Case 1 - two distinct real roots: Worked examples — Hinglish deep dive

4.6.11 · D3 · Maths › Ordinary Differential Equations › Case 1 - two distinct real roots

Yeh page Case 1 ka "no surprises" drill hai. Parent note ne tumhe machine dikhaya tha — Replace, Solve, Superpose. Yahan hum us machine ko har tarah ke input ke through chalate hain jo use mil sakta hai, taaki jab exam ek variation throw kare, tum uska shape pehle se dekh chuke ho.

Kisi bhi algebra se pehle, ek reminder us object ka jo hum solve kar rahe hain:

Yahan $r_{1}, r_{2}$ do growth rates hain; $C_{1}, C_{2}$ free constants hain jo starting data se set hote hain. Neeche sab kuch sirf yahi ek formula hai, har tarah ke disguise mein.

Scenario matrix

Case 1 mein hamesha do real numbers $r_{1}, r_{2}$ milte hain. Jo cheez alag hoti hai woh hai unke signs, kya ek zero hai, kya problem word problem ki tarah dress ki gayi hai, aur constants kaise fix hote hain. Neeche har row ek alag "answer ki shape" hai jo tumhe dekh ke pehchanni chahiye.

Cell	Roots mein kya khaas hai	$y$ ka long-term behaviour	Example
A	Dono roots negative ( $r_{1} < 0, r_{2} < 0$ )	$0$ par decay karta hai	Ex 1
B	Dono roots positive ( $r_{1} > 0, r_{2} > 0$ )	Blow up ho jaata hai	Ex 2
C	Opposite signs ( $r_{1} > 0, r_{2} < 0$ )	Generally blow up (ek term dominate karta hai)	Ex 3
D	Ek root zero hai ( $b = 0$ ? nahi — $c = 0$ )	Nonzero constant par settle ho jaata hai	Ex 4
E	$b = 0$ symmetric roots ( $r = \pm k$ )	Badhta hai jab tak growing part switch off na ho	Ex 5
F	Initial conditions $C_{1}, C_{2}$ pin karte hain	Data par depend karta hai	Ex 6
G	Boundary conditions (do $x$ values par values)	Data par depend karta hai	Ex 7
H	Word problem (real-world decay/mixing)	Physical decay	Ex 8
I	Exam twist — leading coefficient $a \neq = 1$ , quadratic formula chahiye	Depend karta hai	Ex 9
J	Degenerate boundary — koi solution nahi / infinitely many	Failure ke liye watch karo	Ex 10

Neeche figure roots ke chaar sign-combinations plot karta hai taaki tum compute karne se pehle dekh sako ki har row kya karta hai.

Figure — Case 1 - two distinct real roots

Chaar coloured curves dekho: green (dono roots negative) zero par sink karta hai; red (dono positive) upar shoot karta hai; blue (opposite signs) apne growing half se upar pull hota hai; yellow (ek zero root) ek constant shelf par flat ho jaata hai. Neeche har example inhi shapes mein se ek par land karta hai.

Cell A — dono roots negative (pure decay)

Derivatives ko $r$ ki powers se replace karo: $r^{2} + 6 r + 8 = 0$ . Yeh step kyun? Ansatz $y = e^{r x}$ se $y^{''} \to r^{2} e^{r x}$ , $y^{'} \to r e^{r x}$ ban jaata hai; common factor $e^{r x} \neq = 0$ divide ho jaata hai, sirf pure algebra bachta hai.
Discriminant check karo: $b^{2} - 4 a c = 36 - 32 = 4 > 0$ . Yeh step kyun? Positive discriminant Case 1 ka entry ticket hai — do distinct reals.
Quadratic solve karo: $(r + 2) (r + 4) = 0 \Rightarrow r_{1} = - 2, r_{2} = - 4$ . Yeh step kyun? Yeh do growth rates hain jo exponential ko ODE satisfy karate hain.
Superpose: $y = C_{1} e^{- 2 x} + C_{2} e^{- 4 x}$ . Yeh step kyun? Superposition plus 2nd-order ODE mein do free constants ki zaroorat.

Recall Verify

$y = e^{- 2 x}$ lo: $y^{''} + 6 y^{'} + 8 y = 4 e^{- 2 x} - 12 e^{- 2 x} + 8 e^{- 2 x} = 0$ ✔. Dono roots negative hain, isliye jab $x \to \infty$ , $y \to 0$ — forecast se match karta hai.

Cell B — dono roots positive (pure growth)

Characteristic equation: $r^{2} - 7 r + 10 = 0$ . Kyun? Same replacement rule.
Discriminant: $49 - 40 = 9 > 0$ . Case 1 confirm. Kyun? Do distinct real roots guarantee karta hai.
Factor: $(r - 2) (r - 5) = 0 \Rightarrow r_{1} = 2, r_{2} = 5$ . Kyun? Dono positive hain, sum/product prediction se match karta hai.
General solution: $y = C_{1} e^{2 x} + C_{2} e^{5 x}$ .

Recall Verify

Roots ka sum $= 2 + 5 = 7 = - b / a$ ✔; product $= 2 \cdot 5 = 10 = c / a$ ✔ (Vieta's checks). Test $y = e^{5 x}$ : $25 e^{5 x} - 35 e^{5 x} + 10 e^{5 x} = 0$ ✔.

Cell C — opposite-sign roots (saddle)

Characteristic equation: $r^{2} - r - 12 = 0$ .
Discriminant: $1 + 48 = 49 > 0$ . Kyun? Case 1.
Factor: $(r - 4) (r + 3) = 0 \Rightarrow r_{1} = 4, r_{2} = - 3$ . Kyun? Ek positive ( $e^{4 x}$ grow karta hai), ek negative ( $e^{- 3 x}$ decay karta hai).
General solution: $y = C_{1} e^{4 x} + C_{2} e^{- 3 x}$ .

Recall Verify

Product $= 4 \cdot (- 3) = - 12 = c$ ✔, sum $= 4 + (- 3) = 1 = - b$ ✔. Test $y = e^{- 3 x}$ : $9 e^{- 3 x} + 3 e^{- 3 x} - 12 e^{- 3 x} = 0$ ✔.

Cell D — ek root zero hai (constant par settle karta hai)

Characteristic equation: $r^{2} + 3 r = 0$ . Kyun? Same replacement; $c = 0$ se koi constant term nahi bachta.
Factor: $r (r + 3) = 0 \Rightarrow r_{1} = 0, r_{2} = - 3$ . Kyun? Yeh ab bhi do distinct reals hain, isliye genuinely Case 1 hai — zero root bilkul legal hai.
General solution: $y = C_{1} e^{0 \cdot x} + C_{2} e^{- 3 x} = C_{1} + C_{2} e^{- 3 x}$ . Kyun? $e^{0 \cdot x} = 1$ , isliye "growth rate zero" wala solution constant function hai.

Recall Verify

Test $y = 1$ (constant): $y^{''} + 3 y^{'} = 0 + 0 = 0$ ✔. Jab $x \to \infty$ , $y \to C_{1}$ , ek constant shelf — forecast se match karta hai.

Cell E — symmetric roots $r = \pm k$ ( $b = 0$ )

Characteristic equation: $r^{2} - 16 = 0$ .
Solve: $r = \pm 4$ , yaani $r_{1} = 4, r_{2} = - 4$ . Distinct reals — Case 1. Kyun? Discriminant $= 0 - 4 (1) (- 16) = 64 > 0$ .
General solution: $y = C_{1} e^{4 x} + C_{2} e^{- 4 x}$ .

Recall Verify

Test $y = e^{4 x}$ : $16 e^{4 x} - 16 e^{4 x} = 0$ ✔. Test $y = e^{- 4 x}$ : $16 e^{- 4 x} - 16 e^{- 4 x} = 0$ ✔.

Cell F — initial conditions constants fix karte hain

Characteristic equation & roots: $r^{2} - 3 r + 2 = (r - 1) (r - 2) = 0 \Rightarrow r_{1} = 1, r_{2} = 2$ .
General solution: $y = C_{1} e^{x} + C_{2} e^{2 x}$ .
$y (0) = 3$ apply karo: $x = 0$ par dono exponentials $1$ ke barabar hain, isliye $C_{1} + C_{2} = 3$ . Kyun? $e^{0} = 1$ equation ko plain linear relation mein collapse kar deta hai.
Differentiate karo: $y^{'} = C_{1} e^{x} + 2 C_{2} e^{2 x}$ . $y^{'} (0) = 4$ apply karo: $C_{1} + 2 C_{2} = 4$ . Kyun? 2-parameter family ko pin karne ke liye do conditions chahiye.
Pair solve karo: subtract: $(C_{1} + 2 C_{2}) - (C_{1} + C_{2}) = 4 - 3 \Rightarrow C_{2} = 1$ , phir $C_{1} = 2$ . Kyun? Do linear equations, do unknowns.
Answer: $y = 2 e^{x} + e^{2 x}$ .

Recall Verify

$y (0) = 2 + 1 = 3$ ✔. $y^{'} (0) = 2 \cdot 1 + 2 \cdot 1 = 4$ ✔. ODE mein plug karo: $y^{''} - 3 y^{'} + 2 y = (2 e^{x} + 4 e^{2 x}) - 3 (2 e^{x} + 2 e^{2 x}) + 2 (2 e^{x} + e^{2 x}) = 0$ ✔.

Cell G — boundary conditions (do alag $x$ -values)

Roots & general solution: $r = \pm 1$ , isliye $y = C_{1} e^{x} + C_{2} e^{- x}$ .
$y (0) = 1$ apply karo: $C_{1} + C_{2} = 1$ . Kyun? $x = 0$ par dono exponentials $1$ hain.
$y (ln 2) = \frac{5}{2}$ apply karo: $e^{l n 2} = 2$ , $e^{- l n 2} = \frac{1}{2}$ , isliye $2 C_{1} + \frac{1}{2} C_{2} = \frac{5}{2}$ . Kyun? $x = ln 2$ par exponentials exactly evaluate karo.
Solve karo: step 2 se, $C_{2} = 1 - C_{1}$ . Substitute karo: $2 C_{1} + \frac{1}{2} (1 - C_{1}) = \frac{5}{2} \Rightarrow \frac{3}{2} C_{1} = 2 \Rightarrow C_{1} = \frac{4}{3}$ , phir $C_{2} = - \frac{1}{3}$ . Kyun? Phir se linear elimination — boundary conditions sirf do linear equations hain.
Answer: $y = \frac{4}{3} e^{x} - \frac{1}{3} e^{- x}$ .

Recall Verify

$y (0) = \frac{4}{3} - \frac{1}{3} = 1$ ✔. $y (ln 2) = \frac{4}{3} \cdot 2 - \frac{1}{3} \cdot \frac{1}{2} = \frac{8}{3} - \frac{1}{6} = \frac{16 - 1}{6} = \frac{15}{6} = \frac{5}{2}$ ✔.

Cell H — real-world word problem

Characteristic equation: $r^{2} + 5 r + 4 = (r + 1) (r + 4) = 0 \Rightarrow r_{1} = - 1, r_{2} = - 4$ . Kyun? Standard replacement.
General solution: $y = C_{1} e^{- x} + C_{2} e^{- 4 x}$ .
$y (0) = 10$ : $C_{1} + C_{2} = 10$ . Kyun? $e^{0} = 1$ .
$y^{'} = - C_{1} e^{- x} - 4 C_{2} e^{- 4 x}$ , $y^{'} (0) = - 13$ : $- C_{1} - 4 C_{2} = - 13 \Rightarrow C_{1} + 4 C_{2} = 13$ . Kyun? Doosri condition dono constants pin karti hai.
Solve karo: step 3 subtract karo: $3 C_{2} = 3 \Rightarrow C_{2} = 1$ , phir $C_{1} = 9$ .
Answer: $y = 9 e^{- x} + e^{- 4 x}$ . Jab $x \to \infty$ , dono terms $\to 0$ , isliye $y \to 0$ .

Recall Verify

$y (0) = 9 + 1 = 10$ ✔. $y^{'} (0) = - 9 - 4 = - 13$ ✔. Units: $y^{'}$ , $y$ ke units per second share karta hai; rates $- 1, - 4$ ke units $1/ s$ hain ( $r x$ ke exponential ke liye correct hai jahan $x$ seconds mein hai). Long-term $y \to 0$ physical decay se match karta hai ✔.

Cell I — exam twist: $a \neq = 1$ , quadratic formula chahiye

Characteristic equation: $2 r^{2} + 5 r - 3 = 0$ . Kyun? Coefficient $a = 2$ rakho; galti se divide mat karo.
Quadratic formula: $r = \frac{- 5 \pm 25 + 24}{4} = \frac{- 5 \pm 7}{4}$ . Yeh tool kyun? Obvious integer factoring nahi hai, isliye formula $r = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ general route hai.
Roots: $r_{1} = \frac{2}{4} = \frac{1}{2}, r_{2} = \frac{- 12}{4} = - 3$ . Kyun? Discriminant $49 > 0$ , distinct reals — Case 1.
General solution: $y = C_{1} e^{x /2} + C_{2} e^{- 3 x}$ .

Recall Verify

Sum $= \frac{1}{2} + (- 3) = - \frac{5}{2} = - b / a$ ✔. Product $= \frac{1}{2} \cdot (- 3) = - \frac{3}{2} = c / a$ ✔. Test $y = e^{x /2}$ : $2 \cdot \frac{1}{4} e^{x /2} + 5 \cdot \frac{1}{2} e^{x /2} - 3 e^{x /2} = (\frac{1}{2} + \frac{5}{2} - 3) e^{x /2} = 0$ ✔.

Cell J — degenerate boundary problem (ek trap)

General solution: $y = C_{1} e^{x} + C_{2} e^{- x}$ .
$lim_{x \to \infty} y = 0$ impose karo: $e^{x} \to \infty$ , isliye zaroor $C_{1} = 0$ hona chahiye. Kyun? Koi bhi nonzero $C_{1}$ $y$ ko blow up karta hai, jo limit ko contradict karta hai.
Ab $y = C_{2} e^{- x}$ . $y (0) = 1$ impose karo: $C_{2} = 1$ , isliye $y = e^{- x}$ . Ek unique solution exist karta hai.
Doosra version — $y (0) = 0$ impose karo: $C_{2} = 0$ , jo $y \equiv 0$ deta hai. Yahan $0$ se start hone wala sirf trivial zero function decaying solution hai. Kyun dono dikhao? Yeh dekhne ke liye ki boundary conditions (a) ek unique nonzero solution select kar sakti hain, ya (b) trivial solution force kar sakti hain — tumhe check karna padega kaun sa case hai.

Recall Verify

$y = e^{- x}$ : $y^{''} - y = e^{- x} - e^{- x} = 0$ ✔; $y (0) = 1$ ✔; $lim_{x \to \infty} e^{- x} = 0$ ✔. Version two: $y \equiv 0$ trivially sab kuch satisfy karta hai ✔.

Recall check

Kis sign ka

c / a

opposite-sign roots guarantee karta hai?

Negative

c / a

(roots ka product

c / a

ke barabar hota hai).

Jab

c = 0

ho, roots mein se hamesha ek kya hota hai aur woh term kaisi dikhti hai?

r = 0

; woh term constant

C_{1}

hai (kyunki

e^{0 x} = 1

y \to 0

x \to \infty

" kyun ek constant ko vanish karne force karta hai?

Yeh kisi bhi

e^{r x}

ke coefficient ko kill kar deta hai jahan

r > 0

ho, kyunki woh part blow up ho jaata hai.

2 y^{''} + 5 y^{'} - 3 y = 0

ke liye roots kya hain?

r = \frac{1}{2}

aur

r = - 3

Connections

Case 1 - two distinct real roots — parent recipe jo yahan drill ki gayi hai.
Characteristic equation — har example ke peeche wala quadratic.
Superposition principle — isliye har answer $C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x}$ hai.
Wronskian and linear independence — guarantee karta hai ki do exponentials saare solutions span karte hain, tab bhi jab ek root $0$ ho.
Case 2 repeated real root — neighbouring case jab discriminant $0$ hit kare.
Case 3 complex conjugate roots — jab discriminant negative ho jaata hai.
Exponential function and its derivative — wajah ki poora method kaam karta hai.