Second-order linear ODEs — superposition principle, general theory: Worked examples

4.6.9 · D3 · Maths › Ordinary Differential Equations › Second-order linear ODEs — superposition principle, general

Yeh page Second-order Linear ODEs — Superposition Principle & General Theory ki "koi case chhootega nahi" companion hai. Wahan humne machinery banaayi thi; yahan hum use har us tarah ke input ke against run karte hain jo ek second-order linear ODE de sakti hai.

Scenario matrix

Isse ek checklist ki tarah padho. Neeche har cell mein kam se kam ek worked example diya gaya hai.

Cell	Kya cheez isse special banati hai	Solution ka shape	Example
A — real distinct roots ( $r_{1} \neq = r_{2}$ )	discriminant $> 0$	$c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}$	Ex 1
B — repeated real root ( $r_{1} = r_{2}$ )	discriminant $= 0$ , degenerate	$(c_{1} + c_{2} x) e^{r x}$	Ex 2
C — complex roots ( $α \pm i β$ )	discriminant $< 0$ , oscillation	$e^{α x} (c_{1} cos β x + c_{2} sin β x)$	Ex 3
D — non-homogeneous, ordinary forcing	RHS $\neq = 0$ , koi overlap nahi	$y_{h} + y_{p}$	Ex 4
E — resonance (forcing = homogeneous soln)	naive guess fail ho jaata hai, $x$ se multiply karo	$y_{h} + x \cdot (\dots)$	Ex 5
F — variable coefficients (Euler)	$p, q$ , $x$ par depend karte hain	power-law solutions	Ex 6
G — discriminant ki sign boundary par	limiting behaviour jab case B→A/C	—	Ex 7
H — real-world word problem	spring / RLC circuit	case C in disguise	Ex 8

Degeneracy/limits ke liye in cells par dhyan do: B (roots collide karte hain), E (particular guess homogeneous se collide karti hai), G (cases ke beech kya hota hai).

Ex 1 — Cell A: real distinct roots

Worked example

$y^{''} + 5 y^{'} + 6 y = 0$ solve karo, with $y (0) = 1, y^{'} (0) = 0$ . Forecast: aage padhne se pehle apne dimag mein do roots guess karo — $r^{2} + 5 r + 6$ factor karo.

Step 1. Trial $y = e^{r x}$ substitute karo. Tab $y^{'} = r e^{r x}$ , $y^{''} = r^{2} e^{r x}$ , aur equation ban jaati hai $(r^{2} + 5 r + 6) e^{r x} = 0$ . Yeh step kyun? $e^{r x}$ woh ek function hai jiske derivatives sirf apni copies hoti hain, isliye calculus collapse ho jaata hai Characteristic equation — constant coefficient ODEs ki algebra mein. Kyunki $e^{r x} \neq = 0$ , hum isse cancel kar sakte hain.

Step 2. $r^{2} + 5 r + 6 = (r + 2) (r + 3) = 0 \Rightarrow r = - 2, r = - 3$ solve karo. Yeh step kyun? Do distinct real roots matlab do genuinely alag exponential solutions.

Step 3. Fundamental set $y_{1} = e^{- 2 x}, y_{2} = e^{- 3 x}$ ; independence Wronskian se check karo: $W = e^{- 2 x} - 2 e^{- 2 x} e^{- 3 x} - 3 e^{- 3 x} = - 3 e^{- 5 x} - (- 2 e^{- 5 x}) = - e^{- 5 x} \neq = 0.$ Yeh step kyun? $W \neq = 0$ (kahin bhi) ek fundamental set certify karta hai, isliye $y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x}$ har solution hai.

Step 4. ICs apply karo. $y (0) = c_{1} + c_{2} = 1$ . $y^{'} = - 2 c_{1} e^{- 2 x} - 3 c_{2} e^{- 3 x}$ , isliye $y^{'} (0) = - 2 c_{1} - 3 c_{2} = 0$ . Solve karne par: $c_{2} = - 2$ , $c_{1} = 3$ . Yeh step kyun? Do initial conditions do constants ko uniquely pin karte hain (Existence & Uniqueness).

Answer: $y = 3 e^{- 2 x} - 2 e^{- 3 x} .$

Verify: $y (0) = 3 - 2 = 1$ ✔; $y^{'} (0) = - 6 + 6 = 0$ ✔. Aur $y^{''} + 5 y^{'} + 6 y$ : har $e^{- 2 x}$ term $(4 - 10 + 6) = 0$ contribute karta hai, har $e^{- 3 x}$ term $(9 - 15 + 6) = 0$ contribute karta hai. ✔

Ex 2 — Cell B: repeated (degenerate) root

Worked example

$y^{''} - 4 y^{'} + 4 y = 0$ solve karo, with $y (0) = 1, y^{'} (0) = 5$ . Forecast: yahan discriminant exactly zero hai. Jab do roots ek mein merge ho jaayein to doosra solution kya use karte ho?

Step 1. Characteristic equation: $r^{2} - 4 r + 4 = (r - 2)^{2} = 0 \Rightarrow r = 2$ (twice). Yeh step kyun? Discriminant $b^{2} - 4 a c = 16 - 16 = 0$ : ek double root. Yeh woh degenerate case hai jahan naive method sirf ek function $e^{2 x}$ deta hai.

Step 2. Humein ek doosra independent solution chahiye. Claim: $y_{2} = x e^{2 x}$ kaam karta hai. Substitution se check karo: $y_{2}^{'} = (1 + 2 x) e^{2 x}$ , $y_{2}^{''} = (4 + 4 x) e^{2 x}$ , isliye $y_{2}^{''} - 4 y_{2}^{'} + 4 y_{2} = e^{2 x} [(4 + 4 x) - 4 (1 + 2 x) + 4 x] = e^{2 x} \cdot 0 = 0.✔$ Yeh step kyun? Jab roots collide karte hain, Variation of parameters (ya ek limiting argument) humein bolta hai ki repeated solution ko $x$ se multiply karo — yeh $e^{2 x}$ se independent hai kyunki $x$ ek constant nahi hai.

Step 3. General solution $y = (c_{1} + c_{2} x) e^{2 x}$ . ICs apply karo: $y (0) = c_{1} = 1$ . $y^{'} = (c_{2} + 2 c_{1} + 2 c_{2} x) e^{2 x}$ , isliye $y^{'} (0) = c_{2} + 2 c_{1} = 5 \Rightarrow c_{2} = 3$ . Yeh step kyun? Wahi "do ICs → do constants" logic; sirf shape badli.

Answer: $y = (1 + 3 x) e^{2 x} .$

Verify: $y (0) = 1$ ✔; $y^{'} (0) = 3 + 2 = 5$ ✔.

Ex 3 — Cell C: complex roots (oscillation)

Worked example

$y^{''} + 2 y^{'} + 5 y = 0$ solve karo, with $y (0) = 0, y^{'} (0) = 4$ . Forecast: discriminant negative → roots complex hain. Kya answer decay karega, grow karega, ya oscillate karega? Dono, actually.

Step 1. Characteristic: $r^{2} + 2 r + 5 = 0$ . Quadratic formula se $r = \frac{- 2 \pm 4 - 20}{2} = \frac{- 2 \pm - 16}{2} = - 1 \pm 2 i$ . Yeh step kyun? $- 16 = 4 i$ jahan $i^{2} = - 1$ (the imaginary unit). Negative discriminant ek complex conjugate pair $α \pm i β$ force karta hai jahan $α = - 1, β = 2$ .

Step 2. Real form mein convert karo. $e^{(- 1 + 2 i) x} = e^{- x} (cos 2 x + i sin 2 x)$ Euler's formula se. Real aur imaginary parts lene par do real solutions milte hain $e^{- x} cos 2 x$ aur $e^{- x} sin 2 x$ . Isliye $y = e^{- x} (c_{1} cos 2 x + c_{2} sin 2 x) .$ Yeh step kyun? Hum ek real physical quantity ke liye real functions chahte hain. $α = - 1$ decay rate hai; $β = 2$ oscillation frequency hai. Neeche picture dekho: exponential envelope ek sine ko squeeze kar rahi hai.

Step 3. ICs. $y (0) = c_{1} = 0$ . Differentiate karo: $y^{'} = e^{- x} [(- c_{1} + 2 c_{2}) cos 2 x + (- c_{2} - 2 c_{1}) sin 2 x]$ . $x = 0$ par: $y^{'} (0) = - c_{1} + 2 c_{2} = 4$ . $c_{1} = 0$ ke saath, $c_{2} = 2$ . Yeh step kyun? Standard two-IC pinning; derivative sine aur cosine ko mix karta hai, isliye dono ka present hona zaroori hai.

Answer: $y = 2 e^{- x} sin 2 x .$

Verify: $y (0) = 0$ ✔; $y^{'} (0) = 2 \cdot 1 \cdot 1 = 2 cos 0 = 2$ … ruko — $y^{'} = 2 e^{- x} (- sin 2 x + 2 cos 2 x)$ , isliye $y^{'} (0) = 2 (0 + 2) = 4$ ✔.

Ex 4 — Cell D: non-homogeneous, ordinary forcing

Worked example

$y^{''} + y = 3 x$ solve karo (general solution). Forecast: forcing $3 x$ ek straight line hai. Compute karne se pehle ek particular solution ki shape guess karo.

Step 1. Homogeneous part: $r^{2} + 1 = 0 \Rightarrow r = \pm i$ , isliye $y_{h} = c_{1} cos x + c_{2} sin x$ . Yeh step kyun? Structure theorem ke anusaar, har solution $y_{h} + y_{p}$ hai. Pehle $y_{h}$ banao.

Step 2. Method of undetermined coefficients se particular solution: RHS degree-1 polynomial hai, isliye $y_{p} = A x + B$ try karo. Tab $y_{p}^{''} = 0$ , aur $0 + (A x + B) = 3 x$ se $A = 3, B = 0$ force hota hai. Yeh step kyun? $L$ yahaan polynomials ko same degree ke polynomials mein map karta hai, isliye ek linear guess operation ke under closed hai — kuch fancier zaroori nahi. Importantly, $x$ ek homogeneous solution nahi hai (woh sines/cosines hain), isliye koi resonance nahi.

Step 3. Assemble karo: $y = c_{1} cos x + c_{2} sin x + 3 x .$ Yeh step kyun? Structure theorem: general = homogeneous + ek particular.

Verify: $y^{''} = - c_{1} cos x - c_{2} sin x$ , isliye $y^{''} + y = (- c_{1} cos x - c_{2} sin x) + (c_{1} cos x + c_{2} sin x + 3 x) = 3 x$ ✔.

Ex 5 — Cell E: resonance (guess collide kar jaati hai)

Worked example

$y^{''} + y = cos x$ solve karo (general solution). Forecast: "obvious" guess $y_{p} = A cos x + B sin x$ try karo — aur dekho yeh kaise nonsense mein explode hoti hai. Kyun?

Step 1. Homogeneous: $y_{h} = c_{1} cos x + c_{2} sin x$ (Ex 4 jaisa hi).

Step 2. Forcing $cos x$ already ek homogeneous solution hai. $A cos x + B sin x$ ko $L = \frac{d ^{2}}{d x ^{2}} + 1$ mein plug karne par exactly $0$ milta hai — right par $cos x$ produce karna kabhi possible nahi. Yeh resonance hai. Yeh step kyun? Jab input frequency natural frequency se match karti hai, naive family $L$ ke kernel mein hoti hai. Fix (Method of undetermined coefficients se) yeh hai ki trial ko $x$ se multiply karo.

Step 3. $y_{p} = x (A cos x + B sin x)$ try karo. Differentiate karo: $y_{p}^{'} = A cos x + B sin x + x (- A sin x + B cos x),$ $y_{p}^{''} = - 2 A sin x + 2 B cos x + x (- A cos x - B sin x) .$ Tab $y_{p}^{''} + y_{p} = - 2 A sin x + 2 B cos x$ . $cos x$ ke barabar set karo: $- 2 A = 0, 2 B = 1 \Rightarrow A = 0, B = \frac{1}{2}$ . Yeh step kyun? $x$ -multiplied terms $y_{p}$ ke saath cancel ho jaate hain, sirf derivative-of-envelope terms bachte hain — exactly wahi jo forcing match karne ke liye survive karte hain.

Answer: $y = c_{1} cos x + c_{2} sin x + \frac{x}{2} sin x .$

Verify: $y_{p} = \frac{1}{2} x sin x$ ke saath: $y_{p}^{'} = \frac{1}{2} sin x + \frac{1}{2} x cos x$ , $y_{p}^{''} = cos x - \frac{1}{2} x sin x$ , isliye $y_{p}^{''} + y_{p} = cos x$ ✔. Amplitude $x$ ke saath linearly badhti hai — resonance ki physical signature.

Ex 6 — Cell F: variable coefficients (Euler equation)

Worked example

$x^{2} y^{''} - 2 x y^{'} + 2 y = 0$ solve karo $x > 0$ ke liye. Forecast: yahan $p, q$ , $x$ par depend karte hain, isliye exponentials galat guess hain. $x$ ki kaunsi power kaam kar sakti hai?

Step 1. Coefficients aise scale karte hain ki agar $y = x^{m}$ ho to har term degree-matched ho jaati hai: tab $x^{2} y^{''} = m (m - 1) x^{m}$ , $- 2 x y^{'} = - 2 m x^{m}$ , $2 y = 2 x^{m}$ . Yeh step kyun? $x^{r}$ Euler equations ke liye wahi hai jo $e^{r x}$ constant-coefficient equations ke liye hai — har derivative power ko ek se kam karta hai aur ek factor se multiply karta hai, isliye $x^{2} \cdot (stuff)$ wahi power $x^{m}$ restore kar deta hai. Sab kuch phir se algebra mein collapse ho jaata hai.

Step 2. $x^{m}$ factor out karo: $m (m - 1) - 2 m + 2 = m^{2} - 3 m + 2 = (m - 1) (m - 2) = 0 \Rightarrow m = 1, 2$ . Yeh step kyun? Do distinct real exponents → do independent power-law solutions $x^{1}$ aur $x^{2}$ .

Step 3. General solution $y = c_{1} x + c_{2} x^{2}$ . Wronskian se independence check karo: $W = x 1 x^{2} 2 x = 2 x^{2} - x^{2} = x^{2} \neq = 0 (for x > 0) .✔$ Yeh step kyun? Interval $x > 0$ par nonzero Wronskian ek fundamental set confirm karta hai (compare Abel's theorem, jo explain karta hai ki $W$ sneakily zero kyun nahi ho sakta).

Answer: $y = c_{1} x + c_{2} x^{2} .$

Verify: $x^{2}$ ke liye: $x^{2} (2) - 2 x (2 x) + 2 x^{2} = 2 x^{2} - 4 x^{2} + 2 x^{2} = 0$ ✔. $x$ ke liye: $x^{2} (0) - 2 x (1) + 2 x = 0$ ✔.

Ex 7 — Cell G: cases ke beech boundary (limiting behaviour)

Worked example

$y^{''} + 2 ε y^{'} + y = 0$ mein kya hota hai jab damping $ε$ $1$ cross karta hai? Forecast: exactly $ε = 1$ par tum oscillation (Cell C) aur pure decay (Cell A) ke beech fence par baithe ho. Pehle guess karo kaunsa side kaunsa hai.

Step 1. Characteristic roots: $r = - ε \pm ε^{2} - 1$ . Yeh step kyun? Poori kahani discriminant $ε^{2} - 1$ ki sign par hai.

Step 2. Teen regimes:

$ε < 1$ (Cell C): $ε^{2} - 1 < 0$ , roots $- ε \pm i 1 - ε^{2}$ → decaying oscillation.
$ε = 1$ (Cell B): discriminant exactly $0$ , double root $r = - 1$ → critically damped $(c_{1} + c_{2} x) e^{- x}$ .
$ε > 1$ (Cell A): $ε^{2} - 1 > 0$ , do real negative roots → overdamped, koi oscillation nahi.

Yeh step kyun? Yeh exactly woh limit hai jahan Cell C, Cell B mein degenerate hoti hai aur phir Cell A mein split ho jaati hai. Jab $ε \to 1^{-}$ , frequency $1 - ε^{2} \to 0$ : oscillation slow hokar ruk jaati hai, aur do complex roots real axis par slide karke saath aa jaate hain.

Verify ( $ε = 1$ par): $r^{2} + 2 r + 1 = (r + 1)^{2}$ se double root $r = - 1$ ; solution $(c_{1} + c_{2} x) e^{- x}$ ODE satisfy karta hai (Ex 2 jaisa hi check pattern). $ε = 0.5$ par: roots $- 0.5 \pm i 0.866$ , frequency $\approx 0.866 > 0$ (oscillates). $ε = 2$ par: roots $- 2 \pm 3$ , dono real negative (decays, koi wiggle nahi). ✔

Ex 8 — Cell H: real-world spring/RLC word problem

Worked example Ek mass ek spring par

$m \overset{u}{¨} + c \overset{u}{˙} + k u = 0$ follow karta hai jahan $m = 1 kg$ , $c = 2 kg/s$ , $k = 5 N/m$ . Mass $u (0) = 0.1 m$ se start karta hai, rest mein $\overset{u}{˙} (0) = 0$ . $u (t)$ nikalo. Forecast: damping $c = 2$ , stiffness $k = 5$ — kya yeh over-, critically-, ya under-damped hai? Compute karne se pehle guess karo.

Step 1. $m = 1$ se divide karo: $\overset{u}{¨} + 2 \overset{u}{˙} + 5 u = 0$ . Characteristic $r^{2} + 2 r + 5 = 0 \Rightarrow r = - 1 \pm 2 i$ (yeh exactly Ex 3 ki equation hai!). Yeh step kyun? Damped spring ke liye Newton's law ek constant-coefficient second-order linear ODE hai — wahi machine, physical meaning attached. Discriminant $4 - 20 < 0$ ⇒ underdamped (woh decay hote hue oscillate karega).

Step 2. General solution $u (t) = e^{- t} (c_{1} cos 2 t + c_{2} sin 2 t)$ . Decay rate $1 s^{- 1}$ , angular frequency $2 rad/s$ . Yeh step kyun? Real form wahi hai jo ek displacement (ek real, measurable metres value) demand karta hai.

Step 3. ICs. $u (0) = c_{1} = 0.1$ . $\overset{u}{˙} = e^{- t} [(- c_{1} + 2 c_{2}) cos 2 t + (- c_{2} - 2 c_{1}) sin 2 t]$ , isliye $\overset{u}{˙} (0) = - c_{1} + 2 c_{2} = 0 \Rightarrow c_{2} = c_{1} /2 = 0.05$ . Yeh step kyun? Rest se start karna ( $\overset{u}{˙} (0) = 0$ ) aur ek known displacement do zaroori data deta hai.

Answer: $u (t) = e^{- t} (0.1 cos 2 t + 0.05 sin 2 t) m .$

Verify: units — har term metres mein hai (coefficients metres, trig/exp dimensionless), $t$ seconds mein dimensionless $e^{- t}$ aur $cos 2 t$ ke andar ✔. $u (0) = 0.1$ m ✔. $\overset{u}{˙} (0) = - 0.1 + 2 (0.05) = 0$ ✔. Physically: amplitude har second $e^{- 1} \approx 0.37$ factor se shrink hota hai wobbling karte hue — ek plausible damped spring.

Recall Kaunsa root case kaun si shape deta hai?

Do real distinct roots → sum of exponentials ::: $c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}$ (Cell A) Repeated real root $r$ → ::: $(c_{1} + c_{2} x) e^{r x}$ (Cell B, $x$ se multiply karo) Complex $α \pm i β$ → ::: $e^{α x} (c_{1} cos β x + c_{2} sin β x)$ (Cell C) Forcing ek homogeneous solution ke barabar → ::: resonance, trial ko $x$ se multiply karo (Cell E)

Parent par wapas jaao: Second-order Linear ODEs — Superposition Principle & General Theory. Har "yeh saari solutions capture karta hai" ke peeche existence guarantee ke liye, Existence and uniqueness theorems for ODEs dekho; do-constant family ki vector-space framing ke liye, Linear algebra — vector spaces and bases dekho.