Worked examples — Classification — order, degree, linear vs nonlinear, autonomous vs non-autonomous

4.6.1 · D3 · Maths › Ordinary Differential Equations › Classification — order, degree, linear vs nonlinear, autonom

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Scenario matrix

Har ODE-classification problem aslmein ek four-axis space mein ek point hai: un chaar axes mein se do — linear vs nonlinear aur autonomous vs non-autonomous — independent hain, isliye yeh ek genuine 2-D grid of four corners banaate hain. Neeche ki picture woh grid draw karti hai aur har worked example ko uske corner mein rakhti hai (order aur degree phir alag se har example ke liye padhe jaate hain).

Table un distinct case-classes ko list karta hai — woh corners aur edges jahan beginners slip karte hain. Last column us example ka naam deta hai jo us cell ko nail karta hai.

#	Case class (woh "scenario")	Kya tricky banata hai	Covered by
C1	Clean polynomial ODE, saare chaar labels obvious	baseline — kuch chhupa nahi	Ex 1
C2	Radical ek power chhupa raha hai	degree padhne se pehle square karna zaroori	Ex 2
C3	Derivative ka transcendental function ($\sin y^{\prime }$, $\ln y^{\prime \prime }$)	degree undefined hai	Ex 3
C4	Ugly $x$-coefficients jo nonlinearity jaisi dikhaate hain	phir bhi linear hai — sirf $y$ dekho	Ex 4
C5	$y$ ya uske derivatives ke products/powers	genuinely nonlinear	Ex 5
C6	Linear phir bhi non-autonomous (forcing mein explicit $t$)	do axes independent hain	Ex 6
C7	Autonomous par nonlinear (no clock, phir bhi $y^2$)	autonomy ≠ linearity	Ex 7
C8	Degenerate / limiting — degree-zero-power, missing $y$, algebraic (order 0)	"kya yeh ODE bhi hai?" ka boundary	Ex 8
C9	Word problem → ODE banao, phir classify karo	translation, sirf labelling nahi	Ex 9
C10	Exam twist — derivatives ke fractions, denominators clear karne zaroori	cleanup se visible degree badal jaati hai	Ex 10

Matrix ko top to bottom padho: C1 easy warm-up hai, C8 "trap" row hai, C9 application hai, C10 final boss hai.

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Example 1 — clean baseline (cell C1)

Steps.

1. Order. Sabse bada derivative $y^{\prime \prime }$ hai → order $2$. Yeh step kyun? Order hamesha sabse pehli cheez padhi jaati hai; yeh decide karta hai ki general solution mein kitne arbitrary constants honge — yahan do.
2. Degree. Equation already $y,y^{\prime },y^{\prime \prime }$ mein ek polynomial hai; $y^{\prime \prime }$ par power $1$ hai → degree $1$. Yeh step kyun? Koi radical nahi, derivatives ke fractions nahi → kuch clean karna nahi, isliye exponent seedha padh sakte hain.
3. Linear? $y,y^{\prime },y^{\prime \prime }$ mein se har ek first power par appear karta hai, koi multiply nahi, koi nonlinear function ke andar nahi → linear. Yeh step kyun? Linearity superposition aur characteristic-equation method unlock karta hai.
4. Autonomous? Koi explicit independent variable $x$ kahi nahi → autonomous. Yeh step kyun? Autonomy ka matlab hai slope field har horizontal line ke saath repeat hota hai, isliye hum phase line par equilibria study kar sakte hain.

Answer: order 2, degree 1, linear, autonomous.

Verify: general solution $y=c_1{e}^{-2x}+c_2{e}^{-3x}$ hai ($r^2+5r+6=0$ ki roots $-2,-3$ hain). Do arbitrary constants ✓ order 2 se match. $y=e^{-2x}$ plug karke: $4e^{-2x}-10e^{-2x}+6e^{-2x}=0$ ✓.

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Example 2 — ek radical saccha degree chhupa raha hai (cell C2)

Steps.

1. Chhupa hua power dhundho. Cube root ek fractional power hai; degree sirf derivatives mein ek polynomial ke liye define hoti hai. Isliye pehle root hatana zaroori hai. Yeh step kyun? Jab tak ek derivative radical ke neeche baithe, honestly exponent nahi padh sakte — equation abhi tak polynomial nahi hai.
2. Dono taraf cube karo $\sqrt[3]{}$ clear karne ke liye: ${\left(\frac{d^{2}y}{d{x}^2}\right)}^3=1+{\left(\frac{dy}{dx}\right)}^2.$ Yeh step kyun? Cubing cube root ka inverse hai — yeh fractional power ko integer power mein badal deta hai, equation ko $y^{\prime }$ aur $y^{\prime \prime }$ mein genuine polynomial bana deta hai.
3. Order. Highest $y^{\prime \prime }$ hai → order $2$. Yeh step kyun? Order cleanup ke under kabhi nahi badalti — squaring ya cubing powers badhata hai, highest derivative ki identity nahi.
4. Degree. Cleanup ke baad $y^{\prime \prime }$ par power $3$ hai → degree $3$. Yeh step kyun? Sirf abhi highest derivative ka exponent unambiguous hai.
5. Linear? $(y^{\prime \prime }{)}^3$ ek third power hai → nonlinear. Yeh step kyun? Kisi bhi derivative par power 1 se upar linearity rule 1 tod deta hai, superposition forbid karta hai.
6. Autonomous? Koi explicit $x$ nahi → autonomous. Yeh step kyun? Independent variable $x$ apne aap kahi nahi aata, isliye "rules" input value par depend nahi karte.

Answer: order 2, degree 3, nonlinear, autonomous.

Verify: neeche ki figure dono sides ko curvature $y^{\prime \prime }$ ke function ke roop mein plot karti hai jabki slope $y^{\prime }$ fixed rakha jaata hai. Jahan bhi do graphs cross karte hain woh ek curvature hai jo actually equation satisfy karta hai — aur exactly ek crossing hai.

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Example 3 — derivative ka transcendental function (cell C3)

Steps.

1. Polynomialise karne ki koshish karo. Kya $\sin (y^{\prime })$ kabhi $y^{\prime }$ mein finite polynomial ke roop mein likha ja sakta hai? Nahi — iska Taylor series $y^{\prime }-\frac{(y^{\prime }{)}^3}{6}+\dots$ kabhi terminate nahi karta, aur koi algebra isse hata nahi sakti. Yeh step kyun? Degree sirf tab define hoti hai jab equation derivatives mein polynomial ho. Agar yeh possible nahi, toh hum ruk jaate hain.
2. Degree conclude karo. Kyunki isse derivatives mein polynomial nahi banaya ja sakta → degree undefined. Yeh step kyun? Yeh honest answer hai, "1" nahi. $\sin (y^{\prime })$ ki presence hi poora point hai.
3. Order. Highest derivative $y^{\prime \prime }$ → order $2$ (transcendental wrapping order nahi chhupata). Yeh step kyun? Order count karta hai kaunsa derivative highest hai, aur $y^{\prime \prime }$ $\sin$ ke baawajood visibly present hai.
4. Linear? $\sin (y^{\prime })$ ek derivative ko nonlinear function ke andar rakhta hai → nonlinear (rule 3 toota). Yeh step kyun? Rule 3 kisi bhi derivative ko $\sin ,e^{(\cdot )},\ln$, etc. ke andar rehne se mana karta hai.
5. Autonomous? $e^x$ on the right explicit $x$ hai → non-autonomous. Yeh step kyun? Independent variable $x$ forcing mein explicitly appear karta hai, isliye rule input value ke saath badalta hai.

Answer: order 2, degree undefined, nonlinear, non-autonomous.

Verify: contrast karo — $y^{\prime \prime }+(y^{\prime }{)}^2=e^x$ degree 1 hoti (derivatives mein polynomial; highest $y^{\prime \prime }$ power 1 par). $(y^{\prime }{)}^2\to \sin (y^{\prime })$ swap karna exactly wahi hai jo "degree" destroy karta hai. Difference hai polynomial vs transcendental, step 1 confirm karta hai. ✓

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Example 4 — ugly coefficients jo nonlinear lagte hain (cell C4)

Steps.

1. Dekho $y$ kaise enter karta hai. Mentally coefficients cover karo aur sirf underlined unknown parts rakho: $x^2\underline{y^{\prime \prime }}+(\sin x)\underline{y^{\prime }}+e^x\underline{y}=\tan x.$ Har underlined piece — $\underline{y^{\prime \prime }},\underline{y^{\prime }},\underline{y}$ — ek baar, first power par, kisi doosre $y$-cheez se multiply nahi, aur $y$ ke nonlinear function ke andar nahi appear karta hai. Yeh step kyun? Linearity test sirf unknown $y$ aur uske derivatives (underlined parts) ko inspect karta hai — unke around wrapped coefficients ko kabhi nahi.
2. Coefficients check karo. $x^2,\sin x,e^x,\tan x$ sirf $x$ ke functions hain — jaisi marzi wild ho, allowed hain. Yeh step kyun? Linear ODE ki definition: coefficients $a_k(x)$ aur forcing $g(x)$ $x$ ke any functions ho sakte hain.
3. Verdict: teeno linearity rules hold karte hain → linear. Yeh step kyun? Steps 1–2 ke har rule se pass hona exactly linearity ki definition hai, isliye ab confidently label kar sakte hain.
4. Order / degree. Highest $y^{\prime \prime }$, first power → order $2$, degree $1$. Yeh step kyun? Koi radical ya derivative-fractions clean karne nahi hain, isliye order (kaunsa derivative) aur degree (uski power) seedhe padh lete hain.
5. Autonomous? $x$ explicitly har jagah appear karta hai → non-autonomous. Yeh step kyun? Autonomy test sirf poochta hai ki independent variable apne aap aata hai ya nahi — yahan $x$ poori equation mein hai.

Answer: order 2, degree 1, linear, non-autonomous.

Verify: standard linear form $a_2(x)y^{\prime \prime }+a_1(x)y^{\prime }+a_0(x)y=g(x)$ mein rakho jahan $a_2=x^2,a_1=\sin x,a_0=e^x,g=\tan x$. Har slot $x$ ka pure function hai ✓ — ek linear ODE ki defining shape.

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Example 5 — product ki wajah se genuinely nonlinear (cell C5)

Steps.

1. Teeno violations scan karo. Dhundho (i) powers $>1$, (ii) $y$-cheezein ka products, (iii) $y$ ke nonlinear functions. Yeh step kyun? Ek violation bhi nonlinear banane ke liye kaafi hai; kaunsa ek naam lena samajh badhata hai.
2. Dhundho unhe. $y\cdot y^{\prime \prime }$ do $y$-quantities ka product hai (rule 2), aur $(y^{\prime }{)}^2$ second power hai (rule 1). Do violations. Yeh step kyun? Unknown ke products aur powers nonlinearity ki pehchaan hain.
3. Verdict: nonlinear. Yeh step kyun? Step 2 ka koi bhi single violation decisive hai; do hona ise unambiguous banata hai.
4. Order / degree. Highest $y^{\prime \prime }$ power $1$ par → order $2$, degree $1$. Yeh step kyun? Surprise note karo: yeh degree 1 hai ($y^{\prime \prime }$ par power ek hai) phir bhi nonlinear hai — degree aur linearity alag sawaal hain.
5. Autonomous? Koi explicit $x$ nahi → autonomous. Yeh step kyun? Independent variable kabhi akela nahi aata, isliye rule sirf is par depend karta hai ki $y$ abhi kahan hai.

Answer: order 2, degree 1, nonlinear, autonomous.

Verify: notice karo $y{y}^{\prime \prime }+(y^{\prime }{)}^2=\frac{d}{dx}(y{y}^{\prime })$. Isliye equation hai $(y{y}^{\prime }{)}^{\prime }=0\Rightarrow y{y}^{\prime }=c_1\Rightarrow \frac{1}{2}(y^2{)}^{\prime }=c_1\Rightarrow y^2=2c_{1}x+c_2$. Do constants ✓ order 2 se match. Product nonlinearity kyun prove karta hai (superposition test): superposition ka matlab hai "agar $y_1$ aur $y_2$ solutions hain, toh $y_1+y_2$ bhi hai." Hamare equation ke do solutions lo aur unhe add karo: $y{y}^{\prime \prime }$ ki left side $(y_1+y_2)(y_1+y_2{)}^{\prime \prime }$ ho jaati hai $=y_1{y}_1^{\prime \prime }+y_1{y}_2^{\prime \prime }+y_2{y}_1^{\prime \prime }+y_2{y}_2^{\prime \prime }$. Do cross terms $y_1{y}_2^{\prime \prime }+y_2{y}_1^{\prime \prime }$ cancel nahi hote aur kabhi kisi original equation mein present nahi the — sum ODE solve karna fail karta hai. Woh uncancelled cross terms exactly product $y{y}^{\prime \prime }$ se janm lete hain; ek linear equation (koi product nahi) mein koi cross terms nahi hote aur sum survive karta. Yahi wajah hai ki product superposition forbid karta hai.

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Example 6 — linear aur non-autonomous (cell C6)

Steps.

1. Linearity. $y^{\prime \prime }$ aur $y$ first power par hain, multiply nahi, wrapped nahi; constant ${\omega }^2$ sirf ek (constant) coefficient hai → linear. Yeh step kyun? ${\omega }^2$ jaisa constant ek allowed coefficient ka special case hai — fixed number hona linearity kabhi nahi tod sakta.
2. Autonomy. Forcing $\cos (\gamma t)$ explicitly $t$ par depend karta hai → non-autonomous. Yeh step kyun? Autonomy sirf poochta hai ki independent variable $t$ explicitly aata hai ya nahi; ek linear equation phir bhi clock dekh sakta hai. $\gamma$ andar constant hai — $t$ ki presence matter karti hai, $\gamma$ ki nahi.
3. Order / degree. Highest $y^{\prime \prime }$, first power → order $2$, degree $1$. Yeh step kyun? Koi cleanup nahi chahiye, isliye order (highest derivative) aur degree (uski power) seedhe padh lete hain.

Answer: order 2, degree 1, linear, non-autonomous — prove karta hai ki do axes independent hain.

Verify: homogeneous partner $y^{\prime \prime }+{\omega }^{2}y=0$ is autonomous hai; $t$-dependent forcing add karna hi autonomy todata hai. Neeche ki slope-field picture difference visible karti hai.

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Example 7 — autonomous par nonlinear (cell C7)

Steps.

1. Right side expand karo: $y^{\prime }=y-y^2$. Yeh step kyun? Chhupa hua $y^2$ expand karne par dikhna aasaan hai, toh linearity honestly judge ki ja sake.
2. Linearity. $y^2$ term unknown ki second power hai → nonlinear (rule 1 toota). Yeh step kyun? Ek bhi squared unknown superposition forbid karta hai.
3. Autonomy. Right side par koi explicit $t$ nahi → autonomous. Yeh step kyun? Rate sirf current population $y$ par depend karta hai, calendar par nahi — ek classic phase-line candidate.
4. Order / degree. Highest $y^{\prime }$, first power → order $1$, degree $1$. Yeh step kyun? Sirf $y^{\prime }$ appear karta hai (order 1) aur expand karne ke baad kabhi power par raise nahi hota (degree 1).

Answer: order 1, degree 1, nonlinear, autonomous.

Verify: iske equilibria $y(1-y)=0$ solve karte hain $\Rightarrow y=0$ aur $y=1$. First-order autonomous ODE ke liye do equilibria ✓ — exactly wahi jo phase line predict karta hai. Yeh separable hona (ek nonlinear-par-solvable case) yahi wajah hai ki nonlinearity ke bawajood logistic curve ka closed form hai.

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Example 8 — degenerate & limiting cases (cell C8)

Steps.

1. (a) $y^{\prime }=3$. Order 1 (ek derivative), degree 1, linear (RHS ek constant $g(x)=3$ hai, $y$ absent par theek hai), aur autonomous (koi explicit $x$ nahi). Yeh step kyun? Constant right side sabse simple linear forcing hai; missing $y$ linearity nahi todta.
2. (b) $(y^{\prime }{)}^0+y=x$. Koi bhi nonzero quantity power $0$ par $1$ hoti hai, isliye yeh collapse ho jaata hai $1+y=x$ mein, yaani $y=x-1$ — koi derivative survive nahi karta. Yeh ek algebraic equation, order 0 hai: bilkul bhi ODE nahi. Yeh step kyun? Limiting exponent $0$ ek trap hai; degenerate powers derivative ko poori tarah mita sakte hain.
3. (c) $x^2+y^2=4$. Koi derivative nahi → order $0$ → ek algebraic (implicit) relation, ODE nahi. (Ise differentiate karo aur tumhe ek milega: $2x+2y{y}^{\prime }=0$.) Yeh step kyun? "Order 0" order axis ka degenerate floor hai — naam lena zaroori hai taki mislabel na karo.
4. (d) $y^{\prime \prime }=0$. Order 2, degree 1, linear, autonomous. Limiting-simple case: general solution $y=c_{1}x+c_2$ (ek straight line). Yeh step kyun? All-zeros right side homogeneous limit hai; do constants order 2 confirm karte hain.

Answers: (a) order 1, deg 1, linear, autonomous. (b) order 0 par collapse ho jaata hai (algebraic, ODE nahi). (c) order 0 (algebraic). (d) order 2, deg 1, linear, autonomous.

Verify: (a) solution $y=3x+c$, ek constant ✓ order 1. (d) $y=c_{1}x+c_2$, $y^{\prime \prime }=0$ ✓, do constants ✓ order 2.

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Example 9 — word problem: pehle banao phir classify karo (cell C9)

Steps.

1. Words translate karo. "$T$ ka rate of change" $=\frac{dT}{dt}$. "Room se kitna zyada garam uske proportional" $=-k(T-R)$ jahan $k>0$ ek fixed constant hai (minus: thanda hota hai). Yeh step kyun? Har English phrase ek algebraic piece mein map hota hai — yeh modelling half hai, aur ise sahi karna kisi bhi labelling ka prerequisite hai.
2. Fixed-room ODE: $\frac{dT}{dt}=-k(T-20).$ Yeh step kyun? Constant $R=20$ substitute karna us concrete equation deta hai jise hum classify karenge.
3. Classify (fixed room). Order 1; degree 1; linear ($T$ first power par, constant coefficient); autonomous (koi explicit $t$ nahi — $20$ ek constant hai). Yeh step kyun? $R$ constant hone par "landscape" nahi hiltaa → autonomous, aur hum phase line par equilibrium $T=20$ dhundh sakte hain.
4. Moving-room ODE: $\frac{dT}{dt}=-k(T-(20+5\cos t))=-kT+k(20+5\cos t).$ Yeh step kyun? Time-varying $R(t)$ substitute karna aur expand karna explicit $\cos t$ expose karta hai, jo autonomy test dhundh raha hota hai.
5. Re-classify (moving room). Abhi bhi order 1, degree 1, $T$ mein abhi bhi linear — par ab $\cos t$ explicitly appear karta hai → non-autonomous. Yeh step kyun? Sirf autonomy label badalta hai; $t$-dependent room temperature external forcing hai, exactly Example 6 jaisi.

Answers: fixed room — order 1, degree 1, linear, autonomous; moving room — order 1, degree 1, linear, non-autonomous.

Verify (fixed room): solve karo $T^{\prime }=-k(T-20)$. $T=20$ substitute karne par $T^{\prime }=0$ ✓ equilibrium. General solution $T=20+C{e}^{-kt}\to 20$ jaise $t\to \infty$ ✓ coffee room temperature reach karta hai. Ek constant ✓ order 1.

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Example 10 — exam twist: derivatives ke fractions (cell C10)

Steps.

1. Denominator mein derivative spot karo. $y^{\prime }$ ek denominator mein baitha hai; equation abhi derivatives mein polynomial nahi hai. Yeh step kyun? Degree ke liye polynomial form chahiye, isliye denominator mein derivative pehle clear karna zaroori hai (bilkul radical ki tarah).
2. $(1+y^{\prime })$ se multiply karo: $\left(1+\frac{dy}{dx}\right)\frac{d^{2}y}{d{x}^2}=1,$ yaani $y^{\prime \prime }+y^{\prime }{y}^{\prime \prime }=1.$ Yeh step kyun? Fraction clear karna equation ko $y^{\prime }$ aur $y^{\prime \prime }$ mein honest polynomial mein badal deta hai.
3. Order. Highest $y^{\prime \prime }$ hai → order $2$. Yeh step kyun? Order count karta hai kaunsa derivative highest hai, denominator clear karne se unaffected.
4. Degree. Ab highest derivative $y^{\prime \prime }$ ki powers padho: yeh $y^{\prime \prime }$ ke roop mein aur $y^{\prime }{y}^{\prime \prime }$ ke andar appear karta hai — dono $y^{\prime \prime }$ mein first power hain → degree $1$. Yeh step kyun? Degree sirf highest-order derivative ki power dekhta hai; $y^{\prime \prime }$ kabhi squared nahi hai, isliye degree $1$ rehta hai chahe cleanup messy ho.
5. Linear? Term $y^{\prime }{y}^{\prime \prime }$ do derivatives ka product hai → nonlinear (rule 2). Yeh step kyun? Cleanup ne ek product reveal kiya jo fraction chhupa raha tha.
6. Autonomous? Koi explicit $x$ nahi → autonomous. Yeh step kyun? Independent variable $x$ kabhi akele nahi aata — rule sirf current $y^{\prime }$ par depend karta hai.

Answer: order 2, degree 1, nonlinear, autonomous.

Verify: trap tha fraction se degree $>1$ expect karna; instead highest derivative first-power par rehta hai, par ek chhupa product $y^{\prime }{y}^{\prime \prime }$ ise nonlinear banata hai. Isliye cleanup linearity verdict badal sakti hai degree 1 par rehte hue bhi ✓ — exactly woh skill jo yeh cell test karta hai.

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Recall Quick self-test (guess karne ke baad reveal karo)

Har ke liye order, degree, linear?, autonomous? do. $\sqrt{y^{\prime \prime }}=y^{\prime }+x$ ::: order 2; degree 2 (dono taraf square karo → $y^{\prime \prime }=(y^{\prime }+x{)}^2$); nonlinear ($(y^{\prime }+x{)}^2$ mein $y^{\prime }$ squared hai); non-autonomous (explicit $x$). $y^{\prime \prime \prime }+y=0$ ::: order 3; degree 1; linear; autonomous. $y^{\prime }=\frac{y}{t}$ ::: order 1; degree 1; linear; non-autonomous (explicit $t$). $(y^{\prime \prime }{)}^2=y$ ::: order 2; degree 2; nonlinear ($y^{\prime \prime }$ par power 2); autonomous.