Case 1 - two distinct real roots — Hinglish notes

4.6.11 · Maths › Ordinary Differential Equations

Context: homogeneous linear ODE with constant coefficients solve karna $a y^{''} + b y^{'} + cy = 0, a \neq = 0.$ Is note mein woh case cover hota hai jahan characteristic equation ke do alag real roots hote hain.

Derivation from scratch

Hum KYA compute karte hain: $y = e^{r x}, y^{'} = r e^{r x}, y^{''} = r^{2} e^{r x} .$

KAISE substitute karte hain $a y^{''} + b y^{'} + cy = 0$ mein: $a r^{2} e^{r x} + b r e^{r x} + c e^{r x} = 0.$

Yeh step kyun? Har derivative bas $r$ ki ek power neeche le aati hai, isliye $e^{r x}$ ek common factor ban jaata hai.

Usse factor out karo: $e^{r x} (a r^{2} + b r + c) = 0.$

Kyunki $e^{r x} \neq = 0$ sabhi real $x$ ke liye, bracket ka zero hona zaroori hai:

WHY milte hain TWO independent solutions

Har root ek solution deta hai: $y_{1} = e^{r_{1} x}$ aur $y_{2} = e^{r_{2} x}$ .

WHY combine karte hain unhe? ODE linear aur second-order hai. Linear $\Rightarrow$ solutions ka koi bhi sum/scaling phir se ek solution hoga (superposition). Second-order $\Rightarrow$ general solution ko exactly do arbitrary constants chahiye. To:

Figure — Case 1 - two distinct real roots

Worked examples

Characteristic equation: $r^{2} + 5 r + 6 = 0$ . Kyun? $y^{''} \to r^{2}, y^{'} \to r, y \to 1$ replace karo.
Factor karo: $(r + 2) (r + 3) = 0 \Rightarrow r_{1} = - 2, r_{2} = - 3$ . Kyun? Yeh distinct real roots hain; discriminant $= 25 - 24 = 1 > 0$ . ✔ Case 1.
General solution: $y = C_{1} e^{- 2 x} + C_{2} e^{- 3 x}$ . Kyun? Do independent exponentials ko superpose karo.

$r^{2} - r - 2 = 0 \Rightarrow (r - 2) (r + 1) = 0 \Rightarrow r_{1} = 2, r_{2} = - 1$ .
$y = C_{1} e^{2 x} + C_{2} e^{- x}$ .
$y (0) = 1$ apply karo: $C_{1} + C_{2} = 1$ . Kyun? $x = 0$ par, dono exponentials $1$ ke barabar hote hain.
Differentiate karo: $y^{'} = 2 C_{1} e^{2 x} - C_{2} e^{- x}$ . $y^{'} (0) = 4$ apply karo: $2 C_{1} - C_{2} = 4$ . Kyun? Dono constants pin karne ke liye ek doosri equation chahiye.
Dono add karo: $3 C_{1} = 5 \Rightarrow C_{1} = \frac{5}{3}$ , phir $C_{2} = 1 - \frac{5}{3} = - \frac{2}{3}$ .
Answer: $y = \frac{5}{3} e^{2 x} - \frac{2}{3} e^{- x}$ .

$r^{2} - 9 = 0 \Rightarrow r = \pm 3$ . (Yahan $b = 0$ hai, phir bhi do distinct reals.)
$y = C_{1} e^{3 x} + C_{2} e^{- 3 x}$ . Kyun yeh important hai: ek term blow up karta hai, ek decay karta hai — generic solutions badhte hain jab tak $C_{1} = 0$ na ho.

Forecast-then-Verify

Recall Forecast:

$y^{''} + 7 y^{'} + 12 y = 0$ solve karne se pehle, roots ke signs aur long-term behaviour predict karo. Forecast: $b, c > 0$ hain aur $b^{2} > 4 a c$ hai, to dono roots real aur negative hain $\Rightarrow$ solution decay karke 0 ho jaayega. Verify: $r^{2} + 7 r + 12 = (r + 3) (r + 4) = 0$ , roots $- 3, - 4$ (dono negative). $y = C_{1} e^{- 3 x} + C_{2} e^{- 4 x} \to 0$ . ✔

Common mistakes

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho ek function jo apni khud ki size ke proportion mein badhti ya ghatti hai — jaise bank mein paisa interest ke saath, $e^{r x}$ . Equation pooch rahi hai: "woh growth rates $r$ dhundho jisse jab main function ko, uski speed ko aur uski acceleration ko fixed weights ke saath stack karun, to woh zero mein cancel ho jaayein." Woh cancelling bas ek chota sa quadratic puzzle hai. Agar puzzle ke do alag answers $r_{1}$ aur $r_{2}$ hain, to DONO growth patterns kaam karte hain, aur poora answer bas ek mix hai ( $C_{1}$ pehle ka plus $C_{2}$ doosre ka), jahan tum mix isliye choose karte ho taaki woh wahan se start ho jahan se tumhara system shuru hota hai.

Connections

Characteristic equation — sabhi constant-coefficient ODEs ke peeche ka algebra engine.
Case 2 repeated real root — kya hota hai jab $b^{2} - 4 a c = 0$ ho.
Case 3 complex conjugate roots — jab $b^{2} - 4 a c < 0$ ho, Euler's formula ke zariye oscillations milti hain.
Wronskian and linear independence — prove karta hai ki $y_{1}, y_{2}$ solution space span karte hain.
Superposition principle — kyun solutions add karte hain.
Exponential function and its derivative — kyun $e^{r x}$ natural ansatz hai.

a y^{''} + b y^{'} + cy = 0

ke liye hum kaun sa ansatz try karte hain?

y = e^{r x}

, kyunki iske derivatives khud ka scaled copy hote hain.

a y^{''} + b y^{'} + cy = 0

ki characteristic equation kya hai?

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

Case 1 (two distinct real roots) ki condition kya hai?

Discriminant

b^{2} - 4 a c > 0

Jab roots distinct reals

r_{1} \neq = r_{2}

hon to general solution kya hoti hai?

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

Kyun chahiye do constants

C_{1}, C_{2}

ODE 2nd-order hai, isliye iske solution family mein do free parameters hote hain.

Jab

r_{1} \neq = r_{2}

ho to

e^{r_{1} x}

aur

e^{r_{2} x}

linearly independent kyun hain?

Unka ratio

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

non-constant hai; Wronskian

(r_{2} - r_{1}) e^{(r_{1} + r_{2}) x} \neq = 0

y^{''} + 5 y^{'} + 6 y = 0

ke roots kya hain?

r = - 2, - 3

, to

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

Agar

b^{2} - 4 a c = 0

ho to kaun sa case aur kya change hota hai?

Case 2 (repeated root); solution ban jaata hai

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

Case 1 - two distinct real roots

The big idea