Direction fields and Euler's method — visual - numerical intuition first: Worked examples

4.6.2 · D3 · Maths › Ordinary Differential Equations › Direction fields and Euler's method — visual - numerical int

Yeh page parent topic ki drill sheet hai. Hum yahan Euler's method aur direction fields ke har tarah ke situation ko cover karenge — positive slopes, negative slopes, bilkul zero slope, upar bend karta curve, neeche bend karta curve, itna bada step jo tumhe galat answer deta hai, ek real word problem, aur ek exam-style trap.

Koi bhi symbol aane se pehle, ek reminder plain words mein:

Scenario matrix

Har Euler / direction-field problem neeche diye cells ka koi combination hota hai. Jo worked examples follow karte hain, unhe un cell(s) ke saath tag kiya gaya hai jo woh cover karte hain, taaki ant mein koi bhi cell blank na rahe.

#	Cell (case class)	Isme kya khaas hai	Covered by
A	Slope $f > 0$ (rising)	height har step mein upar jaati hai	Ex 1, Ex 4
B	Slope $f < 0$ (falling)	height har step mein neeche jaati hai	Ex 2, Ex 3
C	Slope $f = 0$ (flat point)	ek step jo sirf sideways move karta hai	Ex 3
D	Convex solution $y^{''} > 0$ (bowl)	Euler under-shoots (tangents curve ke neeche)	Ex 1, Ex 3, Ex 6
E	Concave solution $y^{''} < 0$ (dome)	Euler over-shoots (tangents curve ke upar)	Ex 2
F	Equilibrium / degenerate $f \equiv 0$ line	solution flat line hai, Euler hamesha wahan rukta hai	Ex 3b
G	Step-size effect (halve $h$ )	error roughly half ho jaata hai	Ex 4
H	Big $h$ overshoot / instability	Euler blow past karke oscillate kar sakta hai	Ex 5
I	Real-world word problem	$f$ ka naam, units, answer interpret karna	Ex 6
J	Exam twist (backward vs forward, sign trap)	woh classic marks-khoone-wali mistakes	Ex 7

Example 1 — Cell A + D: rising slope, convex (bowl) curve

Worked example

$\frac{d y}{d x} = x + y, y (0) = 1, h = 0.5$ , $y (1)$ nikalo. Forecast: Slope $f = x + y$ start pe $0 + 1 = 1$ (positive) hai aur jaise-jaise $x$ aur $y$ dono badhte hain yeh sirf badhta jaata hai — slope increasing hai, isliye $y^{''} > 0$ : ek bowl (convex). Compute karne se pehle: kya Euler true curve ke upar ya neeche land karega? Abhi guess karo.

Step 1 ( $n = 0$ ): start pe slope machine read karo. $f (0, 1) = 0 + 1 = 1.$ Yeh step kyun? Euler hamesha slope ko us point pe evaluate karta hai jahan woh khara hai — step ke left end pe. Step lo: $y_{1} = 1 + 0.5 \cdot 1 = 1.5, x_{1} = 0.5.$ Yeh step kyun? rise = run × slope = $0.5 \times 1 = 0.5$ , isliye height $1$ se $1.5$ tak chadhi.

Step 2 ( $n = 1$ ): machine ko naye point pe re-read karo. $f (0.5, 1.5) = 0.5 + 1.5 = 2.$ $y_{2} = 1.5 + 0.5 \cdot 2 = 2.5, x_{2} = 1.0.$ Yeh step kyun? Slope ko dobara compute karna padega — yeh $1$ se $2$ ho gayi kyunki hum move kiye.

Answer: $y (1) \approx 2.5$ .

Verify: Exact solution hai $y = 2 e^{x} - x - 1$ (yeh ek separable-adjacent linear ODE hai — Separable First-Order ODEs techniques se compare karo). Phir $y (1) = 2 e - 2 \approx 3.44$ . Euler ne $2.5$ diya, ek under-shoot — forecast se match karta hai, kyunki bowl ke liye ( $y^{''} > 0$ ) har seedha tangent step true curve ke neeche hota hai. Figure dekho: har red step black curve ke neeche end hota hai.

Figure 1 — Euler ke seedhe steps (red) true convex "bowl" curve ke neeche hain: ek undershoot.

Example 2 — Cell E: ek concave (dome) curve, aur ise kaise pehchane

Worked example

$\frac{d y}{d x} = \frac{1}{2 x + 1}, y (0) = 1, h = 1$ , $y (3)$ nikalo. (Ek genuine dome, Cell E.) Forecast: Slope $\frac{1}{2 1} = 0.5$ se start hoti hai aur $x$ badhne se ghatti hai — ek decreasing slope, isliye $y^{''} < 0$ : ek dome (concave). Compute karne se pehle: kya Euler true curve ke upar ya neeche land karega? Abhi guess karo.

Step 1 ( $n = 0$ , $x = 0$ pe): slope sirf $x$ pe depend karti hai. $f (0) = \frac{1}{2 1} = 0.5.$ $y_{1} = 1 + 1 \cdot 0.5 = 1.5, x_{1} = 1.$ Yeh step kyun? rise = run × slope = $1 \times 0.5 = 0.5$ .

Step 2 ( $n = 1$ , $x = 1$ pe): $f (1) = \frac{1}{2 2} \approx 0.3536.$ $y_{2} = 1.5 + 1 \cdot 0.3536 = 1.8536, x_{2} = 2.$ Yeh step kyun? Slope pehle se shallower hai ( $0.354 < 0.5$ ) — slope ghatt raha hai, dome ka signature ( $y^{''} < 0$ ).

Step 3 ( $n = 2$ , $x = 2$ pe): $f (2) = \frac{1}{2 3} \approx 0.2887.$ $y_{3} = 1.8536 + 1 \cdot 0.2887 = 2.1423, x_{3} = 3.$

Answer: $y (3) \approx 2.142$ .

Verify: Exact $y = x + 1$ , isliye $y (3) = 4 = 2$ . Euler ne $2.142$ diya — ek over-shoot, bilkul waise jaisa dome ( $y^{''} < 0$ ) predict karta hai: har tangent line true curve ke upar hoti hai, isliye har step bahut upar land karta hai. Yeh Cell E sahi se done. Sanity: slopes $0.5 \to 0.354 \to 0.289$ tak gire, confirm karta hai curve puri tarah neeche bend ho raha tha.

Figure 2 — Concave "dome" curve ke liye red Euler steps true curve ke upar ride karte hain: ek overshoot.

Example 3 — Cell B + D aur Cell C + F: ek decay, aur uski equilibrium

Worked example Part (a) — ek

falling, convex decay: $\frac{d y}{d x} = - y, y (0) = 2, h = 0.5$ , $y (1)$ nikalo. Forecast: $f = - y < 0$ $y > 0$ ke liye (Cell B, falling). Bowl hai ya dome? Slope-of-slope: $y^{''} = \frac{d}{d x} (- y) = - y^{'} = - (- y) = + y > 0$ — ek bowl (convex). Isliye predict karo ek under-shoot bhale hi curve falling ho: "falling" over/undershoot decide nahi karta; $y^{''}$ karta hai.

Step 1 ( $n = 0$ ): $f (0, 2) = - 2. y_{1} = 2 + 0.5 (- 2) = 1, x_{1} = 0.5.$ Yeh step kyun? Negative slope ⇒ step neeche jaata hai: rise $= 0.5 \times (- 2) = - 1$ .

Step 2 ( $n = 1$ ): $f (0.5, 1) = - 1. y_{2} = 1 + 0.5 (- 1) = 0.5, x_{2} = 1.0.$ Yeh step kyun? Slope neeche wale point pe recompute hui; yeh shallower hai ( $- 1$ vs $- 2$ ) — slope zero ki taraf upar ja raha hai, yaani increasing, confirm karta hai $y^{''} > 0$ .

Answer: $y (1) \approx 0.5$ .

Verify: Exact $y = 2 e^{- x}$ , $y (1) = 2 e^{- 1} \approx 0.736$ . Euler ne $0.5$ diya — ek under-shoot, bowl prediction ( $y^{''} = 2 e^{- x} > 0$ ) se match karta hai. Units/sanity: $y$ positive raha aur ghata, jaise kisi bhi decay mein hona chahiye. (Yeh woh case hai jo logon ko trip karta hai: falling lekin convex.)

Worked example Part (b) — Cell C + F:

equilibrium line pe zero slope. $\frac{d y}{d x} = y (1 - y), y (0) = 1, h = 0.5$ , $y (1.5)$ nikalo. Forecast: $f = y (1 - y) = 0$ exactly $y = 1$ pe. $y = 1$ pe machine zero return karti hai — ek horizontal dash. Kya hoga agar Euler us flat line pe start kare?

Step 1 ( $n = 0$ ): $f (x, 1) = 1 (1 - 1) = 0. y_{1} = 1 + 0.5 \cdot 0 = 1, x_{1} = 0.5.$ Yeh step kyun? Slope exactly $0$ ⇒ step sirf sideways hai — height unchanged.

Steps 2 & 3: ab bhi $y = 1$ hai, isliye $f = 0$ phir: $y_{2} = 1, x_{2} = 1.0; y_{3} = 1, x_{3} = 1.5.$ Yeh step kyun? $y = 1$ ek equilibrium hai (ek degenerate case: $f \equiv 0$ ek puri horizontal line ke saath). Euler isme phansa rehta hai hamesha ke liye — jo exactly sahi hai.

Answer: $y (1.5) = 1$ (flat).

Verify: $y (0) = 1$ ke saath exact solution constant $y \equiv 1$ hai (ek equilibrium stability analysis mein padha jata hai). Euler ise perfectly reproduce karta hai: zero slope ⇒ zero rise. Direction field mein yeh height $y = 1$ pe horizontal dashes ki row hai; nearby curves iske toward draw hoti hain. Figure dekho: red equilibrium line bilkul flat hai.

Figure 3 — Logistic direction field: red line $y = 1$ pe har dash horizontal hai, isliye Euler kabhi use nahi chhodta.

Example 4 — Cell A + G: step halve karo, error ko shrink hote dekho

Worked example Example 1 jaisa hi (

$y^{'} = x + y, y (0) = 1$ ) lekin $h = 0.25$ , $y (1)$ tak char steps. Forecast: Example 1 mein $h = 0.5$ se $2.5$ mila; truth hai $3.44$ . Euler ek first-order method hai — plain words mein, iska matlab hai total error step $h$ ke proportion mein ghatta hai: $h$ halve karo, aur error roughly half ho jaata hai (quarter nahi — proportion, square nahi). Isliye half step se hum expect karte hain answer $3.44$ ke kareeb (around $2.9$ ). Compute karne se pehle guess karo.

Step 1 ( $n = 0$ ): $(0, 1.0000)$ pe, slope $f = 0 + 1 = 1.0000$ ; step $y_{1} = 1.0000 + 0.25 (1.0000) = 1.2500$ , $x_{1} = 0.25$ . Yeh step kyun? Slope start point pe read karo aur run×slope $= 0.25 \times 1 = 0.25$ se rise lo.

Step 2 ( $n = 1$ ): $(0.25, 1.2500)$ pe, slope $f = 0.25 + 1.25 = 1.5000$ ; step $y_{2} = 1.2500 + 0.25 (1.5000) = 1.6250$ , $x_{2} = 0.50$ . Yeh step kyun? Slope naye point pe recompute ki — yeh $1$ se $1.5$ tak baadhi kyunki $x$ aur $y$ dono bade.

Step 3 ( $n = 2$ ): $(0.50, 1.6250)$ pe, slope $f = 0.50 + 1.625 = 2.1250$ ; step $y_{3} = 1.6250 + 0.25 (2.1250) = 2.1563$ , $x_{3} = 0.75$ . Yeh step kyun? Phir se current spot pe fresh slope; aur bhi steep, isliye rise per step badhti rehti hai.

Step 4 ( $n = 3$ ): $(0.75, 2.1563)$ pe, slope $f = 0.75 + 2.1563 = 2.9063$ ; step $y_{4} = 2.1563 + 0.25 (2.9063) = 2.8828$ , $x_{4} = 1.00$ . Yeh step kyun? Final slope wahan read ki jahan hum ab khade hain; ek aur run×slope humein $x = 1$ pe land karta hai.

Table ki tarah collect kiya:

$n$	$x_{n}$	$y_{n}$	$f = x_{n} + y_{n}$	$y_{n + 1} = y_{n} + 0.25 f$
0	0.00	1.0000	1.0000	1.2500
1	0.25	1.2500	1.5000	1.6250
2	0.50	1.6250	2.1250	2.1563
3	0.75	2.1563	2.9063	2.8828

Answer: $y (1) \approx 2.883$ .

Verify: Truth hai $3.44$ . Errors: $h = 0.5$ ne error $3.44 - 2.50 = 0.94$ diya; $h = 0.25$ ne $3.44 - 2.88 = 0.56$ . Ratio $0.56/0.94 \approx 0.60$ — exactly $\frac{1}{2}$ nahi kyunki "error proportional to $h$ " sirf leading behaviour hai jo exact tab banta hai jab $h$ bahut chhota ho; yahan $h$ abhi bhi kaafi bada hai. Lekin error clearly ghata chhote $h$ se, jaisa first-order accuracy ka waada hai. Gap bahut tezi se close karne ke liye tum RK4 switch karoge, jiska error $h^{4}$ ki tarah ghatta hai. Do staircases compare karta figure dekho.

Figure 4 — Do Euler staircases: finer step (red) true curve ke saath coarse wale se zyada closely hugs karta hai.

Example 5 — Cell H: itna bada step ki Euler unstable ho jaata hai

Worked example

$\frac{d y}{d x} = - 5 y, y (0) = 1$ . Char steps ke liye $h = 0.5$ try karo, phir explain karo. Forecast: True solution $y = e^{- 5 x}$ smoothly $0$ ki taraf decay karta hai. Lekin slope steep hai (factor $5$ ). Ek fast-decaying curve ko ek bada step kya kar sakta hai? Guess karo: smooth decay, ya wild oscillation?

Steps $y_{n + 1} = y_{n} + h (- 5 y_{n}) = y_{n} (1 - 5 h) = y_{n} (1 - 2.5) = - 1.5 y_{n}$ use karke: $y_{1} = - 1.5, y_{2} = 2.25, y_{3} = - 3.375, y_{4} = 5.0625.$ Yeh step kyun? Amplification factor $1 - 5 h = 1 - 2.5 = - 1.5$ hai. Iska size $∣ - 1.5∣ = 1.5 > 1$ , isliye har step magnify karta hai aur sign flip karta hai — numbers explode aur oscillate karte hain.

Answer: Euler ${- 1.5, 2.25, - 3.375, 5.0625}$ deta hai — true $y (0.5) = e^{- 2.5} \approx 0.082$ ke comparison mein bilkul bakwaas.

Verify: Explicit Euler ke liye $y^{'} = λ y$ pe stability rule hai $∣1 + hλ ∣ \leq 1$ . Yahan $λ = - 5$ , isliye hamein $∣1 - 5 h ∣ \leq 1$ chahiye, yaani $0 \leq h \leq 0.4$ . Hamara $h = 0.5$ ise violate karta hai, isliye blow-up. $h = 0.1$ se dobara karo: factor $= 1 - 0.5 = 0.5$ , jisse $y_{1} = 0.5, y_{2} = 0.25, \dots \to 0$ — well-behaved. Yeh exactly Stability and Stiff Equations ki "stiff equation" warning hai: chhota $h$ sirf accuracy ke liye nahi, kabhi kabhi yeh survival ke liye hai. Figure dekho: exploding red zig-zag vs tame decay.

Figure 5 — Oversized step se red Euler iterates zig-zag karte hain aur blow up karte hain, jabki true solution quietly decay karta hai.

Example 6 — Cell I + D: real-world word problem

Worked example Ek cup of coffee Newton's law se cool hoti hai: temperature

$T$ (°C) $\frac{d T}{d t} = - k (T - 20)$ follow karta hai jisme $k = 0.2 min^{- 1}$ , room $20°$ C pe, $T (0) = 90°$ C se start. $h = 1$ min use karke $2$ minutes baad $T$ estimate karo. Forecast: Coffee room temperature se bahut upar hai, isliye pehle tezi se cool honi chahiye, phir slow (slope $\propto$ gap $T - 20$ ). $2$ minutes ki cooling mein zyada giregi ya thodi? Ek range guess karo.

$f$ ka naam: yahan $f (t, T) = - 0.2 (T - 20)$ . Units check: $[k] = 1/ min$ , $[T - 20] = °$ C, isliye $[f] = ° C / min$ — ek genuine rate of temperature change. Achha.

Step 1 ( $n = 0$ , $t_{0} = 0, T_{0} = 90$ pe): $f = - 0.2 (90 - 20) = - 0.2 \cdot 70 = - 14 ° C/min .$ $T_{1} = 90 + 1 \cdot (- 14) = 76° C, t_{1} = t_{0} + h = 1.$ Yeh step kyun? rise = run × rate = $1 min \times (- 14°/ min) = - 14°$ .

Step 2 ( $n = 1$ , $t_{1} = 1, T_{1} = 76$ pe): $f = - 0.2 (76 - 20) = - 0.2 \cdot 56 = - 11.2° C/min .$ $T_{2} = 76 + 1 \cdot (- 11.2) = 64.8° C, t_{2} = t_{1} + h = 2.$ Yeh step kyun? Gap ghata, isliye cooling rate bhi ghati ( $- 11.2$ vs $- 14$ ) — woh "slow down" behaviour jo humne forecast kiya tha.

Answer: $T (2) \approx 64.8°$ C.

Verify: Exact solution $T = 20 + 70 e^{- 0.2 t}$ , isliye $T (2) = 20 + 70 e^{- 0.4} \approx 20 + 70 (0.6703) = 66.9°$ C. Euler ka $64.8$ true $66.9$ ko under-shoot karta hai: cooling curve $T$ mein ek bowl hai ( $T^{''} > 0$ ), tangents neeche hoti hain, Cell D se consistent. Physically estimate sahi lagta hai (abhi bhi hot coffee), aur error ki direction exactly convexity ke predict kiye hue hai.

Example 7 — Cell J: exam twist (forward vs backward Euler, sign trap)

Worked example

$\frac{d y}{d x} = 2 x - y, y (1) = 3$ . Ek exam ek forward-Euler step $h = - 0.5$ ke saath $y (0.5)$ estimate karne ke liye poochta hai. (Note karo negative step — hum left chalte hain.) Forecast: Negative $h$ ka matlab hai chhote $x$ ki taraf move karna. Formula unchanged hai; trap hai (a) galat point ka slope use karna, aur (b) $h$ ke sign ko galat handle karna. Dono watch karo.

Step 1 ( $n = 0$ , $x_{0} = 1, y_{0} = 3$ pe): $f (1, 3) = 2 \cdot 1 - 3 = - 1.$ Yeh step kyun? Explicit (forward) Euler us point pe slope use karta hai jahan tum chhod rahe ho, $(x_{0}, y_{0})$ — destination pe nahi. Destination point use karna backward/implicit Euler hoga, ek alag method (aur yahan algebra ke bina solve nahi hoga). $y_{1} = y_{0} + h f = 3 + (- 0.5) (- 1) = 3 + 0.5 = 3.5, x_{1} = 1 + (- 0.5) = 0.5.$ Yeh step kyun? Do negatives multiply hokar plus bante hain: downward slope par left chalte hue $y$ upar jaata hai. Dono sign flips ko respect karna padega.

Answer: $y (0.5) \approx 3.5$ .

Verify: $y^{'} = 2 x - y, y (1) = 3$ ka exact solution $y = 2 x - 2 + 3 e^{1 - x}$ hai. Check: $x = 1$ pe, $y = 2 - 2 + 3 = 3$ ✓. $x = 0.5$ pe: $y = 1 - 2 + 3 e^{0.5} = - 1 + 3 (1.6487) = 3.946$ . Euler ka ek bada backward step $3.5$ diya — sahi ballpark, aur yeh correctly upar move kiya negative slope ke bawajood, kyunki $h < 0$ . Do classic mark-losers (destination-slope trap, $h$ ka sign) dono handle ho gaye.

Recall

Recall Kis cell ne under/overshoot cause kiya?

$f$ ka sign (rising/falling) direction decide karta hai; $y^{''}$ ka sign (bowl/dome) over- ya under-shoot decide karta hai. ::: Convex $y^{''} > 0$ (bowl) ⇒ tangents neeche ⇒ Euler undershoots; concave $y^{''} < 0$ (dome) ⇒ tangents upar ⇒ Euler overshoots.

Recall

$y^{'} = λ y$ ke liye stability threshold? Explicit Euler ke blow up na karne ke liye $h$ kya satisfy karna chahiye? ::: $∣1 + hλ ∣ \leq 1$ ; $λ = - 5$ ke liye yeh $h \leq 0.4$ chahiye.

Recall Negative step size — kya badalta hai?

Kya update formula tab badalta hai jab $h < 0$ ho? ::: Nahi — same formula $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ ; negative $h$ automatically left chalti hai.

Connections

Separable First-Order ODEs — jahan se exact comparison curves aati hain.
Taylor's Theorem — "error proportional to $h$ " (first-order) accuracy ka origin.
Runge-Kutta Methods (RK4) — Euler ki slow first-order accuracy ko kaise beat karein.
Existence and Uniqueness (Picard–Lindelöf) — guarantee karta hai ki woh single curve jishe Euler chase kar raha hai exist karti hai.
Stability and Stiff Equations — Example 5 ka blow-up, poori detail mein.