Worked examples — Case 2 - repeated real root — reduction of order

4.6.12 · D3 · Maths › Ordinary Differential Equations › Case 2 - repeated real root — reduction of order

Yeh page Case 2 (repeated real root) ka drill floor hai. Parent note ne kyun prove kiya tha ki do solutions $e^{rx}$ aur $x{e}^{rx}$ hain. Yahan hum ensure karte hain ki problem ka koi bhi version aisa na ho jo tumne pehle dekha na ho.

Shuru karne se pehle, ek waada: neeche jo bhi symbol dikhe, woh sab parent mein define kiye gaye hain. Agar characteristic equation kahan se aati hai iske baare mein tumhara doubt hai, toh pehle Characteristic equation of linear ODEs dekh lo.

---

Scenario matrix

Ek "repeated real root" problem ek problem nahi hai — yeh ek family hai. Mushkil packaging mein chupi hoti hai: kya $a=1$ hai ya nahi, kya $r$ positive, negative ya zero hai, kya general solution maangi hai ya initial/boundary conditions diye gaye hain, aur kya poori cheez ek word problem ki tarah present ki gayi hai.

Yeh pura grid hai. Har cell ka kam se kam ek fully worked example neeche diya gaya hai.

Cell	Kya alag hai	Kaunsa trap set karta hai	Example
C1	Clean $a=1$, positive root $r>0$	koi nahi — warm-up	Ex 1
C2	$a=1$, negative root $r<0$	$e^{rx}$ mein $r$ ka sign	Ex 2
C3	Repeated root $r=0$ (degenerate)	$e^{0x}=1$, toh solutions hain $1$ aur $x$	Ex 3
C4	Leading coefficient $a\ne 1$	$r=-b/2a$ ke liye carefully divide/factor karna padega	Ex 4
C5	Initial-value problem (ek point par conditions)	$y^{\prime }(0)$ sahi se apply karna	Ex 5
C6	Boundary-value problem (do points par conditions)	$2\times 2$ system solve karna	Ex 6
C7	Limiting / coalescing roots viewpoint	$x{e}^{rx}$ ko limit ke roop mein emerge hote dekhna	Ex 7
C8	Word problem (physics: critical damping)	words → ODE → root translate karna	Ex 8
C9	Exam twist — verify karo ki ek given function isse solve karta hai	$a,b,c$ reverse-engineer karna	Ex 9

Figure s01 — examples se pehle yeh padho. Plot Case 2 general solution $y=(C_1+C_{2}x)e^{rx}$ ki shape dikhata hai repeated root ke teen sign-classes ke liye. Solid black curve positive root ($r>0$) hai: yeh bina bound ke grow karti hai. Red curve negative root ($r<0$) hai: yeh thodi der upar jaati hai phir zero par decay ho jaati hai — yeh critical-damping shape hai jo Ex 5, Ex 6 aur Ex 8 mein milti hai, aur red arrow us monotone decay ki taraf point karta hai. Dashed black line degenerate root $r=0$ hai: exponential $1$ hai, isliye solution sirf straight line $1+x$ hai (Ex 3). Ek nazar se pata chal jaata hai ki tum matrix ke kis cell mein khade ho sirf $r$ ke sign se.

---

Worked Examples

Ex 1 — Cell C1: clean positive root

Forecast: aage padhne se pehle root guess karo — $25$ ek perfect square hai, toh $r$ kya hai?

1. Characteristic equation likho. $y^{\prime \prime }\to r^2$, $y^{\prime }\to r$, $y\to 1$ replace karo: $r^2-10r+25=0.$ Yeh step kyun? Trial $y=e^{rx}$ derivatives ko $r$ ki powers mein badal deta hai (dekho Characteristic equation of linear ODEs).
2. Discriminant check karo. $(-10{)}^2-4(1)(25)=100-100=0$. Yeh step kyun? $b^2-4ac=0$ ek repeated root ka signature hai — yeh batata hai ki hum Case 2 mein hain, Case 1 mein nahi.
3. Perfect square ki tarah factor karo. $r^2-10r+25=(r-5{)}^2=0\Rightarrow r=5$ (double). Yeh step kyun? $(r-r_0{)}^2$ ki tarah factor karne se root sahi sign ke saath directly mil jaata hai, koi formula nahi chahiye.
4. Do solutions assemble karo Case 2 rule use karke: $e^{5x}$ aur $x{e}^{5x}$. Yeh step kyun? 2nd-order ODE ko do independent solutions chahiye; repeated root ek exponential deta hai, aur Case 2 rule missing partner supply karta hai $x$ lagake.
5. General solution: $\boxed{y=(C_1+C_{2}x)e^{5x}}$.

Verify: maano $y_2=x{e}^{5x}$. Tab $y_2^{\prime }=e^{5x}(1+5x)$, $y_2^{\prime \prime }=e^{5x}(10+25x)$. Substitute karo: $e^{5x}[(10+25x)-10(1+5x)+25x]=e^{5x}[10+25x-10-50x+25x]=0$. ✓

---

Ex 2 — Cell C2: negative repeated root

Forecast: beech ka sign $+$ hai. Root positive hoga ya negative?

1. Characteristic equation: $r^2+8r+16=0$. Yeh step kyun? Trial $y=e^{rx}$ ke under $y^{\prime \prime }\to r^2,y^{\prime }\to r,y\to 1$ replace karne se ODE ek algebra problem ban jaata hai jiske roots exponents hain.
2. Discriminant: $8^2-4(16)=64-64=0$ — repeated root confirm. Yeh step kyun? $b^2-4ac=0$ exactly woh condition hai jo humein Case 2 mein daalti hai (do roots nahi, ek root).
3. Factor: $(r+4{)}^2=0\Rightarrow r=-4$. Yeh step kyun? $+8r$ opposite sign ka root produce karta hai, $-4$. Yahan exactly students galti karte hain — sign flip ho jaata hai.
4. General solution: $\boxed{y=(C_1+C_{2}x)e^{-4x}}$. Yeh step kyun? Repeated root $r=-4$ haath mein aane ke baad, Case 2 pair hai $e^{-4x}$ aur $x{e}^{-4x}$, do free constants ke saath combine karo.

Verify: $y=e^{-4x}$ se: $y^{\prime \prime }+8y^{\prime }+16y=16e^{-4x}+8(-4)e^{-4x}+16e^{-4x}=(16-32+16)e^{-4x}=0$. ✓

---

Ex 3 — Cell C3: degenerate root $r=0$

Forecast: yeh bahut simple lagta hai. Iske do solutions kya hain, aur kya Case 2 rule abhi bhi apply hoga?

1. Standard form mein likho: $y^{\prime \prime }+0\cdot y^{\prime }+0\cdot y=0$, toh $a=1,b=0,c=0$. Yeh step kyun? $a,b,c$ explicitly name karne se hum unhe seedha characteristic equation aur discriminant mein daal sakte hain — machinery is shape mein equation expect karti hai.
2. Characteristic equation: $r^2=0\Rightarrow r=0$ (double root). Yeh step kyun? $b^2-4ac=0-0=0$, toh yeh hai ek repeated root — special value $r=0$ hai.
3. Case 2 rule mechanically apply karo: solutions hain $e^{0x}$ aur $x{e}^{0x}$. Yeh step kyun? Hum rule skip NAHI karte sirf isliye ki yeh trivial lagta hai — pair $\{e^{rx},x{e}^{rx}\}$ har repeated root ke liye holds karta hai, including $r=0$.
4. Simplify $e^{0x}=1$ ke saath: solutions hain $1$ aur $x$. Yeh step kyun? $e^{0x}=1$ exponentials ko collapse kar deta hai, solutions ko unke simplest form mein reveal karta hai.
5. General solution: $\boxed{y=C_1+C_{2}x}$ — ek straight line, exactly wahi jo $y^{\prime \prime }=0$ (zero curvature) ko dena chahiye.

Verify: $y=C_1+C_{2}x\Rightarrow y^{\prime \prime }=0$ identically. ✓

---

Ex 4 — Cell C4: leading coefficient $a\ne 1$

Forecast: leading $9$ $r$ extract karne ka tarika badal deta hai. Factor karne se pehle $r=-b/2a$ se $r$ predict karo.

1. Characteristic equation: $9r^2-30r+25=0$. Yeh step kyun? Same trial $y=e^{rx}$ ODE ko ek quadratic mein convert karta hai — lekin ab leading coefficient $9$ hai, isliye hum ise assume nahi kar sakte ki $a=1$.
2. Discriminant: $(-30{)}^2-4(9)(25)=900-900=0$ — repeated root. Yeh step kyun? $b^2-4ac=0$ with $a=9$ confirm karta hai ki hum abhi bhi Case 2 mein hain; non-unit leading coefficient classification nahi badalta, sirf arithmetic badalta hai.
3. $r$ nikalne ke do safe tarike:
   • Formula: $r=-\frac{b}{2a}=-\frac{-30}{2\cdot 9}=\frac{30}{18}=\frac{5}{3}$.
   • Perfect square: $9r^2-30r+25=(3r-5{)}^2=0\Rightarrow 3r=5\Rightarrow r=\frac{5}{3}$. Do tarike kyun? Yeh ek doosre ko cross-check karte hain. $a\ne 1$ ke saath perfect-square factor $(3r-5)$ hai, na ki $(r-5)$ — $3$ bhoolna classic galti hai.
4. General solution: $\boxed{y=(C_1+C_{2}x)e^{5x/3}}$. Yeh step kyun? Repeated root $r=\frac{5}{3}$ milne ke baad, Case 2 pair hai $e^{5x/3}$ aur $x{e}^{5x/3}$; leading $9$ ki aur koi role nahi rahi jab $r$ mil gaya.

Verify: $r=5/3$ ko $9r^2-30r+25=0$ satisfy karna chahiye: $9\cdot \frac{25}{9}-30\cdot \frac{5}{3}+25=25-50+25=0$. ✓

---

Ex 5 — Cell C5: initial-value problem

Forecast: tumhe $y=(C_1+C_{2}x)e^{2x}$ milega. Guess karo: kaunsi condition $C_1$ pin karti hai, aur kaunsi $C_2$ pin karti hai?

1. Characteristic: $r^2-4r+4=(r-2{)}^2=0\Rightarrow r=2$ (double). Yeh step kyun? Pehle general solution solve karna zaroori hai; tabhi conditions ke paas constants hote hain fix karne ke liye.
2. General solution: $y=(C_1+C_{2}x)e^{2x}$. Yeh step kyun? Repeated root $r=2$ Case 2 pair deta hai; do free constants woh hain jo do initial conditions determine karengi.
3. $y(0)=2$ apply karo. $x=0$ par: $e^0=1$ aur $C_{2}x$ term zero ho jaata hai. Toh $y(0)=C_1=2$. Yeh step kyun? $x=0$ par evaluate karne se $C_1$ isolate ho jaata hai kyunki $x$, $C_2$ ko multiply karta hai.
4. Differentiate karo product rule use karke: $y^{\prime }=C_2{e}^{2x}+2(C_1+C_{2}x)e^{2x}=e^{2x}[2C_1+C_2+2C_{2}x].$ Yeh step kyun? Doosri condition use karne ke liye humein $y^{\prime }$ chahiye.
5. $y^{\prime }(0)=5$ apply karo. $x=0$ par: $y^{\prime }(0)=2C_1+C_2=5$. $C_1=2$ ke saath: $4+C_2=5\Rightarrow C_2=1$. Yeh step kyun? Doosri condition $C_2$ solve karne ke liye ek remaining equation supply karti hai.
6. Answer: $\boxed{y=(2+x)e^{2x}}$.

Verify: $y(0)=2$ ✓. $y^{\prime }(x)=e^{2x}(2\cdot 2+1+2\cdot 1\cdot x)=e^{2x}(5+2x)$, toh $y^{\prime }(0)=5$ ✓.

---

Ex 6 — Cell C6: boundary-value problem

Forecast: yahan do conditions alag-alag $x$-values par hain. Matlab ek $2\times 2$ system hai, do-step substitution nahi.

1. Characteristic: $r^2+2r+1=(r+1{)}^2=0\Rightarrow r=-1$ (double). Yeh step kyun? Hamesha ki tarah, boundary values constants pin karne se pehle general solution pehle aani chahiye.
2. General solution: $y=(C_1+C_{2}x)e^{-x}$. Yeh step kyun? Repeated root $r=-1$ Case 2 pair $e^{-x},x{e}^{-x}$ deta hai do constants ke saath jo do boundary conditions fix karengi.
3. $y(0)=1$ apply karo: $C_1{e}^0=C_1=1$. Yeh step kyun? Pehle ki tarah — $x=0$ par $x$-term vanish ho jaata hai, $C_1$ isolate ho jaata hai.
4. $y(1)=0$ apply karo: $y(1)=(C_1+C_2)e^{-1}=0$. Kyunki $e^{-1}\ne 0$, humein $C_1+C_2=0\Rightarrow C_2=-C_1=-1$ chahiye. Yeh step kyun? Ek exponential kabhi zero nahi hota, isliye bracket zero hona chahiye.
5. Answer: $\boxed{y=(1-x)e^{-x}}$.

Verify: $y(0)=1$ ✓. $y(1)=(1-1)e^{-1}=0$ ✓. Aur $y^{\prime \prime }+2y^{\prime }+y$: $y=(1-x)e^{-x}$ ke saath, $y^{\prime }=e^{-x}(-1)-(1-x)e^{-x}=e^{-x}(x-2)$, $y^{\prime \prime }=e^{-x}(1)-(x-2)e^{-x}=e^{-x}(3-x)$; sum $=e^{-x}[(3-x)+2(x-2)+(1-x)]=e^{-x}[3-x+2x-4+1-x]=0$ ✓.

---

Ex 7 — Cell C7: roots coalesce hote dekho

Forecast: target value $x{e}^{3x}$ hai $x=1$ par, yaani $1\cdot e^3=e^3\approx 20.0855$. Difference quotient ko is taraf badhte dekho.

1. Setup. Do distinct roots $3$ aur $3+\epsilon$ solutions $e^{3x}$ aur $e^{(3+\epsilon )x}$ dete hain. Unka difference quotient bhi ek solution hai (linear combination). Yeh step kyun? Yeh parent ka Derivation B hai — $x{e}^{3x}$ root ke respect mein solution family ka derivative hai.
2. $x=1$ par shrinking $\epsilon$ ke liye evaluate karo: $\frac{e^{3+\epsilon }-e^3}{\epsilon }.$
   • $\epsilon =0.1$: $\frac{e^{3.1}-e^3}{0.1}\approx 21.084$.
   • $\epsilon =0.01$: $\approx 20.186$.
   • $\epsilon =0.001$: $\approx 20.095$. Yeh step kyun? $e^3\approx 20.0855$ ki taraf march karte numbers dikhate hain ki limit hai $x{e}^{3x}$.
3. Conclusion: do exponentials $\{e^{3x},x{e}^{3x}\}$ mein fuse ho jaate hain — doosra solution collision se paida hota hai. Isliye factor $x$ ek lucky guess nahi balki ek limit hai.

Figure s02 — numbers kaisi dikhti hain. Red curve target limit function $x{e}^{3x}$ hai. Teen dashed black curves difference quotients $(e^{(3+\epsilon )x}-e^{3x})/\epsilon$ hain $\epsilon =0.6,0.3,0.1$ ke liye. Jaise-jaise $\epsilon$ shrink hota hai, har dashed curve red wale se aur zyada tightly chipakti hai; red arrow mark karta hai jahan woh $x=1$ ke paas close in karte hain. Dashed curves ko red limit par collapse hote dekhna Derivation B ka poora point hai: doosra solution $x{e}^{3x}$ literally emerge hota hai jab do roots slide karke saath aate hain.

Verify: exact limit $=\frac{\partial }{\partial r}{e}^r{|}_{r=3}=e^3$, upar ke numbers se match karta hai. ✓

---

Ex 8 — Cell C8: critical damping (word problem)

Forecast: in numbers se discriminant zero hoga — yeh critical damping hai, bina oscillate kiye rest par fastest return. Guess karo kya $X(t)$ kabhi zero cross karega.

1. Characteristic equation mein translate karo. $m=1,c=4,k=4$ ke saath: $r^2+4r+4=0$. Yeh step kyun? $\ddot{X}\to r^2$, $\dot{X}\to r$, $X\to 1$ — wahi $e^{rt}$ trial hamesha ki tarah, bas independent variable time $t$ hai.
2. Discriminant $=0$: $16-16=0\Rightarrow$ repeated root $(r+2{)}^2=0$, $r=-2$. Yeh step kyun? $c^2=4mk$ exactly critical damping condition hai; maths ise repeated root ki tarah jaanta hai.
3. General solution: $X(t)=(C_1+C_{2}t)e^{-2t}$. Yeh step kyun? Repeated root $r=-2$ Case 2 pair $e^{-2t},t{e}^{-2t}$ deta hai; constants do initial conditions se fix honge.
4. $X(0)=0.5$ apply karo: $C_1=0.5$. Yeh step kyun? $t=0$ par $C_{2}t$ term zero ho jaata hai aur $e^0=1$, $C_1$ isolate ho jaata hai.
5. $\dot{X}(0)=0$ apply karo. $\dot{X}=e^{-2t}[C_2-2(C_1+C_{2}t)]=e^{-2t}[C_2-2C_1-2C_{2}t]$. $t=0$ par: $C_2-2C_1=0\Rightarrow C_2=2(0.5)=1$. Yeh step kyun? "Rest se release" matlab zero initial velocity, jo exactly condition $\dot{X}(0)=0$ hai $C_2$ pin karne ke liye.
6. Answer: $\boxed{X(t)=(0.5+t)e^{-2t}\text{m}}$.

Interpretation: kyunki $0.5+t>0$ sabhi $t\ge 0$ ke liye aur $e^{-2t}>0$, mass kabhi zero cross nahi karta — yeh monotonically equilibrium par wapas slide karta hai bina overshoot ke. Yeh critical damping ka fingerprint hai (contrast karo oscillating Case 3 complex roots se).

Verify: $X(0)=0.5$ ✓. $\dot{X}(t)=e^{-2t}(1-2(0.5+t))=e^{-2t}(-2t)$, toh $\dot{X}(0)=0$ ✓. Units: $[t]$ seconds mein, $X$ metres mein. ✓

---

Ex 9 — Cell C9: exam twist (reverse-engineer)

Forecast: exponent $7x$ chilla raha hai $r=7$. Agar yeh repeated root hai, toh $b$ aur $c$ kya hone chahiye?

1. Root exponent se padho. Factor $e^{7x}$ aur $x$-term $(-2x)$ ki presence ek repeated root $r=7$ signal karti hai. Yeh step kyun? Sirf ek repeated root $(A+Bx)e^{rx}$ shape produce karta hai — isliye given function ki packaging root aur uski multiplicity batati hai.
2. Repeated root $\Rightarrow$ characteristic ek perfect square hai: $(r-7{)}^2=r^2-14r+49=0$. Yeh step kyun? $7$ par double root matlab quadratic exactly $(r-7{)}^2$ hai.
3. $r^2+br+c$ se match karo: $b=-14$, $c=49$. Yeh step kyun? Expanded square ke coefficients ko target polynomial se compare karne se seedha $b$ aur $c$ mil jaate hain.
4. Discriminant $=0$ confirm karo: $b^2-4c=196-196=0$ ✓ — genuinely repeated. Yeh step kyun? Yeh loop close karta hai: reconstruct kiye gaye coefficients khud Case 2 condition satisfy karne chahiye.
5. Answer: $\boxed{y^{\prime \prime }-14y^{\prime }+49y=0}$, root $r=7$ (double).

Verify: $y=(3-2x)e^{7x}$ plug back karo. $y^{\prime }=e^{7x}(-2)+7(3-2x)e^{7x}=e^{7x}(19-14x)$. $y^{\prime \prime }=e^{7x}(-14)+7(19-14x)e^{7x}=e^{7x}(119-98x)$. Tab $y^{\prime \prime }-14y^{\prime }+49y=e^{7x}[(119-98x)-14(19-14x)+49(3-2x)]=e^{7x}[119-98x-266+196x+147-98x]=e^{7x}[0]=0$ ✓.

---

Coverage self-check

Recall Kya humne har cell hit kiya?

Positive root ::: Ex 1 (C1) Negative root ::: Ex 2 (C2) Degenerate root $r=0$ ::: Ex 3 (C3) Leading coefficient $\ne 1$ ::: Ex 4 (C4) Initial-value problem ::: Ex 5 (C5) Boundary-value problem ::: Ex 6 (C6) Coalescing-roots limit ::: Ex 7 (C7) Physics word problem ::: Ex 8 (C8) Reverse-engineering twist ::: Ex 9 (C9)

---

Connections

• Case 2 - repeated real root — reduction of order (index 4.6.12) — parent theory jise yeh examples drill karte hain.
• Characteristic equation of linear ODEs — har example yahan se start hota hai.
• Case 1 distinct real roots — contrast karo jab discriminant $>0$ ho.
• Case 3 complex roots — Ex 8 ka oscillating cousin jab $c^2<4mk$ ho.
• Reduction of order (general method) — engine jo $x{e}^{rx}$ factor ke peeche hai.
• Wronskian and linear independence — prove karta hai ki yahan har pair genuinely do solutions hain.
• Higher-order repeated roots — agla tier: multiplicity $m$ deta hai $x^{m-1}{e}^{rx}$.