Case 1 - two distinct real roots: Question bank — Hinglish deep dive

4.6.11 · D5 · Maths › Ordinary Differential Equations › Case 1 - two distinct real roots

Shuru karne se pehle, kaam aane wale objects — taaki koi bhi symbol bina parichay ke na aaye:

$y$ wo unknown function hai jise hum solve kar rahe hain; $y^{'}$ uska slope (first derivative) hai, $y^{''}$ uska curvature (second derivative) hai.
$a y^{''} + b y^{'} + cy = 0$ hamaara ODE hai, jisme fixed number weights $a, b, c$ hain aur $a \neq = 0$ hai.
$e^{r x}$ exponential ansatz hai: wo ek aisi shape jiska derivative sirf khud ka ek number se multiply hota hai, $\frac{d}{d x} e^{r x} = r e^{r x}$ (dekho Exponential function and its derivative).
$a r^{2} + b r + c = 0$ Characteristic equation hai; iske roots $r_{1}, r_{2}$ quadratic formula se aate hain, aur $b^{2} - 4 a c$ wo discriminant hai jo decide karta hai ki hum kis case mein hain.
Case 1 ka matlab hai $b^{2} - 4 a c > 0$ : do alag real roots $r_{1} \neq = r_{2}$ .

True or false — justify karo

TF1. Agar

y_{1}

aur

y_{2}

dono ODE solve karte hain, toh

C_{1} y_{1} + C_{2} y_{2}

bhi use solve karta hai.

True — equation linear hai, isliye solutions ko scale karna aur add karna tumhe solution set mein hi rakhta hai; yahi exactly Superposition principle hai.

TF2.

y = e^{r_{1} x} + e^{r_{2} x}

(bina constants ke) Case 1 mein general solution hai.

False — yeh ek particular solution hai. General solution ko do free constants

C_{1}, C_{2}

chahiye kyunki ek 2nd-order ODE mein two-parameter family hoti hai.

TF3. Do distinct real roots ka matlab hamesha yeh hota hai ki solution

x \to \infty

ke saath zero pe decay karta hai.

False — decay ke liye dono roots negative hone chahiye. Agar koi bhi root positive hai (jaise

r = \pm 3

), toh ek generic solution bina kisi bound ke grow karta hai.

TF4.

e^{r_{1} x}

aur

e^{r_{2} x}

linearly independent hote hain jab bhi

r_{1} \neq = r_{2}

ho.

True — unka ratio

e^{(r_{1} - r_{2}) x}

non-constant hai, isliye koi bhi doosre ka fixed multiple nahi hai (iska confirmation ek nonzero Wronskian se hota hai).

TF5. Kyunki

e^{r x} \neq = 0

har real

x

ke liye, hum ise substituted equation se divide kar sakte hain.

True — exponential kabhi zero nahi hota, isliye ise cancel karne se koi solution nahi khoota aur pure algebra

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

bach jaata hai.

TF6. Agar roots

r_{1} = 0

aur

r_{2} = - 4

hain, toh do basis solutions mein se ek constant function

1

hai.

True —

e^{0 \cdot x} = 1

, isliye

y = C_{1} + C_{2} e^{- 4 x}

; constant term ek bilkul valid, non-trivial solution hai.

TF7. Discriminant ka exactly

0

hona phir bhi Case 1 mein count hota hai kyunki "quadratic formula ek root deta hai."

False —

b^{2} - 4 a c = 0

se ek repeated root milta hai; do exponentials ek saath aa jaate hain aur independence kho dete hain. Yeh Case 2 repeated real root hai, Case 1 nahi.

TF8. Labels

r_{1} \leftrightarrow r_{2}

ko swap karne se general solution badal jaata hai.

False — isse sirf constants rename hote hain;

C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x}

bilkul usi family of functions ko describe karta hai.

TF9. Characteristic equation initial conditions

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

par depend karta hai.

False — roots sirf

a, b, c

se aate hain. Initial conditions baad mein aate hain,

C_{1}, C_{2}

ko general solution se fix karne ke liye.

Galti dhundho

SE1. "

r^{2} - r - 2 = 0

, toh factor karne par

(r - 2) (r + 1)

se roots

- 2

aur

1

hain."

Roots wahan hain jahan har factor zero ho:

r - 2 = 0 \Rightarrow r = 2

aur

r + 1 = 0 \Rightarrow r = - 1

. Roots

+ 2

aur

- 1

hain, brackets ke andar ke numbers nahi.

SE2. "Discriminant hai

b^{2} - 4 a c = 25 - 24 = 1

, jo positive nahi hai, isliye yeh Case 1 nahi hai."

1

positive hai (

1 > 0

), isliye yeh Case 1 hai. Trap yeh hai ki ek chhote positive number ko strict inequality fail karte hue padhna.

SE3. "

y^{''} - 9 y = 0

ke liye hume

r^{2} - 9 = 0

milta hai, aur kyunki

b = 0

hai, sirf ek root hai."

r^{2} - 9 = 0

r = \pm 3

milta hai — do distinct real roots. Zero middle coefficient se root count kam nahi hota; bas roots symmetric ho jaate hain.

SE4. "

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

, isliye

y^{'} = C_{1} r_{1} e^{r_{1} x} + C_{2} e^{r_{2} x}

Chain rule doosre term se bhi

r_{2}

neeche laata hai:

y^{'} = C_{1} r_{1} e^{r_{1} x} + C_{2} r_{2} e^{r_{2} x}

. Ek

r

ka factor bhoolne se baad ke har step mein galti aa jaati hai.

SE5. "Dono roots ODE solve karte hain, isliye unka product

e^{r_{1} x} \cdot e^{r_{2} x} = e^{(r_{1} + r_{2}) x}

bhi ek solution hai."

Superposition sirf sums aur scalings allow karta hai, products nahi.

e^{(r_{1} + r_{2}) x}

ki generally growth rate

r_{1} + r_{2}

hoti hai jo ek root nahi hoti, isliye yeh ODE satisfy nahi bhi kar sakta.

SE6. "Kyunki

e^{r x} (a r^{2} + b r + c) = 0

hai, ya toh

e^{r x} = 0

hai ya bracket zero hai — chalo

e^{r x} = 0

try karte hain."

e^{r x}

real

x

ke liye kabhi zero nahi hota, isliye woh branch empty hai; sirf bracket hi vanish ho sakta hai, aur yahi wajah hai ki hume characteristic equation milti hai.

Why questions

WQ1. Hum

e^{r x}

guess kyun karte hain aur

sin (r x)

ya koi polynomial kyun nahi?

Kyunki

e^{r x}

ki yeh unique property hai ki uske derivatives sirf khud ke scaled copies hote hain, isliye

e^{r x}

factor out ho jaata hai aur ODE

r

mein simple algebra mein collapse ho jaata hai.

WQ2. Algebra quadratic kyun nikalta hai, koi aur equation kyun nahi?

Sabse highest derivative second order hai (

y^{''}

), jo

r^{2}

contribute karta hai; har derivative ko

r

ki ek power se match karne par

a r^{2} + b r + c

milta hai, jo degree-2 polynomial hai.

WQ3. Ek second-order ODE mein exactly do arbitrary constants kyun hone chahiye?

y^{''}

y

recover karne ke liye do integrations chahiye, har ek ek constant add karta hai; equivalently tumhe ek unique solution pin karne ke liye do initial data (

y

aur

y^{'}

) chahiye.

WQ4.

r_{1} \neq = r_{2}

independence guarantee kyun karta hai, jabki

r_{1} = r_{2}

use kyon khatam kar deta hai?

Jab roots alag hote hain, ratio

e^{(r_{1} - r_{2}) x}

x

ke saath change hota hai (independent). Jab roots equal hote hain toh ratio

1

hota hai — ek hi function do baar — isliye tumne solution space ki sirf ek direction dhundhi hai.

WQ5. Hum Wronskian check kyun karte hain agar hum already dekh sakte hain ki ratio non-constant hai?

Wronskian ek single, rigorous nonzero test deta hai jo har point par valid hai, aur yeh un cases mein bhi generalize hota hai jahan "aankhon se ratio dekho" wali trick obvious nahi hoti.

WQ6. Long-term behaviour roots ka sign kyun decide karta hai, unka size kyun nahi?

e^{r x}

mein, positive

r

x \to \infty

ke saath grow karta hai aur negative

r

decay karta hai; exponent ka sign direction set karta hai, jabki uski magnitude sirf speed set karti hai.

Edge cases

EC1. Agar exactly ek root

0

ho toh general solution ka kya hota hai?

Woh term

C_{1} e^{0 \cdot x} = C_{1}

ban jaata hai, yaani ek constant, isliye

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

— solution zero ki taraf nahi balki

C_{1}

ki taraf level off karta hai.

EC2. Agar dono distinct roots negative hain lekin bahut close hain, jaise

r_{1} = - 3, r_{2} = - 3.0001

Phir bhi Case 1 hai (woh alag hain), isliye

y = C_{1} e^{- 3 x} + C_{2} e^{- 3.0001 x}

; dono decay karte hain. Formulas break nahi karte, halaanki numerically do exponentials almost identical lagte hain.

EC3. Agar roots magnitude mein equal hain lekin sign mein opposite hain,

r = \pm k

Tumhe milta hai

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

— ek growing, ek decaying "saddle" behaviour; solution blow up karta hai jab tak initial data

C_{1} = 0

force na kare.

EC4. Jab

b^{2} - 4 a c \to 0^{+}

, do roots aur Case 1 ka kya hota hai?

Roots ek doosre ki taraf slide karte hain aur merge ho jaate hain; limit mein woh coincide ho jaate hain, Case 1 break down ho jaata hai, aur tumhe extra

x

factor ke saath Case 2 repeated real root par switch karna padta hai.

EC5. Agar

b^{2} - 4 a c < 0

ho toh hum Case 1 mein kyun nahi reh sakte?

Negative number ka square root complex roots deta hai, isliye koi real distinct roots exist nahi karte; yeh Case 3 complex conjugate roots hai, jahan solutions oscillate karte hain.

EC6. Kya koi genuine Case-1 solution kabhi zero function

y \equiv 0

ho sakta hai?

Haan —

C_{1} = C_{2} = 0

choose karne par trivial solution milta hai, jo hamesha ek homogeneous ODE satisfy karta hai lekin koi information nahi deta; yeh family ka woh ek member hai jise sabhi bhool jaate hain.

EC7. Agar koi tumhe

r_{1} = 2, r_{2} = - 1

deta hai lekin

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

likhta hai, toh kya yeh galat hai?

Nahi — terms ka order aur naming free hai, isliye yeh identical general solution hai; sirf exponentials ka set matter karta hai, unka sequence nahi.

Connections

Case 1 - two distinct real roots (index 4.6.11) — woh parent note jise yeh traps stress-test karte hain.
Characteristic equation — woh algebra jiske roots par har card depend karta hai.
Case 2 repeated real root / Case 3 complex conjugate roots — wo do "next door" cases jinhe bahut saare traps disguise karte hain.
Wronskian and linear independence — TF4/WQ5 ke peeche ka rigorous independence test.
Superposition principle — kyun solutions ke sums (products nahi) allowed hain (TF1, SE5).
Exponential function and its derivative — kyun $e^{r x}$ natural guess hai (WQ1).