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

4.6.11 · D4 · Maths › Ordinary Differential Equations › Case 1 - two distinct real roots

Is page ko kaise use karein: har problem ko solution kholne se pehle try karo. Har answer ek collapsible [!recall]- callout ke andar rakha hai — click karo reveal karne ke liye. Problems case ko pehchanne se lekar apna khud ka ODE banana tak step-by-step hard hote jaate hain. Saare numeric answers machine-checked hain.

Yahan sab kuch ek machine par tika hai: $a y^{''} + b y^{'} + cy = 0 y = e^{r x} a r^{2} + b r + c = 0 b^{2} - 4 a c > 0 y = C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x} .$ the parent note mein dekho ki kyun. Agar samajhna ho ki exponential natural guess kyun hai, toh Exponential function and its derivative revisit karo; solutions kyun add karte hain, iske liye Superposition principle dekho.

Level 1 — Recognition

Goal: discriminant padho aur case ka naam batao. Roots spot karne ke aage koi solving nahi chahiye.

Recall Solution 1.1

Har baar $a = 1$ read karo. (a) $b^{2} - 4 a c = 25 - 24 = 1 > 0$ → Case 1. (b) $16 - 16 = 0$ → Case 2 (repeated root), dekho Case 2 repeated real root. (c) $4 - 20 = - 16 < 0$ → Case 3 (complex roots), dekho Case 3 complex conjugate roots. Sign hi kyun sab kuch decide karta hai: $r_{1, 2} = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ . Square root ke andar wali cheez decide karti hai ki $\pm$ do real numbers mein split hoga ( $> 0$ ), ek mein collapse hoga ( $= 0$ ), ya imaginary ho jaayega ( $< 0$ ).

Recall Solution 1.2

Factor karo: $(r - 2) (r - 5) = 0 \Rightarrow r_{1} = 2, r_{2} = 5$ . Distinct reals hain, isliye $y = C_{1} e^{2 x} + C_{2} e^{5 x} .$

Level 2 — Application

Goal: ek ODE ko uski general solution mein convert karo.

Recall Solution 2.1

Characteristic equation: $r^{2} + r - 6 = 0$ . Factor karo: $(r + 3) (r - 2) = 0 \Rightarrow r_{1} = - 3, r_{2} = 2$ . Check karo: discriminant $= 1 + 24 = 25 > 0$ ✔ Case 1. $y = C_{1} e^{- 3 x} + C_{2} e^{2 x} .$

Recall Solution 2.2

Characteristic: $2 r^{2} - 3 r - 2 = 0$ . Quadratic formula: $r = \frac{3 \pm 9 + 16}{4} = \frac{3 \pm 5}{4}$ . Toh $r_{1} = \frac{8}{4} = 2$ aur $r_{2} = \frac{- 2}{4} = - \frac{1}{2}$ . $y = C_{1} e^{2 x} + C_{2} e^{- x /2} .$ $a = 2$ ko quadratic ke andar kyun rakhte hain: denominator mein $2 a = 4$ us leading coefficient se aata hai. Ise silently drop karne se galat roots milenge.

Recall Solution 2.3

Pehla: $r^{2} - 9 = 0 \Rightarrow r = \pm 3$ . $y = C_{1} e^{3 x} + C_{2} e^{- 3 x}$ . Doosra: $r^{2} - 3 r = 0 \Rightarrow r (r - 3) = 0 \Rightarrow r_{1} = 0, r_{2} = 3$ . Kyunki $e^{0 \cdot x} = 1$ , $y = C_{1} + C_{2} e^{3 x} .$ Comparison: $r = 0$ ka root phir bhi ek perfectly valid distinct real root hai — bas yeh constant solution $C_{1}$ produce karta hai. Yahan zero koi degenerate case nahi hai; yeh Case 1 ka ek honest member hai.

Level 3 — Analysis

Goal: initial/boundary conditions use karo constants pin karne ke liye, aur long-term behaviour padho.

Recall Solution 3.1

Roots: $r^{2} - r - 2 = (r - 2) (r + 1) = 0 \Rightarrow r_{1} = 2, r_{2} = - 1$ . General: $y = C_{1} e^{2 x} + C_{2} e^{- x}$ , aur $y^{'} = 2 C_{1} e^{2 x} - C_{2} e^{- x}$ . $x = 0$ par conditions (dono exponentials $= 1$ ): $C_{1} + C_{2} = 0, 2 C_{1} - C_{2} = 3.$ Add karo: $3 C_{1} = 3 \Rightarrow C_{1} = 1$ , tab $C_{2} = - 1$ . $y = e^{2 x} - e^{- x} .$

Recall Solution 3.2

Roots: $r^{2} + 3 r + 2 = (r + 1) (r + 2) = 0 \Rightarrow r_{1} = - 1, r_{2} = - 2$ . $y = C_{1} e^{- x} + C_{2} e^{- 2 x}$ , $y^{'} = - C_{1} e^{- x} - 2 C_{2} e^{- 2 x}$ . $x = 0$ par: $C_{1} + C_{2} = 2$ , $- C_{1} - 2 C_{2} = - 5 \Rightarrow C_{1} + 2 C_{2} = 5$ . Pehli ko doosri se subtract karo: $C_{2} = 3$ , tab $C_{1} = - 1$ . $y = - e^{- x} + 3 e^{- 2 x} .$ Long term: dono exponents negative hain → $y \to 0$ as $x \to \infty$ . Neeche figure dekho.

Figure — Case 1 - two distinct real roots

Recall Solution 3.3

Roots $r = \pm 2$ : $y = C_{1} e^{2 x} + C_{2} e^{- 2 x}$ . $y (0) = 1$ : $C_{1} + C_{2} = 1$ . $y (1) = 0$ : $C_{1} e^{2} + C_{2} e^{- 2} = 0 \Rightarrow C_{1} = - C_{2} e^{- 4}$ . Substitute karo: $- C_{2} e^{- 4} + C_{2} = 1 \Rightarrow C_{2} (1 - e^{- 4}) = 1 \Rightarrow C_{2} = \frac{1}{1 - e ^{- 4}}$ . Tab $C_{1} = \frac{- e ^{- 4}}{1 - e ^{- 4}}$ . Numerically $C_{2} \approx 1.01866$ , $C_{1} \approx - 0.018659$ . $y = \frac{- e ^{- 4}}{1 - e ^{- 4}} e^{2 x} + \frac{1}{1 - e ^{- 4}} e^{- 2 x} .$ Method identical kyun hai: boundary conditions phir bhi $C_{1}, C_{2}$ mein do linear equations dete hain. Bas kahan hum $y$ sample karte hain woh change hota hai.

Level 4 — Synthesis

Goal: diye gaye data se ODE, ya solution, ulta banao.

Recall Solution 4.1

Exponents humein roots batate hain: $r_{1} = 4, r_{2} = - 1$ . Unse characteristic equation banao: $(r - 4) (r + 1) = r^{2} - 3 r - 4 = 0.$ $r$ ki powers ko wapas derivatives mein padho ( $r^{2} \to y^{''}, r \to y^{'}, 1 \to y$ ): $y^{''} - 3 y^{'} - 4 y = 0 .$ Yeh kyun kaam karta hai: ODE $\leftrightarrow$ characteristic polynomial ka map ek two-way street hai. Roots $\to$ factors $\to$ polynomial $\to$ ODE.

Recall Solution 4.2

Constant $1 = e^{0 \cdot x}$ ka matlab hai $r_{1} = 0$ ; $e^{5 x}$ ka matlab hai $r_{2} = 5$ . $(r - 0) (r - 5) = r^{2} - 5 r = 0 \Rightarrow y^{''} - 5 y^{'} = 0 .$ Check karo: $c = 0$ , yeh match karta hai ki ek root $r = 0$ par hai (koi constant term nahi $\Rightarrow$ zero ek root hai).

Recall Solution 4.3

$y_{1}^{'} = 2 e^{2 x}$ , $y_{2}^{'} = - 3 e^{- 3 x}$ . $W = e^{2 x} (- 3 e^{- 3 x}) - (2 e^{2 x}) e^{- 3 x} = (- 3 - 2) e^{- x} = - 5 e^{- x} .$ Kyunki $e^{- x} \neq = 0$ har $x$ ke liye, $W \neq = 0$ everywhere → linearly independent. Note karo $- 5 = r_{2} - r_{1} = - 3 - 2$ , yeh general formula $W = (r_{2} - r_{1}) e^{(r_{1} + r_{2}) x}$ se match karta hai jo Wronskian and linear independence mein hai.

Level 5 — Mastery

Goal: sab kuch combine karo — behaviour design karo, system-like constraint handle karo, signs ke baare mein reason karo.

Recall Solution 5.1

" $0$ par decay" ke liye dono roots negative chahiye. "Ek twice as fast" ka matlab hai ek root doosre ka double ho: $r_{1} = - 1, r_{2} = - 2$ choose karo. $(r + 1) (r + 2) = r^{2} + 3 r + 2 = 0 \Rightarrow y^{''} + 3 y^{'} + 2 y = 0 .$ General solution $y = C_{1} e^{- x} + C_{2} e^{- 2 x}$ ; $e^{- 2 x}$ mode $e^{- x}$ se twice as fast decay karta hai. ✔ Sign rules se sanity check: dono roots negative $\Leftrightarrow$ sum $= - b < 0$ (toh $b > 0$ ) aur product $= c > 0$ . Yahan $b = 3 > 0, c = 2 > 0$ . ✔

Recall Solution 5.2

Roots: $r^{2} - r - 6 = (r - 3) (r + 2) = 0 \Rightarrow r_{1} = 3, r_{2} = - 2$ . $y = C_{1} e^{3 x} + C_{2} e^{- 2 x}$ , $y^{'} = 3 C_{1} e^{3 x} - 2 C_{2} e^{- 2 x}$ . Conditions: $C_{1} + C_{2} = A$ , $3 C_{1} - 2 C_{2} = 0 \Rightarrow C_{1} = \frac{2}{3} C_{2}$ . Tab $\frac{2}{3} C_{2} + C_{2} = A \Rightarrow \frac{5}{3} C_{2} = A \Rightarrow C_{2} = \frac{3 A}{5}, C_{1} = \frac{2 A}{5}$ . $y = \frac{2 A}{5} e^{3 x} + \frac{3 A}{5} e^{- 2 x} .$ $e^{3 x}$ term blow up karta hai jab tak uska coefficient zero na ho: $C_{1} = \frac{2 A}{5} = 0 \Rightarrow A = 0$ . Interpretation: $r_{1} > 0 > r_{2}$ ke saath yeh ek saddle hai. Sirf exact starting choice $A = 0$ (jisse $y \equiv 0$ milta hai) stable direction par land karta hai; koi bhi $A \neq = 0$ growing mode ko excite karta hai. Figure dekho.

Recall Solution 5.3

Characteristic: $3 r^{2} - r - 2 = 0$ . Quadratic formula: $r = \frac{1 \pm 1 + 24}{6} = \frac{1 \pm 5}{6}$ . $r_{1} = 1, r_{2} = - \frac{2}{3}$ . $y = C_{1} e^{x} + C_{2} e^{- 2 x /3}$ , $y^{'} = C_{1} e^{x} - \frac{2}{3} C_{2} e^{- 2 x /3}$ . $x = 0$ par: $C_{1} + C_{2} = 1$ , $C_{1} - \frac{2}{3} C_{2} = 0 \Rightarrow C_{1} = \frac{2}{3} C_{2}$ . Tab $\frac{2}{3} C_{2} + C_{2} = 1 \Rightarrow \frac{5}{3} C_{2} = 1 \Rightarrow C_{2} = \frac{3}{5}, C_{1} = \frac{2}{5}$ . $y = \frac{2}{5} e^{x} + \frac{3}{5} e^{- 2 x /3} .$

Recall Self-test recap (cloze)

Ek distinct root $r = 0$ solution ==the constant $1$ == contribute karta hai. Dono roots negative $\Leftrightarrow$ $b > 0$ aur == $c > 0$ == (monic). $e^{r_{1} x}, e^{r_{2} x}$ ka Wronskian == $(r_{2} - r_{1}) e^{(r_{1} + r_{2}) x}$ == ke barabar hota hai, nonzero iff $r_{1} \neq = r_{2}$ . Roots $r_{1}, r_{2}$ se monic characteristic polynomial == $(r - r_{1}) (r - r_{2})$ == hai.

Connections

Parent: Case 1 — woh theory jisko yeh exercises drill karte hain.
Characteristic equation — L4/L5 mein use hua two-way bridge.
Wronskian and linear independence — Exercise 4.3 ka tool.
Superposition principle — isliye har general solution $C_{1} y_{1} + C_{2} y_{2}$ hoti hai.
Exponential function and its derivative — isliye $e^{r x}$ ansatz hai.
Case 2 repeated real root / Case 3 complex conjugate roots — L1 ke discriminant siblings.