Non-homogeneous — method of undetermined coefficients (annihilator method): Worked examples

4.6.14 · D3 · Maths › Ordinary Differential Equations › Non-homogeneous — method of undetermined coefficients (annih

Intuition Ye page kis kaam ki hai

Parent note ne tumhe teen saaf examples diye the. Real problems thodi messy hoti hain: right-hand side $g (x)$ ek polynomial ho sakti hai, ek decaying exponential, chezon ka product, chezon ka sum, ya aisi function jo homogeneous solution ke saath collide kare. Ye page har tarah ke collision aur shape ka ek-ek example walk through karta hai, taaki koi bhi case aisa na ho jo tumne pehle na dekha ho.

Shuru karne se pehle, un symbols ki ek reminder jo hum use karte hain, seedhi baat mein:

$D$ matlab "ek derivative lo": $D [y] = \frac{d y}{d x}$ . Toh $D^{2} [y] = y^{''}$ . Ye D-operator hai.
$L$ hamaari machine hai, $D$ mein ek polynomial jisme constant numbers hain, jaise $L = D^{2} - 3 D + 2$ matlab $L [y] = y^{''} - 3 y^{'} + 2 y$ .
$g (x)$ forcing term hai — jo bhi $L [y] = g (x)$ ke right side par baithe.
$A$ (curly capital A) $g$ ka annihilator hai: sabse simple polynomial-in- $D$ machine jo $g$ ko pure $0$ mein badal de. Plain letters $A, B, C$ hum trial solution ke unknown coefficients ke liye reserve karte hain, taaki operator ko kabhi kisi number jaisa badge na mile.

Poora answer hamesha $y = y_{h} + y_{p}$ hota hai, jahan $y_{h}$ $L [y] = 0$ solve karta hai (homogeneous piece) aur $y_{p}$ ek particular solution hai.

Scenario matrix

Is topic ke har problem ka answer in cells mein se exactly ek mein aata hai. Neeche ke examples cell number se label hain taaki tum dekh sako ki poora space cover ho gaya hai.

#	Case class	Kya mushkil hai	Example
C1	Pure polynomial $g$	root $r = 0$ ; $D^{n + 1}$ chahiye; dekho kahin $r = 0$ pehle se $L$ mein toh nahi	Ex 1
C2	Exponential, koi overlap nahi	shape copy karo, sabse simple case	Ex 2
C3	Exponential, overlap (resonance)	$g$ ka root pehle se $L$ mein hai → $x$ se multiply karo	Ex 3
C4	Trig, no overlap	dono $cos$ aur $sin$ include karne chahiye	Ex 4
C5	Trig, overlap (pure resonance)	$\pm i β$ pehle se $L$ mein hai → $x$ se multiply karo; hamesha ke liye badhta hai	Ex 5
C6	Product $x^{n} e^{α x}$ (repeated-root annihilator)	$(D - α)^{n + 1}$ chahiye, sirf $(D - α)$ nahi	Ex 6
C7	Sum $g_{1} + g_{2}$ (superposition)	$A_{1} A_{2}$ se annihilate karo, trial forms add karo	Ex 7
C8	Word problem (real-world, initial conditions ke saath)	physics → ODE mein translate karo, phir method apply karo	Ex 8
C9	Exam twist: double overlap ( $x^{2}$ appears)	root already doubled in $L$ → $x^{2}$ se multiply karo	Ex 9
C10	Polynomial × trig: $g = x sin x$	product ko $(D^{2} + β^{2})^{2}$ chahiye; $x$ -times-both-trig carry karo	Ex 10
C11	$e^{α x} sin β x$ (damped oscillation drive)	complex roots $α \pm i β$ ; annihilator $(D - α)^{2} + β^{2}$	Ex 11

Hum in degenerate / limiting corners ko bhi cover karte hain: $g$ ek constant (Ex 1 ka sabse simple sub-case), $β \to$ natural frequency (Ex 5, resonance limit), aur $α = 0$ jo $x^{n} e^{α x}$ ko wapas polynomial bana deta hai (Ex 6 mein note kiya gaya).

C1 — Pure polynomial forcing

Step 1 — homogeneous part. Ye step kyun? Hum hamesha pehle $y_{h}$ nikalte hain, kyunki $y_{p}$ ka trial jo bhi $y_{h}$ already cover karta hai use avoid karna chahiye. $D^{2} - 3 D + 2 = (D - 1) (D - 2)$ , roots $1, 2$ : $y_{h} = c_{1} e^{x} + c_{2} e^{2 x} .$

Step 2 — $g$ ko annihilate karo. Ye step kyun? Yahan $n$ polynomial forcing term ki degree hai — $g = 2 x^{2}$ ke liye woh degree $n = 2$ hai. Degree $n$ ka polynomial root $r = 0$ ke $n + 1$ baar repeat hone se banta hai (root $0$ deta hai $1$ ; doubled deta hai $x$ ; tripled deta hai $x^{2}$ ). Killer hai $D^{n + 1}$ ; $n = 2$ ke saath $A = D^{3}$ . Check karo: $D^{3} [2 x^{2}] = 0$ kyunki degree-2 polynomial ko teen baar differentiate karne se woh wipe out ho jaata hai. ✓

Step 3 — bada homogeneous ODE. Ye step kyun? Dono taraf $A$ apply karne se right side $0$ ho jaata hai, toh ab hume sirf ek homogeneous equation solve karni hai jiske roots hum read off kar sakte hain. $D^{3} (D - 1) (D - 2) y = 0, roots new 0, 0, 0, 1, 2.$

Step 4 — sirf naye terms rakho. Ye step kyun? Roots $1, 2$ sirf $y_{h}$ rebuild karte hain aur $g$ mein kuch contribute nahi karte. Naya root $0$ (tripled) contribute karta hai $1, x, x^{2}$ . Toh: $y_{p} = A x^{2} + B x + C .$ (Ye forecast ka jawab hai: teeno chahiye, chahe $g$ mein sirf $x^{2}$ dikhta ho — kyunki $y^{''}$ aur $y^{'}$ lower powers ko neeche kheenchte hain.)

Step 5 — coefficients determine karo. Ye step kyun? Form fix ho gayi hai lekin numbers $A, B, C$ abhi nahi; original ODE mein substitute karne se woh pin down ho jaate hain. $y_{p}^{'} = 2 A x + B$ , $y_{p}^{''} = 2 A$ . $y^{''} - 3 y^{'} + 2 y$ mein substitute karo: $2 A - 3 (2 A x + B) + 2 (A x^{2} + B x + C) = 2 x^{2} .$ $x$ ki powers ke hisaab se group karo: $2 A x^{2} + (- 6 A + 2 B) x + (2 A - 3 B + 2 C) = 2 x^{2} + 0 x + 0.$ Match karo:

$x^{2}$ : $2 A = 2 \Rightarrow A = 1$ .
$x^{1}$ : $- 6 A + 2 B = 0 \Rightarrow B = 3$ .
$x^{0}$ : $2 A - 3 B + 2 C = 0 \Rightarrow 2 - 9 + 2 C = 0 \Rightarrow C = \frac{7}{2}$ .

Step 6 — combine karo. $y = c_{1} e^{x} + c_{2} e^{2 x} + x^{2} + 3 x + \frac{7}{2} .$

Recall Verify karo:

$y_{p} = x^{2} + 3 x + \frac{7}{2}$ wapas plug karo. $y_{p}^{''} = 2$ , $y_{p}^{'} = 2 x + 3$ . Phir $2 - 3 (2 x + 3) + 2 (x^{2} + 3 x + \frac{7}{2}) = 2 - 6 x - 9 + 2 x^{2} + 6 x + 7 = 2 x^{2}$ . ✓ $g$ se match karta hai.

C2 — Exponential, no overlap

Step 1 — homogeneous part. Ye step kyun? Same wajah jaise hamesha: trial choose karne se pehle hume $y_{h}$ pata hona chahiye, taaki koi bhi overlap detect ho sake. $D^{2} + 4 = 0 \Rightarrow r = \pm 2 i$ (ek pure imaginary pair, jo oscillation produce karta hai), toh $y_{h} = c_{1} cos 2 x + c_{2} sin 2 x$ .

Step 2 — $g$ ko annihilate karo. $(D + 1)$ kyun? $e^{- x}$ ka root $- 1$ hai, toh $A = D - (- 1) = D + 1$ use annihilate karta hai: $(D + 1) e^{- x} = - e^{- x} + e^{- x} = 0$ . ✓ Aur $- 1 \neq = \pm 2 i$ , toh koi overlap nahi — $x$ ki zaroorat nahi.

Step 3–4 — banao aur prune karo. Ye step kyun? $A L [y] = 0$ ke roots $- 1, \pm 2 i$ hain; $\pm 2 i$ ki pair sirf $y_{h}$ rebuild karti hai, toh sirf naya root $- 1$ hi $g$ produce kar sakta hai: $y_{p} = C e^{- x} .$

Step 5 — coefficient determine karo. Ye step kyun? Original ODE mein substitute karne se $C$ fix hota hai. $y_{p}^{''} = C e^{- x}$ , toh $y_{p}^{''} + 4 y_{p} = C e^{- x} + 4 C e^{- x} = 5 C e^{- x} = 3 e^{- x} \Rightarrow C = \frac{3}{5} .$

Step 6 — combine karo. $y = c_{1} cos 2 x + c_{2} sin 2 x + \frac{3}{5} e^{- x} .$

Recall Verify karo:

$y_{p} = \frac{3}{5} e^{- x}$ . $y_{p}^{''} = \frac{3}{5} e^{- x}$ , toh $y_{p}^{''} + 4 y_{p} = \frac{3}{5} e^{- x} + \frac{12}{5} e^{- x} = \frac{15}{5} e^{- x} = 3 e^{- x}$ . ✓

C3 — Exponential overlap (single resonance)

Step 1 — homogeneous part. Ye step kyun? Hume $y_{h}$ shuru mein chahiye exactly isliye kyunki $g = e^{2 x}$ wahan already rehne ka suspicious lagta hai. $D^{2} - 4 D + 4 = (D - 2)^{2}$ , root $2$ doubled: $y_{h} = c_{1} e^{2 x} + c_{2} x e^{2 x} .$

Step 2 — $g$ ko annihilate karo. Collision kyun matter karti hai: $g = e^{2 x}$ ka root $2$ hai, jo $L$ mein pehle se double root hai. Annihilator $A = (D - 2)$ : $(D - 2) (D - 2)^{2} y = (D - 2)^{3} y = 0.$

Step 3–4 — banao aur prune karo. Ye step kyun? Root $2$ ab multiplicity teen rakhta hai, deta hai $e^{2 x}, x e^{2 x}, x^{2} e^{2 x}$ . Pehle do $y_{h}$ hain; genuinely naya ek $x^{2} e^{2 x}$ hai: $y_{p} = A x^{2} e^{2 x} .$ (Double overlap ne $x^{2}$ force kiya — exam mein usi effect ke liye Ex 9 dekho.)

Step 5 — coefficient determine karo. Ye step kyun? $A$ fix karne ke liye trial ko original ODE mein substitute karo. $y_{p} = A x^{2} e^{2 x}$ ke saath: $y_{p}^{'} = A e^{2 x} (2 x + 2 x^{2})$ , $y_{p}^{''} = A e^{2 x} (2 + 8 x + 4 x^{2})$ . Phir $y_{p}^{''} - 4 y_{p}^{'} + 4 y_{p} = A e^{2 x} [(2 + 8 x + 4 x^{2}) - 4 (2 x + 2 x^{2}) + 4 x^{2}] = A e^{2 x} \cdot 2 = e^{2 x} .$ Toh $A = \frac{1}{2}$ .

Step 6 — combine karo. $y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{2} x^{2} e^{2 x} .$

Ye resonance hai apne exponential form mein: forcing system ke apne decaying mode se match karti hai, toh response usi mode ka ek bada multiple hota hai.

Recall Verify karo:

$y_{p} = \frac{1}{2} x^{2} e^{2 x}$ . Coefficient bracket $= 2 A = 2 \cdot \frac{1}{2} = 1$ , exactly $e^{2 x}$ deta hai. ✓

C4 — Trig, no overlap

Step 1 — homogeneous part. Ye step kyun? Hum $y_{h}$ pehle list karte hain taaki check ho sake ki drive frequency $3$ , $y_{h}$ mein chhipi natural frequency se clash karta hai ya nahi. $D^{2} + 1 = 0 \Rightarrow r = \pm i$ , $y_{h} = c_{1} cos x + c_{2} sin x$ (natural frequency $1$ ).

Step 2 — $g$ ko annihilate karo. $D^{2} + 9$ kyun? $sin 3 x$ roots $\pm 3 i$ se aata hai; $cos β x, sin β x$ ke liye annihilator $D^{2} + β^{2}$ hai jahan $β = 3$ . Kyunki $\pm 3 i \neq = \pm i$ , koi overlap nahi.

Step 3–4 — banao aur prune karo. Ye step kyun? $(D^{2} + 9) (D^{2} + 1) y = 0$ ke roots $\pm i, \pm 3 i$ hain; $\pm i$ pair $y_{h}$ rebuild karta hai, toh sirf naya $\pm 3 i$ pair matter karta hai: $y_{p} = A cos 3 x + B sin 3 x .$

Step 5 — coefficients determine karo. Ye step kyun? Original ODE mein substitute karo aur $cos$ aur $sin$ parts ko alag-alag match karo. $y_{p}^{''} = - 9 A cos 3 x - 9 B sin 3 x$ , toh $y_{p}^{''} + y_{p} = - 8 A cos 3 x - 8 B sin 3 x = sin 3 x .$ Match karo: $- 8 A = 0 \Rightarrow A = 0$ ; $- 8 B = 1 \Rightarrow B = - \frac{1}{8}$ .

Step 6 — combine karo. $y = c_{1} cos x + c_{2} sin x - \frac{1}{8} sin 3 x .$

Recall Verify karo:

$y_{p} = - \frac{1}{8} sin 3 x$ . $y_{p}^{''} = \frac{9}{8} sin 3 x$ ; $y_{p}^{''} + y_{p} = \frac{9}{8} sin 3 x - \frac{1}{8} sin 3 x = sin 3 x$ . ✓

C5 — Trig overlap (pure resonance) ek figure ke saath

Step 1 — homogeneous part. Ye step kyun? Hume $y_{h}$ ki frequency chahiye taaki collision dekh sakein: drive $cos 2 x$ frequency $2$ par hai, exactly natural wali. $D^{2} + 4 = 0 \Rightarrow r = \pm 2 i$ , $y_{h} = c_{1} cos 2 x + c_{2} sin 2 x$ .

Step 2 — $g$ ko annihilate karo. Collision kyun matter karti hai: $cos 2 x$ ke roots $\pm 2 i$ hain — exactly $L$ ke roots. Annihilator $A = D^{2} + 4$ : $(D^{2} + 4) (D^{2} + 4) y = (D^{2} + 4)^{2} y = 0.$

Step 3–4 — banao aur prune karo. Ye step kyun? Roots $\pm 2 i$ ab multiplicity do rakhte hain, dete hain $cos 2 x, sin 2 x, x cos 2 x, x sin 2 x$ . Pehle do $y_{h}$ hain; naye wale extra $x$ carry karte hain: $y_{p} = x (A cos 2 x + B sin 2 x) .$

Step 5 — coefficients determine karo. Ye step kyun? Original ODE mein substitute karne se $A, B$ fix hote hain; dekho ki un-shifted $x$ -terms cancel ho jaate hain — exactly wahi cancellation ki wajah se extra $x$ ki zaroorat thi. $y_{p} = A x cos 2 x + B x sin 2 x$ differentiate karo: $y_{p}^{'} = A (cos 2 x - 2 x sin 2 x) + B (sin 2 x + 2 x cos 2 x),$ $y_{p}^{''} = A (- 4 sin 2 x - 4 x cos 2 x) + B (4 cos 2 x - 4 x sin 2 x) .$ Phir $y_{p}^{''} + 4 y_{p}$ : $x cos 2 x$ aur $x sin 2 x$ terms cancel ho jaate hain, bacha rehta hai $- 4 A sin 2 x + 4 B cos 2 x = cos 2 x .$ Match karo: $- 4 A = 0 \Rightarrow A = 0$ ; $4 B = 1 \Rightarrow B = \frac{1}{4}$ .

Step 6 — combine karo. $y = c_{1} cos 2 x + c_{2} sin 2 x + \frac{1}{4} x sin 2 x .$

Neeche ki figure is particular solution $y_{p} = \frac{1}{4} x sin 2 x$ ko plot karti hai. Horizontal axis $x$ (time) hai, vertical axis displacement $y$ hai. Chalk blue mein hai khud wiggling solution; do pink dashed lines hain envelope $\pm \frac{1}{4} x$ — seedhi-line ceiling aur floor jiske against oscillation bar-bar dhabba karti hai. Kyunki woh envelope $\frac{1}{4}$ slope ke saath origin se seedhi line hai, amplitude hamesha ke liye badhta rehta hai jab $x$ badhta hai: har swing thodi pichli se unchi hoti hai. Ye linear-growth envelope resonance ka visual signature hai — woh limiting behaviour jab drive frequency exactly natural frequency ke barabar ho.

Recall Verify karo:

$y_{p} = \frac{1}{4} x sin 2 x$ . $A = 0, B = \frac{1}{4}$ ke saath: $y_{p}^{''} + 4 y_{p} = 4 B cos 2 x = cos 2 x$ . ✓

C6 — Product $x e^{α x}$ (repeated-root annihilator)

Step 1 — homogeneous part. Ye step kyun? Hume $y_{h}$ pata hona chahiye taaki detect ho sake ki drive ka root $1$ pehle se wahan hai. $D^{2} - 1 = (D - 1) (D + 1)$ , roots $\pm 1$ : $y_{h} = c_{1} e^{x} + c_{2} e^{- x} .$

Step 2 — $g$ ko annihilate karo. $(D - 1)^{2}$ kyun? Factor $x^{n} e^{α x}$ ko root $α$ ki $n + 1$ repetitions chahiye; yahan $n = 1, α = 1$ , toh $A = (D - 1)^{2}$ . (Agar $α$ $0$ hota, toh ye polynomial case C1 mein collapse ho jaata — woh degenerate corner hai.) Kyunki root $1$ pehle se $L$ mein ek baar hai, $A$ apply karne se total multiplicity $1 + 2 = 3$ ho jaati hai: $(D - 1)^{2} (D - 1) (D + 1) y = (D - 1)^{3} (D + 1) y = 0.$

Step 3–4 — banao aur prune karo. Ye step kyun? Root $1$ triple → $e^{x}, x e^{x}, x^{2} e^{x}$ ; root $- 1$ → $e^{- x}$ . Inme se $e^{x}$ aur $e^{- x}$ $y_{h}$ hain. Naye terms hain $x e^{x}, x^{2} e^{x}$ : $y_{p} = x (A x + B) e^{x} = (A x^{2} + B x) e^{x} .$

Step 5 — coefficients determine karo. Ye step kyun? Original ODE mein substitute karne se $A, B$ fix hote hain. Product $p (x) e^{x}$ (jahan $p$ koi bhi polynomial hai) differentiate karne ke liye, product rule deta hai $D [p (x) e^{x}] = p^{'} (x) e^{x} + p (x) e^{x} = (p^{'} (x) + p (x)) e^{x},$ kyunki $e^{x}$ khud differentiate hoke $e^{x}$ hi deta hai. Toh har derivative sirf polynomial $p$ ko $p^{'} + p$ se replace karta hai. Yahan $p = A x^{2} + B x$ , toh $p^{'} = 2 A x + B$ aur: $y_{p}^{'} = (p^{'} + p) e^{x} = (A x^{2} + (B + 2 A) x + B) e^{x}$ , $y_{p}^{''} = ((p^{'} + p)^{'} + (p^{'} + p)) e^{x} = (A x^{2} + (B + 4 A) x + (2 B + 2 A)) e^{x}$ . Phir $y_{p}^{''} - y_{p} = [(B + 4 A) x + (2 B + 2 A) - B x] e^{x} = (4 A x + 2 A + 2 B) e^{x} = x e^{x} .$ Match karo: $4 A = 1 \Rightarrow A = \frac{1}{4}$ ; $2 A + 2 B = 0 \Rightarrow B = - \frac{1}{4}$ .

Step 6 — combine karo. $y = c_{1} e^{x} + c_{2} e^{- x} + (\frac{1}{4} x^{2} - \frac{1}{4} x) e^{x} .$

Recall Verify karo:

$y_{p} = (\frac{1}{4} x^{2} - \frac{1}{4} x) e^{x}$ . $4 A x + 2 A + 2 B = 4 (\frac{1}{4}) x + 2 (\frac{1}{4}) + 2 (- \frac{1}{4}) = x + 0 = x$ , toh $y_{p}^{''} - y_{p} = x e^{x}$ . ✓

C7 — Sum of forcings (superposition)

Step 1 — homogeneous part. Ye step kyun? Hum $y_{h}$ pehle list karte hain taaki pata chale do forcing pieces mein se kaun sa isse collide karta hai. $D^{2} + 1$ , roots $\pm i$ : $y_{h} = c_{1} cos x + c_{2} sin x$ .

Step 2 — superposition use karke $g$ ko annihilate karo. Ye step kyun? Linearity hume har summand ke root ko independently treat karne deti hai, phir annihilators multiply karo. $g = g_{1} + g_{2}$ split karo jahan $g_{1} = e^{x}$ (root $1$ , annihilator $D - 1$ , koi overlap nahi) aur $g_{2} = 2 cos x$ (roots $\pm i$ , annihilator $D^{2} + 1$ , $y_{h}$ se overlaps). Combined annihilator $A = (D - 1) (D^{2} + 1)$ , apply karo: $(D - 1) (D^{2} + 1) (D^{2} + 1) y = (D - 1) (D^{2} + 1)^{2} y = 0.$

Step 3–4 — banao aur prune karo. Ye step kyun? Roots: $1$ (naya), $\pm i$ doubled. Naye terms: root $1$ se $e^{x}$ ; aur doubled $\pm i$ se naye pieces $x cos x, x sin x$ . Toh: $y_{p} = g_{1} ke liye C e^{x} + g_{2} ke liye x (A cos x + B sin x) .$

Step 5 — coefficients determine karo. Ye step kyun? Substitute karo; linearity hume do pieces alag-alag solve karne deti hai. Piece 1: $C e^{x}$ : $y^{''} + y = C e^{x} + C e^{x} = 2 C e^{x} = e^{x} \Rightarrow C = \frac{1}{2} .$ Piece 2: $x (A cos x + B sin x)$ . Exactly C5 ki tarah $β = 1$ ke saath, surviving terms hain $- 2 A sin x + 2 B cos x = 2 cos x \Rightarrow A = 0, B = 1.$

Step 6 — combine karo. $y = c_{1} cos x + c_{2} sin x + \frac{1}{2} e^{x} + x sin x .$

Recall Verify karo:

$y_{p} = \frac{1}{2} e^{x} + x sin x$ symbolically substitute karo. Piece 1: $C e^{x}$ ke liye, $y^{''} + y = 2 C e^{x}$ ; $C = \frac{1}{2}$ ke saath ye $e^{x}$ hai. ✓ Piece 2: $x sin x$ ke liye, differentiate karo: $(x sin x)^{'} = sin x + x cos x$ , $(x sin x)^{''} = 2 cos x - x sin x$ . Toh $(x sin x)^{''} + x sin x = 2 cos x - x sin x + x sin x = 2 cos x$ . ✓ Dono pieces add karne par $y_{p}^{''} + y_{p} = e^{x} + 2 cos x = g$ . ✓

C8 — Word problem (mass–spring, real world)

Step 1 — homogeneous part. Ye step kyun? $y_{h}$ spring ki natural oscillation encode karta hai; iske frequency ko drive frequency se compare karne par pata chalta hai ki resonance hit hoga ya nahi. $D^{2} + 9$ , roots $\pm 3 i$ : $y_{h} = c_{1} cos 3 t + c_{2} sin 3 t$ . (Natural angular frequency $3 rad/s$ .)

Step 2 — $g$ ko annihilate karo. $x$ kyun nahi: drive $cos t$ ke roots $\pm i \neq = \pm 3 i$ hain — koi overlap nahi, koi resonance nahi. Annihilator $A = D^{2} + 1$ .

Step 3–4 — banao aur prune karo. Ye step kyun? $A L [y] = 0$ ke roots $\pm 3 i$ ( $y_{h}$ rebuild karte hain) aur naye $\pm i$ hain; sirf naya pair $g$ produce karta hai: $y_{p} = A cos t + B sin t .$

Step 5 — coefficients determine karo. Ye step kyun? $A, B$ fix karne ke liye original ODE mein substitute karo. $y_{p}^{''} = - A cos t - B sin t$ : $y_{p}^{''} + 9 y_{p} = 8 A cos t + 8 B sin t = 6 cos t \Rightarrow A = \frac{3}{4}, B = 0.$ Toh $y_{p} = \frac{3}{4} cos t$ .

Step 6 — initial conditions apply karo. Ye step kyun? Word problem mein ek specific motion hoti hai; do constants $c_{1}, c_{2}$ diye gaye start state se fix hote hain. $y = c_{1} cos 3 t + c_{2} sin 3 t + \frac{3}{4} cos t$ .

$y (0) = 0$ : $c_{1} + \frac{3}{4} = 0 \Rightarrow c_{1} = - \frac{3}{4} .$
$y^{'} (t) = - 3 c_{1} sin 3 t + 3 c_{2} cos 3 t - \frac{3}{4} sin t$ ; $y^{'} (0) = 3 c_{2} = 0 \Rightarrow c_{2} = 0.$

$y (t) = - \frac{3}{4} cos 3 t + \frac{3}{4} cos t .$

Verify karo (units + sanity): $y (0) = - \frac{3}{4} + \frac{3}{4} = 0$ m ✓; $y^{'} (0) = 0$ ✓. Response ek bounded beat hai do frequencies ke beech — koi runaway growth nahi kyunki hum off resonance hain. Units: $y^{''}$ ka $m/s^{2}$ hai, $9 y$ ka (rad/s) $^{2} \cdot$ m $=$ m/s $^{2}$ hai, $6 cos t$ m/s $^{2}$ mein — consistent.

Recall Verify karo

$t = π$ par. $y (π) = - \frac{3}{4} cos 3 π + \frac{3}{4} cos π = - \frac{3}{4} (- 1) + \frac{3}{4} (- 1) = \frac{3}{4} - \frac{3}{4} = 0$ . ✓

C9 — Exam twist: double overlap $x^{2}$ force karta hai

Step 1 — homogeneous part. Ye step kyun? Hume us root ki multiplicity count karni hai jo drive se match karta hai — woh count $x$ ki power control karta hai. $D^{2} - 2 D + 1 = (D - 1)^{2}$ , root $1$ doubled: $y_{h} = c_{1} e^{x} + c_{2} x e^{x}$ .

Step 2 — $g$ ko annihilate karo. $x^{2}$ kyun aa raha hai: $g = e^{x}$ ka root $1$ hai, jo $L$ mein pehle se multiplicity $2$ rakhta hai. Annihilator $A = (D - 1)$ total multiplicity ko $3$ push karta hai: $(D - 1) (D - 1)^{2} y = (D - 1)^{3} y = 0.$

Step 3–4 — banao aur prune karo. Ye step kyun? Root $1$ triple → $e^{x}, x e^{x}, x^{2} e^{x}$ . Pehle do $y_{h}$ hain; naya term $x^{2} e^{x}$ hai: $y_{p} = A x^{2} e^{x} .$

Step 5 — coefficient determine karo. Ye step kyun? $A$ fix karne ke liye original ODE mein substitute karo; C6 ka product-rule shortcut $D [p (x) e^{x}] = (p^{'} + p) e^{x}$ use karo. $y_{p} = A x^{2} e^{x}$ , jahan $p = A x^{2}$ toh $p^{'} = 2 A x$ : $y_{p}^{'} = A (x^{2} + 2 x) e^{x}$ , $y_{p}^{''} = A (x^{2} + 4 x + 2) e^{x}$ . Phir $y_{p}^{''} - 2 y_{p}^{'} + y_{p} = A [(x^{2} + 4 x + 2) - 2 (x^{2} + 2 x) + x^{2}] e^{x} = A \cdot 2 e^{x} = e^{x} .$ Toh $A = \frac{1}{2}$ .

Step 6 — combine karo. $y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{2} x^{2} e^{x} .$

Notice karo ladder: single overlap ne $x$ diya (Ex 3 exponential ke liye / Ex 5 trig ke liye), double overlap ne $x^{2}$ diya. Rule " $L$ mein multiplicity $m$ + tumhara annihilator ⟹ trial ko $x^{m}$ se multiply karo" yahan fully expose ho jaata hai.

Recall Verify karo:

$y_{p} = \frac{1}{2} x^{2} e^{x}$ . Bracket $= 2 A = 2 \cdot \frac{1}{2} = 1$ , $e^{x}$ deta hai. ✓

C10 — Polynomial × trigonometric forcing

Step 1 — homogeneous part. Ye step kyun? Hum check karte hain ki $g$ mein trig frequency (yahan $β = 1$ ) $y_{h}$ ki frequency se match karti hai ya nahi — karti hai, toh overlap aa raha hai. $D^{2} + 1$ , roots $\pm i$ : $y_{h} = c_{1} cos x + c_{2} sin x$ .

Step 2 — $g$ ko annihilate karo. $(D^{2} + 1)^{2}$ kyun? Pattern $x^{n} e^{α x} cos β x$ ( $α = 0, β = 1, n = 1$ ke saath) roots $\pm i$ ke $n + 1 = 2$ baar repeat hone se aata hai. Annihilator hai $[(D - 0)^{2} + 1]^{2} = (D^{2} + 1)^{2}$ . Check karo: $(D^{2} + 1)$ ek baar $cos x, sin x$ ko kill karta hai; doosra application $x$ ke extra factor ki wajah se produce hue $x cos x, x sin x$ ko bhi kill karne ke liye chahiye.

Step 3–4 — banao aur prune karo. Ye step kyun? $A = (D^{2} + 1)^{2}$ ODE par apply karo, jiska $L$ pehle se ek factor $(D^{2} + 1)$ contribute karta hai: $(D^{2} + 1)^{2} (D^{2} + 1) y = (D^{2} + 1)^{3} y = 0.$ Roots $\pm i$ ab multiplicity teen rakhte hain, dete hain $cos x, sin x, x cos x, x sin x, x^{2} cos x, x^{2} sin x$ . Pehli pair $y_{h}$ hai; chaar naye terms trial banate hain. Algebra clean rakhne ke liye hum ise $x$ times both trig functions ke full first-degree-polynomial-in- $x$ combination ki tarah likhte hain: $y_{p} = x [(A x + B) cos x + (C x + E) sin x] .$ (Hum chauthe coefficient ke liye $E$ use karte hain taaki $D$ derivative operator ke roop mein free rahe.)

Step 5 — coefficients determine karo. Ye step kyun? Original ODE $y^{''} + y = x sin x$ mein substitute karo aur chaar independent shapes $x cos x, cos x, x sin x, sin x$ match karo. Differentiation (product rule do baar) karke aur $y_{p}^{''} + y_{p}$ simplify karne par milta hai $y_{p}^{''} + y_{p} = 4 C x cos x + (2 A + 2 E) cos x - 4 A x sin x + (2 C - 2 B) sin x .$ $x sin x = 0 \cdot x cos x + 0 \cdot cos x + 1 \cdot x sin x + 0 \cdot sin x$ se match karo:

$x cos x$ : $4 C = 0 \Rightarrow C = 0$ .
$x sin x$ : $- 4 A = 1 \Rightarrow A = - \frac{1}{4}$ .
$cos x$ : $2 A + 2 E = 0 \Rightarrow E = \frac{1}{4}$ .
$sin x$ : $2 C - 2 B = 0 \Rightarrow B = 0$ .

Step 6 — combine karo. $y = c_{1} cos x + c_{2} sin x - \frac{1}{4} x^{2} cos x + \frac{1}{4} x sin x .$

Recall Verify karo:

$y_{p} = - \frac{1}{4} x^{2} cos x + \frac{1}{4} x sin x$ . $y^{''} + y$ mein plug karne par exactly $x sin x$ milta hai. ✓ Note karo dono $x^{2}$ (overlap se multiplicity badhne ki wajah se) aur $sin$ -driven problem mein $cos$ ka mixing.

C11 — Damped-oscillation forcing $e^{α x} sin β x$

Step 1 — homogeneous part. Ye step kyun? Hum $y_{h}$ ke roots ( $\pm i$ ) list karte hain taaki drive ke root $1 \pm i$ se compare ho sake aur overlap check ho. $D^{2} + 1$ , roots $\pm i$ : $y_{h} = c_{1} cos x + c_{2} sin x$ .

Step 2 — $g$ ko annihilate karo. $(D - 1)^{2} + 1$ kyun? Pattern $e^{α x} cos β x$ ya $e^{α x} sin β x$ complex roots $α \pm i β$ se aata hai; annihilator hai $(D - α)^{2} + β^{2}$ . Yahan $α = 1, β = 1$ , deta hai $A = (D - 1)^{2} + 1 = D^{2} - 2 D + 2$ . Iske roots $1 \pm i$ hain jo $\pm i$ se alag hain (real part $1 \neq = 0$ ), toh koi overlap nahi — extra $x$ nahi.

Step 3–4 — banao aur prune karo. Ye step kyun? $A L [y] = 0$ ke roots $\pm i$ ( $y_{h}$ rebuild karte hain) aur naya pair $1 \pm i$ hain. Naya pair $e^{x} cos x, e^{x} sin x$ produce karta hai: $y_{p} = e^{x} (A cos x + B sin x) .$

Step 5 — coefficients determine karo. Ye step kyun? Original ODE $y^{''} + y = e^{x} sin x$ mein substitute karo. $y_{p} = e^{x} (A cos x + B sin x)$ differentiate karo: $y_{p}^{'} = e^{x} [(A + B) cos x + (B - A) sin x],$ $y_{p}^{''} = e^{x} [2 B cos x - 2 A sin x] .$ Phir $y_{p}^{''} + y_{p} = e^{x} [(A + 2 B) cos x + (B - 2 A) sin x] = e^{x} sin x .$ Match karo: $A + 2 B = 0$ aur $B - 2 A = 1$ . Pehle se, $A = - 2 B$ ; substitute karo: $B - 2 (- 2 B) = 5 B = 1 \Rightarrow B = \frac{1}{5}$ , $A = - \frac{2}{5}$ .

Step 6 — combine karo. $y = c_{1} cos x + c_{2} sin x + e^{x} (- \frac{2}{5} cos x + \frac{1}{5} sin x) .$

Recall Verify karo:

$y_{p} = e^{x} (- \frac{2}{5} cos x + \frac{1}{5} sin x)$ . $A + 2 B = - \frac{2}{5} + \frac{2}{5} = 0$ (koi $cos$ nahi) aur $B - 2 A = \frac{1}{5} + \frac{4}{5} = 1$ ( $sin$ se match). Toh $y_{p}^{''} + y_{p} = e^{x} sin x$ . ✓

Wrap-up: deciding questions

Recall Kaun sa cell

$y_{p}$ mein $x^{2}$ force karta hai, aur kyun? C9: forcing root pehle se $L$ mein multiplicity $2$ rakhta hai; annihilator use $3$ tak raise karta hai, toh do sabse chhote terms $y_{h}$ hain aur