Classification — order, degree, linear vs nonlinear, autonomous vs non-autonomous: Exercises

4.6.1 · D4 · Maths › Ordinary Differential Equations › Classification — order, degree, linear vs nonlinear, autonom

Exact tests ke reminders (taaki kabhi guess na karo):

Level 1 — Recognition

L1.1

Har ek ka order batao: (a) $\frac{d y}{d x} = 3 y$ (b) $y^{''} + 4 y = 0$ (c) $(\frac{d ^{4} y}{d x ^{4}})^{7} + y = 0$ .

Recall Solution L1.1

HUM KYA DHUNDH RAHE HAIN: sabse highest derivative, uski power ignore karke.

(a) sabse highest $y^{'}$ hai → order 1.
(b) sabse highest $y^{''}$ hai → order 2.
(c) sabse highest $y^{(4)}$ hai; exponent $7$ ek degree clue hai, order clue nahi → order 4.

L1.2

Har ek ki degree batao: (a) $y^{''} + 5 y^{'} + 6 y = 0$ (b) $(y^{'''})^{2} + y^{'} = x$ (c) $(\frac{d y}{d x})^{3} = y$ .

Recall Solution L1.2

HUM KYA DHUNDH RAHE HAIN: har equation pehle se apne derivatives mein polynomial hai, isliye highest derivative par power padho.

(a) highest $y^{''}$ , power $1$ → degree 1.
(b) highest $y^{'''}$ , power $2$ → degree 2 ( $y^{'}$ term ki power irrelevant hai).
(c) highest derivative $y^{'}$ hai, power $3$ par raised hai → degree 3 (order sirf 1 hai!).

Level 2 — Application

L2.1

Linear vs nonlinear classify karo aur ek-word reason do: (a) $y^{'} + x^{2} y = cos x$ (b) $y^{'} + y^{2} = x$ (c) $y y^{''} = 1$ (d) $y^{''} + sin (x) y = 0$ (e) $y^{'} = sin y$ .

Recall Solution L2.1

Teen linearity rules ko $y$ par apply karo ( $x$ mein coefficients hamesha allowed hain).

(a) $y, y^{'}$ first power mein, coefficient $x^{2}$ theek hai → linear.
(b) $y^{2}$ → $y$ par power $2$ rule 1 todata hai → nonlinear.
(c) $y \cdot y^{''}$ → do unknowns multiply hue rule 2 todata hai → nonlinear.
(d) $sin x$ ek coefficient hai independent variable mein, $y$ first power mein hai → linear.
(e) $sin y$ mein $y$ ek nonlinear function ke andar hai, rule 3 todata hai → nonlinear.

L2.2

Autonomous vs non-autonomous classify karo: (a) $y^{'} = y (1 - y)$ (b) $y^{'} = y + e^{t}$ (c) $y^{''} + 9 y = 0$ (d) $y^{''} + 9 y = cos (2 t)$ .

Recall Solution L2.2

Test: kya independent variable ( $t$ ) explicitly dikha?

(a) RHS sirf $y$ ka function hai → autonomous (logistic — dekho Phase Line and Equilibria).
(b) $e^{t}$ mein $t$ explicitly hai → non-autonomous.
(c) kahin bhi $t$ nahi → autonomous.
(d) forcing $cos (2 t)$ mein $t$ hai → non-autonomous.

Level 3 — Analysis

L3.1

$\frac{d ^{2} y}{d x ^{2}} = 1 + (\frac{d y}{d x})^{2}$ ke liye order aur degree batao, cleanup dikhate hue.

Recall Solution L3.1

SQUARE PEHLE KYUN: radical derivatives par ek power chhupata hai; jab tak hai tab tak honest exponent nahi padh sakte. Dono sides square karo: $(\frac{d ^{2} y}{d x ^{2}})^{2} = 1 + (\frac{d y}{d x})^{2} .$ Ab yeh $y^{'}, y^{''}$ mein polynomial hai. Highest derivative $y^{''}$ ki power $2$ hai. Order 2, degree 2. (Note: order squaring se pehle bhi 2 tha — squaring kabhi order nahi badlta.)

L3.2

$y^{''} + sin (y^{'}) = 0$ ke liye order aur degree justification ke saath batao.

Recall Solution L3.2

Sabse highest derivative $y^{''}$ hai → order 2. Degree: $sin (y^{'})$ derivative $y^{'}$ mein transcendental hai; koi bhi algebra equation ko $y^{'}, y^{''}$ mein polynomial nahi bana sakta. Isliye degree undefined hai. (Contrast L3.1 se karo, jahan ek radical ko square karke hataya ja sakta tha — ek genuine transcendental function of a derivative ko nahi hataya ja sakta.)

L3.3

Neeche diya graph ek autonomous equation aur ek non-autonomous equation ka slope field hai. Kaun sa kaun sa hai, aur kyun?

Figure — Classification — order, degree, linear vs nonlinear, autonomous vs non-autonomous

Recall Solution L3.3

KYA DEKHNA HAI: apni aankh horizontally slide karo (fixed $y$ , changing $t$ ).

Left panel: har horizontal line par arrows ka slope identical hai — field time-shift ke under invariant hai. Yeh ek autonomous equation $y^{'} = f (y)$ ki signature hai: rule clock ko ignore karta hai, isliye ek solution sideways slide karke bhi solution rehta hai.
Right panel: fixed $y$ par $t$ badhne ke saath arrows change hote hain (woh wave ke saath tilt karte hain). Rule clock dekhta hai → non-autonomous $y^{'} = f (t, y)$ .

Level 4 — Synthesis

L4.1

$x (\frac{d ^{3} y}{d x ^{3}})^{2} + (\frac{d y}{d x})^{4} = e^{x}$ ke liye chaar labels do.

Recall Solution L4.1

Order: highest derivative $y^{'''}$ → 3.
Degree: derivatives mein pehle se polynomial hai; $y^{'''}$ par power $2$ hai → 2 ( $(y^{'})^{4}$ highest derivative nahi hai).
Linear? $(y^{'''})^{2}$ aur $(y^{'})^{4}$ powers $> 1$ hain → nonlinear.
Autonomous? coefficient $x$ aur forcing $e^{x}$ independent variable mein explicitly hain → non-autonomous.

L4.2

Aisi ODE banao (ya impossible batao) jo simultaneously linear, non-autonomous, order 2, degree 1 ho. Phir ek jo nonlinear ho phir bhi autonomous ho.

Recall Solution L4.2

Linear + non-autonomous + order 2 + degree 1: $y^{''} + 3 y^{'} + 2 y = cos t .$ $y, y^{'}, y^{''}$ first power mein hain (linear, degree 1), highest derivative $y^{''}$ (order 2), aur $cos t$ ise non-autonomous banata hai. Valid — linearity aur autonomy independent axes hain (compare karo Second-Order Linear ODEs with Constant Coefficients se).

Nonlinear + autonomous: $y^{'} = y (1 - y) .$ Koi $t$ nahi (autonomous) lekin $y^{2}$ term ise nonlinear banata hai. Logistic equation iska classic witness hai.

Level 5 — Mastery

L5.1

Poori tarah classify karo aur explain karo kyun superposition apply ho sakta hai ya nahi: (a) $y^{''} + ω^{2} y = 0$ (b) $y^{''} + ω^{2} sin y = 0$ (pendulum).

Recall Solution L5.1

(a) Order 2, degree 1, linear, autonomous. Kyunki yeh linear aur homogeneous hai, agar $y_{1}, y_{2}$ ise solve karte hain toh $c_{1} y_{1} + c_{2} y_{2}$ bhi karta hai — Superposition Principle apply hota hai. Yahi reason hai ki solution space do basis functions se span hota hai.

(b) Order 2, degree — $sin y$ ek transcendental function hai $y$ mein (kisi derivative mein nahi, isliye order theek hai, lekin equation $y$ mein polynomial nahi hai; degree derivatives ki property hai, isliye agar hum $sin y$ ko coefficient-free term maante hain toh $y^{''}$ mein degree 1 kehte hain — conventionally equation nonlinear hai, highest derivative $y^{''}$ mein degree 1). Nonlinear, autonomous. Superposition fail hota hai: $sin (y_{1} + y_{2}) \neq = sin y_{1} + sin y_{2}$ , isliye solutions ka sum solution nahi hota. Yahi wajah hai ki large-swing pendulum ka koi simple closed form nahi hai.

L5.2

Har ek ke liye decide karo ki kya Existence–Uniqueness machinery ise ek initial-value problem treat karegi, aur order = required initial conditions ki number batao: (a) $y^{'} = y^{2}, y (0) = 1$ (b) $y^{'''} + y = 0, y (0) = y^{'} (0) = y^{''} (0) = 0$ .

Recall Solution L5.2

General solution mein arbitrary constants ki number order ke barabar hoti hai, jo unique solution ke liye required initial conditions ki number ke barabar hoti hai (dekho Picard–Lindelöf).

(a) order 1 → ek initial condition $y (0) = 1$ . Yeh autonomous, nonlinear ( $y^{2}$ ) hai. Separable Equations se solve karne par $y = \frac{1}{1 - x}$ milta hai, sirf $x < 1$ ke liye valid (finite-time blow-up).
(b) order 3 → teen initial conditions, jaise diye gaye hain. Linear, autonomous; yahan sab-zero data $y \equiv 0$ force karta hai.

L5.3

L5.2(a) ka solution claim verify karo: dikhao ki $y = \frac{1}{1 - x}$ $y^{'} = y^{2}$ ko $y (0) = 1$ ke saath solve karta hai, aur uska maximal interval do.

Recall Solution L5.3

ODE check karo: $y = \frac{1}{1 - x} = (1 - x)^{- 1}$ , toh $y^{'} = - (1 - x)^{- 2} \cdot (- 1) = (1 - x)^{- 2} = (\frac{1}{1 - x})^{2} = y^{2} . ✓$ Datum check karo: $y (0) = \frac{1}{1 - 0} = 1. ✓$ Maximal interval: formula $x = 1$ par blast hota hai; $x = 0$ se start karke solution $(- \infty, 1)$ par jeeta hai. Finite time mein yeh blow-up nonlinear growth ki pehchaan hai — ek linear equation yeh kabhi nahi kar sakta.

Recall Ek-line self-test

Chaar classifications mein se kaun se do completely independent axes hain? ::: Linearity aur autonomy — ek equation chaar combinations mein se koi bhi ho sakta hai (linear/nonlinear × autonomous/non-autonomous).