double root

r = 3

\pm 3

nahi; sahi solution hai

(C_{1} + C_{2} x) e^{3 x}

"Reduction se

v = C_{1} + C_{2} x

milta hai, isliye

y = C_{1} + C_{2} x e^{r x}

Error: poora

v

y_{1} = e^{r x}

ko multiply karta hai, isliye

y = (C_{1} + C_{2} x) e^{r x}

;

C_{1}

term ke saath bhi

e^{r x}

factor hona chahiye.

"Kyunki

v^{''} = 0

ka matlab hai

v

constant hai, doosra solution sirf ek aur

e^{r x}

hai."

Error:

v^{''} = 0

v = C_{1} + C_{2} x

milta hai, jo ek linear function hai, constant nahi; naya piece

C_{2} x

term hai jo

x e^{r x}

produce karta hai.

4 y^{''} + 12 y^{'} + 9 y = 0

ke liye repeated root

r = - b /2 a = - 12/2 = - 6

hai."

Error:

a = 4

hai, isliye

r = - b /2 a = - 12/8 = - 3/2

; student

a

se divide karna bhool gaya, sirf

2

se nahi.

"Kyunki dono roots

r

ke barabar hain, do exponentials ka Wronskian zero hai, isliye koi independent pair exist nahi karta."

Error: sirf ek exponential

e^{r x}

hai; sahi partner

x e^{r x}

hai, aur

e^{r x}

aur

x e^{r x}

ka Wronskian

e^{2 r x} \neq = 0

hai.

y = (C_{1} + C_{2} x) e^{- 2 x}

y (0) = 1

apply karne se

C_{1} + C_{2} = 1

milta hai."

Error:

x = 0

C_{2} x

term vanish ho jaata hai aur

e^{0} = 1

hai, isliye

y (0) = C_{1} = 1

;

C_{2}

baad mein

y^{'} (0)

se fix hota hai.

y^{''} + 2 y^{'} + y = 0

ka characteristic equation

(r + 1) (r - 1)

mein factor hota hai, jisse distinct roots milte hain."

Error: yeh

(r + 1)^{2}

mein factor hota hai, ek repeated root

r = - 1

; student ne constant term ka sign galat liya.

Why questions

Ek second-order ODE ko exactly do independent solutions kyun chahiye?

Iska solution set ek 2-dimensional vector space hai, isliye koi bhi solution do basis functions ka unique combination hota hai — do initial data

y (x_{0}), y^{'} (x_{0})

match karne ke liye itni freedom chahiye.

Repeated-root reduction mein DONO

v

aur

v^{'}

terms kyun disappear ho jaate hain?

v

-coefficient

a r^{2} + b r + c = 0

hai kyunki

r

ek root hai, aur

v^{'}

-coefficient

2 a r + b = 0

hai kyunki

r = - b /2 a

; double root dono ko simultaneously hold karata hai.

Hum bas

y_{2} = e^{r_{2} x}

kyun nahi le sakte ek "second" root ke saath jaise Case 1 mein?

Koi doosra distinct root nahi hai — discriminant zero hai isliye dono roots same

r

pe collapse ho jaate hain, sirf ek function bacha ke Case 1 ki recipe.

e^{r x}

x

lagana (kisi aur factor ki jagah) missing solution kyun produce karta hai?

Reduction of order

v^{''} = 0

force karta hai, jiska non-constant solution linear

v = x

hai; double root ki algebra

x

ko single out karti hai, aur coalescing-roots limit ise

\partial_{r} e^{r x}

ke roop mein confirm karta hai.

Characteristic equation method humein yahan do solutions kyun nahi deta, unlike distinct-root case mein?

Trial

y = e^{r x}

ODE ko ek quadratic mein convert karta hai; ek double root ek hi number hai, isliye ek hi exponential milta hai — "do coincident roots" ki geometry algebra ko ek solution short chhod deti hai.

r

ka sign is baat ke liye irrelevant kyun hai ki

{e^{r x}, x e^{r x}}

independent hain ya nahi?

Independence Wronskian

e^{2 r x}

ke nonzero hone pe depend karta hai, aur kisi bhi real exponent ke exponentials strictly positive hote hain, isliye koi bhi

r

unhe dependent nahi bana sakta.

General solution automatically is ODE ke kisi bhi initial-value problem ko satisfy kyun karta hai?

Do free constants

C_{1}, C_{2}

exactly do degrees of freedom dete hain, do conditions

y (x_{0}), y^{'} (x_{0})

match karne ke liye; nonzero Wronskian guarantee karta hai ki resulting system solvable hai.

Edge cases

Jab repeated root

r = 0

ho (jaise

y^{''} = 0

) toh solutions kya hain?

r = 0

ke saath,

e^{r x} = 1

aur

x e^{r x} = x

, isliye

y = C_{1} + C_{2} x

— straight lines, phir bhi do independent functions.

Agar

a \to 0

ho isliye equation degenerate ho ke

b y^{'} + cy = 0

ban jaaye, toh kya repeated-root picture ab bhi apply hoti hai?

Nahi — ODE first order mein drop ho jaata hai, ek 1-dimensional solution space hai, aur sirf ek exponential hai; poori "missing partner" story ke liye genuine second-order term chahiye.

Third-order ODE mein triple root

r

ke liye, solution basis kya hai?

{e^{r x}, x e^{r x}, x^{2} e^{r x}}

— Higher-order repeated roots: multiplicity

m

se powers

x^{0}

through

x^{m - 1}

times

e^{r x}

milte hain.

Jab do nearly-equal roots

r

aur

r + ε

merge hote hain, exponential solutions ke pair ka kya hota hai?

Ve nearly parallel ho jaate hain; unka normalized difference

\frac{e ^{(r + ε) x} - e ^{r x}}{ε}

x e^{r x}

ke paas pahunchta hai, isliye basis smoothly do exponentials se

{e^{r x}, x e^{r x}}

mein deform ho jaata hai.

y = (C_{1} + C_{2} x) e^{r x}

ka long-run behaviour kya hai jab

r < 0

ho?

Yeh

0

ki taraf tend karta hai, kyunki exponential decay

e^{r x}

eventually

C_{2} x

ki linear growth ko beat kar deta hai — exponential polynomial ko hara deta hai.

Jab

r = 0

ho lekin

C_{2} \neq = 0

ho toh long-run behaviour kya hai?

y = C_{1} + C_{2} x

linearly bina bound ke grow (ya fall) karta hai, kyunki

x

term ko tame karne ke liye koi decaying exponential nahi hai.

Kya

e^{r x}

aur

x e^{r x}

kabhi simultaneously kisi finite

x

pe zero ho sakte hain?

Nahi —

e^{r x} > 0

hamesha rehta hai, isliye

e^{r x}

kabhi zero nahi hota aur

x e^{r x}

sirf

x = 0

pe vanish hota hai, jahan dono Wronskian ke through phir bhi independent hain.

Recall Har trap ki ek-line summary

Ek double root sirf ek exponential deta hai; honest fix $x e^{r x}$ hai, $v^{''} = 0$ se force hota hai aur nonzero Wronskian $e^{2 r x}$ se confirm hota hai. Constants kabhi collapse mat karo, $x$ kabhi mat chhodo, $r = - b /2 a$ mein hamesha $a$ se divide karo, aur yaad rakho decay ke liye $r < 0$ chahiye.

Connections

Case 2 - repeated real root — reduction of order (index 4.6.12) — parent derivation jinhe ye traps guard karte hain.
Case 1 distinct real roots — woh contrast jisse zyaadatar errors confuse karti hain.
Case 3 complex roots — discriminant ka doosra sign.
Reduction of order (general method) — $v^{''} = 0$ kyun appear hota hai.
Wronskian and linear independence — independence ka certificate.
Higher-order repeated roots — triple-root edge case.
Characteristic equation of linear ODEs — root $r$ ka origin.

Question bank — Case 2 - repeated real root — reduction of order

True ya false — justify karo

Error dhundo

Why questions

Edge cases

Connections