Worked examples — Power series solutions — ordinary points

4.6.17 · D3 · Maths › Ordinary Differential Equations › Power series solutions — ordinary points

Yeh drill page hai Power series solutions at an ordinary point ke liye. Parent note ne dikhaya kaise machine chalti hai. Yahan hum isse har tarah ke case ke saath chalate hain jo topic tumhare saath kar sakta hai — taaki jab exam tumhe koi strange wala de, tum uske cousin se pehle hi mil chuke ho.

Sab kuch zero se build kiya gaya hai. Agar koi symbol aata hai, pehle define kiya gaya hai.

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The scenario matrix

Power-series problem ko kai independent "dials" wala socho. Har dial kuch positions mein ho sakta hai, aur har position se kuch alag hota hai. Yeh hai har dial aur har position jo hum cover karenge.

Cell	Kya vary karta hai	Positions jo tumhe dekhni chahiye
A. Leading coefficient	$y^{\prime \prime }$ ke aage wala number	constant ($1$) vs. ek polynomial ($1-x^2$) jo andar rehna chahiye
B. Series behaviour	kya koi branch rukti hai?	infinite (kabhi nahi rukti) vs. terminating (polynomial ban jaati hai)
C. Recurrence gap	kaun sa earlier $a$ feed karta hai $a_k$ ko	$a_{k+2}\leftarrow a_k$ (step 2) vs. $a_{k+3}\leftarrow a_k$ (step 3, Airy)
D. Extra term	kya first-derivative term $y^{\prime }$ hai?	present ($-2x{y}^{\prime }$) vs. absent ($y^{\prime \prime }+y=0$)
E. Starting point	kya $x_0=0$ hai?	$x_0=0$ vs. nonzero $x_0$ ko origin par shift karo
F. Initial conditions	kya $a_0,a_1$ fixed hain?	free (general solution) vs. diye gaye $y(0),y^{\prime }(0)$ se fix
G. Non-homogeneous	kya right side zero hai?	$=0$ vs. ek forcing term jaise $+x$
H. Word / applied	physical wrapper	ek spring / diffusion phrasing jo ek ODE chhupata hai

Coverage map (kaun sa example kaun sa cell cover karta hai):

Example	Cells covered
1	A(const), B(infinite), C(step-2), D(no $y^{\prime }$)
2	A(poly), B(terminating), D(has $y^{\prime }$)
3	A(poly), C(step-2), degenerate check at $k=0$
4	C(step-3, Airy), B(infinite)
5	E(shift $x_0=1$)
6	F(initial conditions pin $a_0,a_1$)
7	G(non-homogeneous forcing)
8	H(word problem), F, B(terminating)

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Example 1 — constant leading coefficient, no $y^{\prime }$ term (Cells A, B∞, C-step2, D-none)

Forecast: $y^{\prime \prime }+y=0$ se almost identical, bas ek plus sign flip hai. Guess: coefficients mein sign alternate nahi karenge, toh $\cos /\sin$ ki jagah hume $\cosh /\sinh$ milna chahiye. Predict $a_{k+2}=+a_k/[(k+2)(k+1)]$.

Steps.

1. Maano $y=\sum _{n=0}^{\infty }{a}_n{x}^n$, toh $y^{\prime \prime }=\sum _{n=2}^{\infty }n(n-1)a_n{x}^{n-2}$. Yeh step kyun? Har power-series solution unknown ko infinite polynomial ki tarah likhke term by term differentiate karke start hota hai.
2. Substitute karo: $\sum _{n=2}^{\infty }n(n-1)a_n{x}^{n-2}-\sum _{n=0}^{\infty }{a}_n{x}^n=0.$
3. Pehle sum ko $k=n-2$ se re-index karo: $\sum _{k=0}^{\infty }(k+2)(k+1)a_{k+2}{x}^k-\sum _{k=0}^{\infty }{a}_k{x}^k=0.$ Yeh step kyun? Dono sums mein same power $x^k$ hona chahiye tab hum unke coefficients add kar sakte hain.
4. $x^k$ ka coefficient zero set karo: $(k+2)(k+1)a_{k+2}-a_k=0\Rightarrow \boxed{a_{k+2}=\frac{a_k}{(k+2)(k+1)}}.$ Yeh step kyun? Ek power series sabhi $x$ ke liye $0$ equals karta hai iff har coefficient $0$ ho.
5. Calculate karo. Even: $a_2=\frac{a_0}{2!},a_4=\frac{a_0}{4!},\dots$ Odd: $a_3=\frac{a_1}{3!},a_5=\frac{a_1}{5!},\dots$ $y=a_0(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots )+a_1(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots ).$

Verify: Dono bracketed series exactly $\cosh x$ aur $\sinh x$ hain. Numerically $x=1$ par: $a_0$-series $\approx 1.5431=\cosh 1$, $a_1$-series $\approx 1.1752=\sinh 1$. Forecast confirmed — $+$ sign ne alternation khatam kar diya. ✔

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Example 2 — polynomial leading coefficient, terminating branch (Cells A-poly, B-terminate, D-has $y^{\prime }$)

Forecast: Leading coefficient ek plain $1$ hai (kahi bhi koi singular point nahi), toh dono branches har jagah converge karti hain. Constant term $+4y$ lagta hai ki koi branch terminate kar sakti hai. Guess: even branch jaldi ruk jaayegi.

Steps.

1. $y=\sum a_n{x}^n$, $y^{\prime }={\sum }_{n\ge 1}n{a}_n{x}^{n-1}$, $y^{\prime \prime }={\sum }_{n\ge 2}n(n-1)a_n{x}^{n-2}$.
2. Substitute karo, $x$ ko jo $y^{\prime }$ multiply karta hai andar rakhte hue: $\sum n(n-1)a_n{x}^{n-2}-2\sum n{a}_n{x}^n+4\sum a_n{x}^n=0.$ $2x$ ko andar kyun rakhen? $x$ se multiply karna power ko ek se badhata hai; agar hum ise bahar nikalen toh hum track kho dete hain ki hum kaun sa power match kar rahe hain.
3. Pehle sum ko re-index karo ($k=n-2$); baaki mein $n\to k$ rename karo: ${\sum }_k[(k+2)(k+1)a_{k+2}-2k{a}_k+4a_k]x^k=0.$
4. Coefficient $=0$: $(k+2)(k+1)a_{k+2}=(2k-4)a_k\Rightarrow \boxed{a_{k+2}=\frac{2(k-2)}{(k+2)(k+1)}{a}_k}.$ Yeh step kyun? Teen terms ko collect karo jo sabhi $a_k{x}^k$ multiply karte hain: $-2k+4=2(2-k)$… carefully $-2k+4=-2(k-2)$, aur dusri side move karne par $+2(k-2)$ milta hai.
5. Zero test. $k=2$ par: numerator $2(k-2)=0\Rightarrow a_4=0$, aur har higher even term khatam ho jaata hai. Yeh kyun matter karta hai? Yahi exactly cell B-terminate hai: even branch ek finite polynomial ban jaati hai.
   • $a_2=\frac{2(0-2)}{2\cdot 1}{a}_0=\frac{-4}{2}{a}_0=-2a_0$, phir $a_4=0$.
   • Even branch: $y_{\text{even}}=a_0(1-2x^2)$.
6. Odd branch nahi rukti (numerator $2(k-2)\ne 0$ odd $k\ge 1$ ke liye):
   • $a_3=\frac{2(1-2)}{3\cdot 2}{a}_1=\frac{-2}{6}{a}_1=-\frac{1}{3}{a}_1.$
   • $a_5=\frac{2(3-2)}{5\cdot 4}{a}_3=\frac{2}{20}(-\frac{1}{3})a_1=-\frac{1}{30}{a}_1.$

$y=a_0(1-2x^2)+a_1(x-\frac{1}{3}{x}^3-\frac{1}{30}{x}^5-\cdots ).$

Verify: Polynomial $1-2x^2$ hai (ek scale tak) Hermite polynomial $H_2(x)=4x^2-2$: wakai $-2(1-2x^2)=4x^2-2=H_2(x)$. $y=1-2x^2$ back substitute karo: $y^{\prime \prime }=-4$, $-2x{y}^{\prime }=-2x(-4x)=8x^2$, $4y=4-8x^2$; sum $=-4+8x^2+4-8x^2=0$. ✔

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Example 3 — $k=0$ par degenerate check (Cells A-poly, C-step2, zero input)

Forecast: Yeh parent ka Example 2 hai lekin $6y$ ki jagah $2y$ hai. Jo number matter karta tha woh constant term tha. Guess: termination ab alag $k$ par hogi — shayad $k=1$, is baar odd branch ko khatam kar raha ho.

Steps.

1. Substitute karo (parent jaisi hi shape): $\sum n(n-1)a_n{x}^{n-2}-\sum n(n-1)a_n{x}^n-2\sum n{a}_n{x}^n+2\sum a_n{x}^n=0.$
2. Re-index karo aur $x^k$ ka coefficient collect karo: $(k+2)(k+1)a_{k+2}-[k(k-1)+2k-2]a_k=0.$ Yeh step kyun? Teen saare "$x^n$" sums power share karte hain; unke $a_k$ multipliers combine karo.
3. Bracket simplify karo: $k(k-1)+2k-2=k^2+k-2=(k+2)(k-1)$. $\boxed{a_{k+2}=\frac{(k+2)(k-1)}{(k+2)(k+1)}{a}_k=\frac{k-1}{k+1}{a}_k}.$
4. Smallest-index (degenerate) test, $k=0$: $a_2=\frac{0-1}{0+1}{a}_0=-a_0$. Phir $k=2$: $a_4=\frac{1}{3}{a}_2=-\frac{1}{3}{a}_0$ — even branch nahi rukti. $k=0$ alag se kyun check karein? General formula mein koi $0$ ya division ho sakti hai jo boundary par galat behave kare; yahan $k=0$ perfectly fine hai, lekin discipline yahi hai ki hamesha ise test karo.
5. Odd test, $k=1$: numerator $k-1=0\Rightarrow a_3=0$ aur saare higher odd terms gayab ho jaate hain.
   • Odd branch: $y_{\text{odd}}=a_1\cdot x$ — ek linear polynomial.

$y=a_0(1-x^2-\frac{1}{3}{x}^4-\cdots )+a_{1}x.$

Verify: $y_2=x$ ek solution hai? $y^{\prime \prime }=0$, $-2x{y}^{\prime }=-2x$, $2y=2x$; $(1-x^2)\cdot 0-2x+2x=0$. ✔ (Forecast sahi tha: is baar odd branch terminate hui.)

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Example 4 — three-term recurrence, Airy's equation (Cells C-step3, B∞)

Forecast: $y$ ko $x$ se multiply karna uski power ko ek se upar shift karta hai, toh dono sums odd gap waale powers par land karenge. Guess: $a_k$ links karta hai $a_{k-3}$ se (ek step-3 recurrence), jo naya hai — pichle examples ne sab step 2 kiya.

Steps.

1. $y^{\prime \prime }={\sum }_{n\ge 2}n(n-1)a_n{x}^{n-2}$, aur $xy={\sum }_{n\ge 0}{a}_n{x}^{n+1}$.
2. Substitute karo: $\sum _{n\ge 2}n(n-1)a_n{x}^{n-2}-\sum _{n\ge 0}{a}_n{x}^{n+1}=0.$
3. Dono ko power $x^k$ par re-index karo: pehle $k=n-2$ se, doosre ko $k=n+1$ se (toh $n=k-1$): ${\sum }_{k\ge 0}(k+2)(k+1)a_{k+2}{x}^k-{\sum }_{k\ge 1}{a}_{k-1}{x}^k=0.$ Yeh step kyun? Figure dekho: $y^{\prime \prime }$ contributes $x^k$ mein coefficient $a_{k+2}$ se, jabki $-xy$ contributes $a_{k-1}$ se — indices $k+2$ aur $k-1$ ke beech 3 ka gap hai.
4. $k=0$ wala term sirf pehle sum se aata hai: $(2)(1)a_2=0\Rightarrow a_2=0.$ $k\ge 1$ ke liye: $(k+2)(k+1)a_{k+2}=a_{k-1}\Rightarrow \boxed{a_{k+2}=\frac{a_{k-1}}{(k+2)(k+1)}}.$
5. Teeni mein calculate karo. Kyunki $a_2=0$ hai, "$\equiv 2(\mathrm{mod}3)$" wali family zero rehti hai.
   • $a_0$ se: $a_3=\frac{a_0}{3\cdot 2}=\frac{a_0}{6}$, $a_6=\frac{a_3}{6\cdot 5}=\frac{a_0}{180}$.
   • $a_1$ se: $a_4=\frac{a_1}{4\cdot 3}=\frac{a_1}{12}$, $a_7=\frac{a_4}{7\cdot 6}=\frac{a_1}{504}$.

$y=a_0(1+\frac{x^3}{6}+\frac{x^6}{180}+\cdots )+a_1(x+\frac{x^4}{12}+\frac{x^7}{504}+\cdots ).$

Verify: $y_1=1+\frac{x^3}{6}+\frac{x^6}{180}$ (truncated) plug karo: $y_1^{\prime \prime }=x+\frac{x^4}{6}$; $x{y}_1=x+\frac{x^4}{6}+\frac{x^7}{180}$. Difference $y_1^{\prime \prime }-x{y}_1=-\frac{x^7}{180}$ — jitna order humne rakha, utak zero. ✔

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Example 5 — nonzero point ke baare mein expand karna (Cell E, shift)

Forecast: $x_0=1$ origin nahi hai, toh series $\sum a_n{x}^n$ directly kaam nahi karegi. Guess: $t=x-1$ substitute karo problem ko origin par slide karne ke liye, aur yeh Airy-jaisi ban jaayegi ($y^{\prime \prime }+ty=0$).

Steps.

1. $t=x-1$ lo, toh $x-1=t$ aur derivatives unchanged hain ($\frac{d}{dx}=\frac{d}{dt}$ kyunki $\frac{dt}{dx}=1$). Yeh step kyun? Power-series machinery ko expansion centre $0$ par chahiye; variable shift karna $x_0=1$ ko $t=0$ par le jaata hai. (Yahi cell-E move hai.)
2. ODE ban jaati hai $y^{\prime \prime }+ty=0$ with $y={\sum }_n{a}_n{t}^n$ — yeh hai Airy plus sign ke saath.
3. Same re-indexing jaise Example 4: $k\ge 1$ ke liye, $(k+2)(k+1)a_{k+2}=-a_{k-1}$, aur $a_2=0$. $\boxed{a_{k+2}=-\frac{a_{k-1}}{(k+2)(k+1)}}.$
4. Calculate karo: $a_3=-\frac{a_0}{6}$, $a_4=-\frac{a_1}{12}$, $a_6=-\frac{a_3}{30}=\frac{a_0}{180}$.
5. $t=x-1$ restore karo: $y=a_0(1-\frac{(x-1{)}^3}{6}+\frac{(x-1{)}^6}{180}-\cdots )+a_1((x-1)-\frac{(x-1{)}^4}{12}+\cdots ).$

Verify: $u=x-1$ ke saath, $y_1=1-\frac{u^3}{6}$: $y_1^{\prime \prime }=-u$, $(x-1)y_1=u(1-\frac{u^3}{6})=u-\frac{u^4}{6}$; sum $=-u+u-\frac{u^4}{6}=-\frac{u^4}{6}$, leading order tak zero. ✔ (Forecast sahi: shift ne ise $+t$ Airy mein convert kar diya.)

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Example 6 — initial conditions $a_0,a_1$ pin karte hain (Cell F)

Forecast: Parent se, $a_0$ multiply karta hai $\cos x$ ko aur $a_1$ multiply karta hai $\sin x$ ko. Kyunki $y(0)=a_0$ aur $y^{\prime }(0)=a_1$, guess $y=2\cos x-3\sin x$.

Steps.

1. General series yaad karo (parent Example 1 se): $y=a_0(1-\frac{x^2}{2!}+\cdots )+a_1(x-\frac{x^3}{3!}+\cdots )$.
2. $x=0$ par evaluate karo: sirf constant terms survive karte hain, toh $y(0)=a_0$. Yeh step kyun? $n\ge 1$ wala har $x^n$ $x=0$ par vanish ho jaata hai, $a_0$ isolate ho jaata hai.
3. Toh $a_0=2$.
4. Differentiate karo aur $x=0$ set karo: $y^{\prime }(0)=a_1$ ($a_1$ series ka sirf first-power term survive karta hai). Toh $a_1=-3$. Yeh step kyun? $y^{\prime }=a_1(1-\frac{x^2}{2!}+\cdots )+\dots$; $x=0$ par sirf $a_1\cdot 1$ bachta hai.
5. Substitute karo: $y=2(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots )-3(x-\frac{x^3}{6}+\cdots )=2\cos x-3\sin x.$

Verify: $y(0)=2\cdot 1-3\cdot 0=2$ ✔; $y^{\prime }(x)=-2\sin x-3\cos x$, toh $y^{\prime }(0)=-3$ ✔. Bhi $y^{\prime \prime }+y=(-2\cos x+3\sin x)+(2\cos x-3\sin x)=0$ ✔.

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Example 7 — non-homogeneous forcing (Cell G)

Forecast: Homogeneous part deta hai $\cos x,\sin x$. Right side $x$ ek first-degree polynomial hai, toh ek particular series guess karo jo exactly ek leftover $x^1$ produce kare — shayad $y_p=x$ khud hi.

Steps.

1. Maano $y=\sum a_n{x}^n$. Toh (Example 1 ki algebra se) left side hai ${\sum }_k[(k+2)(k+1)a_{k+2}+a_k]x^k$.
2. Right side hai $x=1\cdot x^1$. Har power ka coefficient match karo:
   • $x^0$: $2a_2+a_0=0$.
   • $x^1$: $6a_3+a_1=1$ — forcing sirf yahan aata hai.
   • $x^k,k\ge 2$: $(k+2)(k+1)a_{k+2}+a_k=0$. Yeh step kyun? Equation ab kehti hai "left side par $x^k$ ka coefficient = right side par $x^k$ ka coefficient"; right side sirf $x^1$ par nonzero hai.
3. Solve karo: $a_2=-\frac{a_0}{2}$; $x^1$ row se $a_3=\frac{1-a_1}{6}$; phir $a_4=-\frac{a_2}{12}=\frac{a_0}{24}$, $a_5=-\frac{a_3}{20}=-\frac{1-a_1}{120}$.
4. Notice karo ki "$1$" odd chain mein neeche jaata hai. Forcing wala piece collect karte hue: $y_p=x(\text{lo }{a}_0=0,a_1=1\text{-driven part isolate hokar }{y}_p=x\text{ deta hai}).$ General solution: $y=a_0\cos x+{\tilde{a}}_1\sin x+x$, jahan ${\tilde{a}}_1$ homogeneous freedom absorb karta hai.

Verify: $y_p=x\Rightarrow y_p^{\prime \prime }=0$, $y_p^{\prime \prime }+y_p=0+x=x$ ✔. Full: $y=A\cos x+B\sin x+x$ deta hai $y^{\prime \prime }+y=x$ kisi bhi $A,B$ ke liye. ✔

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Example 8 — word problem, terminating polynomial (Cells H, F, B-terminate)

Forecast: "Symmetric aur centre par flat" ka matlab hai $a_1=0$, toh sirf even branch bachti hai. Parent ne dikhaya ki exactly is equation ki even branch terminate hoti hai $1-3x^2$ ke roop mein. Guess: $y=1-3x^2$.

Steps.

1. Parent ke Example 2 se, recurrence hai $a_{k+2}=\frac{(k+3)(k-2)}{(k+2)(k+1)}{a}_k$, aur $y(0)=a_0$, $y^{\prime }(0)=a_1$. Yeh step kyun? Already-derived recurrence reuse karo — yeh wahi ODE hai, sirf constants alag hain.
2. Conditions apply karo: $y(0)=1\Rightarrow a_0=1$; $y^{\prime }(0)=0\Rightarrow a_1=0$. Yeh step kyun? Cell-F rule: $a_0=y(0)$, $a_1=y^{\prime }(0)$.
3. $a_1=0$ ke saath puri odd branch zero hai (sab $a_1$ se build hoti hai). Sirf even branch bachti hai.
4. Even branch: $a_2=\frac{(3)(-2)}{2\cdot 1}{a}_0=-3$, aur $a_4=0$ (numerator $(k-2)=0$ at $k=2$), toh yeh terminate ho jaati hai. Bounded kyun? Non-terminating odd branch $x\to \pm 1$ ke paas blow up ho jaati (singular points ke paas); polynomial demand karna hi "rod par physically bounded" ka matlab hai. Yahi reason hai ki physics select karta hai Legendre polynomials — dekho Legendre's equation and Legendre polynomials.
5. $y=1-3x^2.$

Verify: $y(0)=1$ ✔, $y^{\prime }(0)=0$ ✔ (kyunki $y^{\prime }=-6x$). ODE: $y^{\prime \prime }=-6$, $-2x{y}^{\prime }=-2x(-6x)=12x^2$, $6y=6-18x^2$; $(1-x^2)(-6)+12x^2+6-18x^2=-6+6x^2+12x^2+6-18x^2=0$ ✔.

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Wrap-up recall

Recall Har example ne kaun sa cell test kiya?

Ex1 :::: constant leading coeff, infinite step-2 series, no $y^{\prime }$. Ex2 :::: polynomial coeff, terminating even branch (Hermite). Ex3 :::: degenerate $k=0$ check, odd branch terminates. Ex4 :::: three-term (step-3) recurrence, Airy. Ex5 :::: $x_0=1$ ko origin par shifting. Ex6 :::: initial conditions pin karte hain $a_0=y(0),a_1=y^{\prime }(0)$. Ex7 :::: non-homogeneous forcing right side par. Ex8 :::: applied word problem jo ek bounded Legendre polynomial select karta hai.

Review ke liye linked prerequisites: Recurrence relations, Taylor series and analyticity, Existence and uniqueness for linear ODEs, Singular points and Frobenius method.