Case 2 - repeated real root — reduction of order: Visual walkthrough

4.6.12 · D2 · Maths › Ordinary Differential Equations › Case 2 - repeated real root — reduction of order

Step 1 — Woh machine kya hai jo hum solve kar rahe hain?

KYA HAI. Humare paas ek equation hai jo ek function $y (x)$ , uski slope $y^{'}$ , aur slope ke jhukne ke tarike $y^{''}$ ko aapas mein jodhti hai: $stiffness a bending y^{''} + damping b slope y^{'} + spring c height y = 0.$

Har symbol padho:

$y$ position $x$ par curve ki height hai — socho ek mass spring par hai, $y$ batata hai woh kitna displaced hai.
$y^{'}$ ("y-prime") slope hai — height kitni tezi se change ho rahi hai. Yeh wahi ordinary rate-of-change hai jo tum speed ke roop mein jaante ho.
$y^{''}$ ("y-double-prime") bending hai — slope khud kitni tezi se change hoti hai. Yeh graph ko upar ya neeche curl karta hai.
$a, b, c$ fixed numbers hain (yeh kabhi $x$ ke saath change nahi hote) — iska matlab yahi hai "constant-coefficient."

KYUN. Ek second-order equation (sabse zyada tick $y^{''}$ hai, do primes) ko ek unique curve pin down karne ke liye do independent starting facts chahiye — usually starting height $y (0)$ aur starting slope $y^{'} (0)$ . Toh humein do independent building-block solutions chahiye honge. Yeh number yaad rakho: do.

PICTURE. Neeche, ek hilti hui curve $y (x)$ jisme uski height, slope arrow, aur bending arc mark kiya hua hai.

Figure — Case 2 - repeated real root — reduction of order

Step 2 — Trial $e^{r x}$ aur characteristic equation

KYA HAI. Hum guess karte hain ki solution ek exponential $y = e^{r x}$ hai, jahan $r$ ek number hai jo humein dhundhna hai. Letter $e \approx 2.718$ woh special base hai jiska exponential apni khud ki slope hai.

Yeh guess itni achhi kyun hai? Kyunki $y = e^{r x}$ ke liye: $y^{'} = r e^{r x}, y^{''} = r^{2} e^{r x} .$ Har derivative sirf ek factor of $r$ neeche kheench leti hai aur $e^{r x}$ ko bina chhuye chhod deti hai. Toh differentiate karne se kabhi koi naya type ka function nahi banta — sab kuch $e^{r x}$ ke proportional rehta hai.

$a y^{''} + b y^{'} + cy = 0$ mein substitute karo aur common $e^{r x}$ (jo kabhi zero nahi hota) factor out karo: $e^{r x} (yeh 0 hona chahiye a r^{2} + b r + c) = 0 \Rightarrow a r^{2} + b r + c = 0.$

Yeh aakhri line characteristic equation hai. Dekho Characteristic equation of linear ODEs.

KYUN yeh tool (ek quadratic)? Kyunki yeh ek calculus problem (derivatives) ko ek plain algebra problem (quadratic ke roots dhundho) mein convert karta hai. Yeh ek bahut bada simplification hai, aur yeh sirf isliye kaam karta hai kyunki exponentials differentiation ke under khud ko reproduce karte hain.

PICTURE. Exponential curve aur "root-machine" jo derivatives ko $r$ ki powers mein convert karti hai.

Step 3 — Repeated root: jahan do roots aapas mein takraate hain

KYA HAI. Quadratic $a r^{2} + b r + c = 0$ ke roots jaane-paehchaane formula se milte hain: $r = \frac{- b \pm b ^{2} - 4 a c}{2 a} .$ Root ke neeche ki quantity, $b^{2} - 4 a c$ , discriminant $Δ$ hai. Yeh decide karta hai ki kitne roots milenge:

$Δ > 0$ → do alag roots (Case 1 distinct real roots),
$Δ < 0$ → do complex roots (Case 3 complex roots),
$Δ = 0$ → $\pm Δ$ gayab ho jaata hai, toh do roots ek mein merge ho jaate hain: $r = - \frac{b}{2 a} .$

KYUN yeh matter karta hai. Jab $Δ = 0$ toh do roots same number hain. Trial $y = e^{r x}$ isliye humein sirf ek building block deta hai — lekin Step 1 ne maanga tha ki do chahiye. Yahi Case 2 ka poora crisis hai: hum ek solution short hain.

PICTURE. Ek number line jisme do roots $Δ \to 0$ hone par ek doosre ki taraf slide karte hain aur ek single dot mein fuse ho jaate hain.

Step 4 — Reduction of order partner manufacture karta hai

KYA HAI. Humare paas already ek solution $y_{1} = e^{r x}$ hai. Doosre ko us same solution ke roop mein guess karo jo ek unknown wobble $v (x)$ se scale kiya gaya ho: $y_{2} = v (x) e^{r x} .$ Yahan $v (x)$ ek mystery function hai jise hum solve karenge. Yeh Reduction of order (general method) hai.

Product rule se differentiate karo (neeche har term label kiya gaya hai kahan se aaya): $y_{2}^{'} = e^{r x} (wobble changes v^{'} + exp changes r v),$ $y_{2}^{''} = e^{r x} (wobble bends v^{''} + cross term 2 r v^{'} + exp bends r^{2} v) .$

$a y^{''} + b y^{'} + cy = 0$ mein substitute karo, $e^{r x}$ se divide karo, aur $v^{''}, v^{'}, v$ ke hisaab se gather karo: $a v^{''} + coeff of v^{'} (2 a r + b) v^{'} + coeff of v (a r^{2} + b r + c) v = 0.$

Ab repeated root ka magic do baar fire karta hai:

$a r^{2} + b r + c = 0$ kyunki $r$ ek root hai → poora $v$ -term mar jaata hai.
$2 a r + b = 2 a (- \frac{b}{2 a}) + b = 0$ kyunki $r = - b /2 a$ → poora $v^{'}$ -term bhi mar jaata hai.

Do coefficients ek saath vanish ho jaate hain — yeh sirf repeated root ke liye possible hai. Jo bachta hai woh hai: $a v^{''} = 0 \Rightarrow v^{''} = 0.$

KYUN yeh tool (reduction of order)? Kyunki hum pehle se ek solution jaante the aur use leverage ki tarah use karna chahte the. $y_{2} = v y_{1}$ assume karna guarantee karta hai ki $v$ ke liye bacha hua equation simpler hoga ("order reduced"). Ek distinct root ke liye, $2 a r + b \neq = 0$ , $v^{'}$ -term zinda rehta hai, aur $v$ exponential nikalta hai — iska matlab doosra root recover ho jaata hai. Sirf collision case mein ultra-simple $v^{''} = 0$ bachta hai.

PICTURE. Teeno grouped terms jisme do doomed ones strike out hain, sirf $a v^{''} = 0$ bachta hai.

Step 5 — $v^{''} = 0$ solve karna aur naya block padhna

KYA HAI. " $v^{''} = 0$ " kehta hai wobble ki zero bending hai — ek straight line: $v^{''} = 0 \Rightarrow v^{'} = C_{2} \Rightarrow v = C_{1} + C_{2} x .$ $e^{r x}$ se wapas multiply karo: $y = v e^{r x} = (C_{1} + C_{2} x) e^{r x} = purana block C_{1} e^{r x} + bilkul naya block C_{2} x e^{r x} .$

Genuinely naya function $== x e^{r x} ==$ hai — purana exponential jisme saamne $x$ staple ho gaya hai.

KYUN yeh ek doosra block count hota hai: yeh $e^{r x}$ se linearly independent hona chahiye — sirf uski rescaling nahi. Hum Wronskian $W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}$ se test karte hain. $y_{2}^{'} = e^{r x} (1 + r x)$ ke saath: $W = e^{r x} \cdot e^{r x} (1 + r x) - r e^{r x} \cdot x e^{r x} = e^{2 r x} [(1 + r x) - r x] = e^{2 r x} \neq = 0.$ Kabhi zero nahi → sach mein har jagah independent hai. ✓

PICTURE. $e^{r x}$ (ek pure curve) aur $x e^{r x}$ (rise karta hai, peak aata hai, phir exponential jeet jaata hai) ek saath plot kiye — visibly alag shapes, scalings nahi.

Step 6 — $x$ kyun? Coalescing-roots picture (degenerate limit)

KYA HAI. Ek second ke liye pretend karo ki do roots almost equal hain: $r$ aur $r + ε$ ek tiny $ε$ ke liye. Toh $e^{r x}$ aur $e^{(r + ε) x}$ dono solutions hain, toh unka difference $ε$ se divide karke bhi solution hai: $\frac{e ^{(r + ε) x} - e ^{r x}}{ε} .$ Yeh exactly derivative ki definition hai — lekin $x$ ke respect se nahi, root $r$ ke respect se. $ε \to 0$ let karo (roots fuse ho jaate hain): $lim_{ε \to 0} \frac{e ^{(r + ε) x} - e ^{r x}}{ε} = \frac{\partial}{\partial r} e^{r x} = x e^{r x} .$

KYUN yeh dikhana? Kyunki yeh explain karta hai ki $x$ physically kahan se aata hai: yeh solution family ki slope hai jab tum root ko thoda nudge karte ho. Jis instant do roots crash karte hain, woh slope direction — $x e^{r x}$ — naya doosra solution ban jaata hai. Step 4 ki algebra aur yeh limit perfectly agree karte hain.

PICTURE. Do nearby exponentials, unka scaled difference, aur limiting $x e^{r x}$ jisme woh converge karte hain.

Ek-picture summary

Poora pipeline compressed: ODE → uska quadratic → discriminant zero ho jaata hai → ek root → reduction of order do terms ko khatam karta hai → $v^{''} = 0$ → do blocks ${e^{r x}, x e^{r x}}$ .

Recall Feynman retelling — poori walkthrough plain words mein

Hum ek spring-jaisi equation solve kar rahe hain jo ek curve ki height, slope, aur bend ko mix karti hai. Ek neat trick: exponential $e^{r x}$ guess karo, kyunki exponentials slopes lene par apni shape maintain karte hain — har derivative sirf $r$ se multiply kar deti hai. Yeh calculus ko ek plain quadratic $a r^{2} + b r + c = 0$ mein convert kar deta hai. Usually quadratic ke do answers hote hain → do building blocks → kaam khatam. Lekin jab do answers same number mein crash kar jaate hain (discriminant zero), toh sirf ek block milta hai, aur ek second-order equation do ki maang karta hai. Toh hum apna ek block $e^{r x}$ lete hain, use ek unknown wobble $v (x)$ se multiply karte hain, aur wapas andar daal dete hain. Kyunki root repeated hai, do poore terms cancel ho jaate hain aur hum baby equation ke saath bache rehte hain " $v$ ki koi bend nahi hai," yani $v$ ek straight line $C_{1} + C_{2} x$ hai. Wapas multiply karne par doosra block milta hai: purana exponential jisme $x$ glue ho gaya hai, $x e^{r x}$ . Wronskian confirm karta hai ki yeh genuinely naya hai. Aur agar tum soch rahe ho ki $x$ kyun aata hai — imagine karo do roots almost-but-not-quite equal hain aur unhe slide karke together laao; unke do solutions ke beech ka gap, rescaled, exactly $x e^{r x}$ ban jaata hai. Yahi missing partner hai, aur ab tum koi bhi solution build kar sakte ho.

Connections

Case 2 - repeated real root — reduction of order — parent topic (woh result jise yeh page draw karta hai).
Characteristic equation of linear ODEs — Step 2, jahan $r$ paida hota hai.
Case 1 distinct real roots — $Δ > 0$ sibling (roots alag rehte hain).
Case 3 complex roots — $Δ < 0$ sibling.
Reduction of order (general method) — Step 4 ka engine.
Wronskian and linear independence — Step 5 mein independence check.
Higher-order repeated roots — $x$ -trick repeat karo: $e^{r x}, x e^{r x}, x^{2} e^{r x}, \dots$