4.6.2 · Maths › Ordinary Differential Equations
Ek first-order ODE d x d y = f ( x , y ) directly nahi batata ki y kya hai.
Yeh batata hai ki plane mein har point ( x , y ) par solution curve ki slope kya hai.
WHY this matters: Agar har point par slope pata ho, to tum "apni naak ki seedh mein chal sakte ho" —
kahin se shuru karo aur us direction mein chalte raho jis taraf slope point kar raha hai. Yahi solution hai.
Ek direction field (slope field) un tamam choti slope arrows ki picture hoti hai.
Euler's method unke saath step-by-step chalte rehne ka numerical recipe hai.
Definition Direction field (slope field)
Diya gaya d x d y = f ( x , y ) , har grid point ( x , y ) par ek choti line segment draw karo jis ki
slope f ( x , y ) ho. In tamam segments ka collection direction field kehlata hai.
Kisi point se guzarne wali solution curve woh curve hoti hai jo har point par segment ke saath tangent ho.
WHY it works: ODE ek constraint hai — "yahan teri slope f ( x , y ) ke barabar honi chahiye." Segment
us constraint ko sirf visualise karta hai. Solution koi bhi aisi curve hai jo har jagah constraint obey kare.
d x d y = x ko padhna
( 0 , 0 ) par: slope = 0 → horizontal dash.
( 2 , 5 ) par: slope = 2 → steep up dash.
( − 1 , 7 ) par: slope = − 1 → down dash.
Why this step? Yahan slope sirf x par depend karta hai, isliye ek vertical column mein har dash parallel hota hai.
Solutions parabolas y = 2 x 2 + C hain — aur wakai sab ki ek given x par same slope hoti hai. ✔
Intuition Isoclines — sketch karne ka aasaan tarika
Ek isocline woh curve hoti hai f ( x , y ) = m jahan slope ek fixed value m ke barabar ho.
Iske upar, har dash ki same tilt hoti hai. Kuch isoclines sketch karo, har ek par parallel dashes banao,
aur field almost free mein bann jata hai. (Yeh haath se sketch karne ka 80/20 trick hai.)
Hum ODE ko formulas se hamesha solve nahi kar sakte. Lekin hum arrows ko numerically follow kar sakte hain.
Ek jaane-pehchane point ( x 0 , y 0 ) (initial condition) se shuru karo. Baad ke x values
x 1 , x 2 , … par y nikalo jo ek chote step h se spaced hain, yaani x n + 1 = x n + h .
Derivation (first principles — local linear approximation):
Derivative hi tangent line ki slope hai. x n ke kareeb, true solution apni tangent se achhi tarah approximate hoti hai:
y ( x n + h ) ≈ y ( x n ) + h ⋅ y ′ ( x n ) .
Yeh sirf first-order Taylor expansion hai:
y ( x n + h ) = y ( x n ) + h y ′ ( x n ) + 2 h 2 y ′′ ( ξ ) .
Lekin ODE hamein y ′ ( x n ) = f ( x n , y n ) deta hai. Substitute karo aur h 2 remainder ko drop karo:
Intuition WHY global error
O ( h ) hai, O ( h 2 ) nahi
Har step mein error ∼ h 2 hoti hai. L length ka interval cross karne ke liye N = L / h steps lagte hain.
Total error ∼ N ⋅ h 2 = h L ⋅ h 2 = L h = O ( h ) .
To h ko half karne se total error roughly half ho jati hai. Euler ek first-order method hai.
d x d y = x + y , y ( 0 ) = 1 , h = 0.5 ke saath solve karo, y ( 1 ) nikalo.
Yahan f ( x , y ) = x + y , x 0 = 0 , y 0 = 1 .
Step 1 (n = 0 ): slope = f ( 0 , 1 ) = 0 + 1 = 1 .
y 1 = y 0 + h f = 1 + 0.5 ( 1 ) = 1.5 , x 1 = 0.5
Why this step? Hum slope 1 ke saath h = 0.5 daayein chalte hain, to rise = 0.5 ⋅ 1 = 0.5 .
Step 2 (n = 1 ): slope = f ( 0.5 , 1.5 ) = 0.5 + 1.5 = 2 .
y 2 = 1.5 + 0.5 ( 2 ) = 2.5 , x 2 = 1.0
Why this step? Naya slope naye point par recompute hota hai — Euler hamesha wahan ki slope use karta hai jahan woh abhi khada hai.
Answer: y ( 1 ) ≈ 2.5 .
Exact se check karo: true solution y = 2 e x − x − 1 hai, to y ( 1 ) = 2 e − 2 ≈ 3.44 .
Euler under-shoot karta hai kyunki curve upar ki taraf bend karti hai (y ′′ > 0 ) aur tangents uske neeche rehti hain.
Chota h gap ko kam karta hai.
Worked example Forecast-then-Verify
Forecast: Usi problem ke liye h = 0.25 (4 steps) ke saath, kya y ( 1 ) bada ya chota hoga, 3.44 ke kareeb ya door?
Verify: Steps dete hain y ≈ 1 , 1.25 , 1.5625 , 1.953 , 2.441 → y ( 1 ) ≈ 2.44 ? Dhyaan se recompute karo:
y 1 = 1 + 0.25 ( 1 ) = 1.25 ; f = 0.25 + 1.25 = 1.5 ⇒ y 2 = 1.625 ; f = 0.5 + 1.625 = 2.125 ⇒ y 3 = 2.156 ;
f = 0.75 + 2.156 = 2.906 ⇒ y 4 = 2.883 . To y ( 1 ) ≈ 2.88 — coarse 2.50 se 3.44 ke kareeb . ✔ Chota h , behtar.
Common mistake "Naye point
f ( x n + 1 , y n + 1 ) ki slope use karo."
Why it feels right: Lagta hai jahan ja rahe ho wahan ki slope "zyada current" hai.
Fix: Plain Euler explicit hai — yeh us point ki slope use karta hai jahan se tum ja rahe ho , f ( x n , y n ) ,
kyunki y n + 1 abhi pata nahi. Naya point use karna implicit (backward) Euler hai, ek alag method.
h se multiply karna bhool jana."
Why it feels right: Tumne f ( x n , y n ) likh liya aur add kar diya — lagta hai poora ho gaya.
Fix: f ek slope hai, rise nahi. Rise = slope × run = h f . Units match karni chahiye: Δ y = ( Δ y /Δ x ) ⋅ Δ x .
Common mistake "Direction field arrows time mein motion ki direction dikhate hain / unki fixed length ka matlab hota hai."
Why it feels right: Arrows velocity vectors jaisi lagti hain.
Fix: Sirf har segment ki slope (tilt) mein meaning hai — length cosmetic hai. Yeh tangent direction mark karte hain, kuch nahi.
Common mistake "Zyada steps matlab hamesha sahi answer."
Why it feels right: Upar chota h ne error reduce ki.
Fix: Truncation error → 0 jab h → 0 , lekin zyada steps ke saath rounding error badhti hai, aur bahut stiff equations blow up ho sakti hain. Euler convergent hai lekin sirf first-order accurate.
Recall Quick self-test (answers hide karo)
Q: y ′ = f ( x , y ) mein f ( x , y ) physically kya deta hai? → ( x , y ) par solution ki slope .
Q: Euler update formula? → y n + 1 = y n + h f ( x n , y n ) .
Q: Global error ka order? → O ( h ) (first order).
Q: Euler slope kahan evaluate karta hai? → current/left point par.
Recall Feynman: 12-saal ke bacche ko samjhao
Socho tum fog mein chal rahe ho aur ek magic compass tumhe batata hai ki "neeche ki taraf" exactly kidhar hai jahan tum khade ho .
Tum poori pahadi nahi dekh sakte, lekin tum us direction mein ek chota step lete ho, compass se phir poochte ho, ek
aur chota step lete ho, aur aage bhi yahi karte rehte ho. Jo path tum trace karte ho woh trail ki tumhari best guess hai. ODE magic
compass hai; Euler's method chote steps lena hai. Chote steps = asli trail ke kareeb, lekin zyada kaam.
"Stand, Slope, Step." ( x n , y n ) par khade raho → Slope f padho → Step y + h f .
Aur formula ke liye: "New = Old + run × rise-rate."
What does f ( x , y ) in y ′ = f ( x , y ) tell you geometrically? ( x , y ) point par solution curve ki slope.
Define a direction (slope) field. Har point ( x , y ) par slope f ( x , y ) ka ek chota segment; solutions woh curves hain jo in segments ke har jagah tangent hain.
What is an isocline? f ( x , y ) = m ka locus jahan sab field segments ek hi slope m share karte hain.
State Euler's method update. y n + 1 = y n + h f ( x n , y n ) with x n + 1 = x n + h .
Derive Euler from Taylor. y ( x n + h ) = y n + h y n ′ + 2 h 2 y ′′ ( ξ ) ; y n ′ = f ( x n , y n ) replace karo aur O ( h 2 ) term drop karo.
What is the local truncation error of Euler? 2 h 2 y ′′ ( ξ ) , yaani O ( h 2 ) per step.
What is the global error order of Euler? O ( h ) — first order; h half karne se total error roughly half hoti hai.
Which point's slope does explicit Euler use? Current (left) point ( x n , y n ) , naye point ki nahi.
Why does Euler undershoot a convex (y ′′ > 0 ) solution? Tangent lines ek convex curve ke neeche rehti hain, to har step thoda neeche land karta hai.
Common error: omitting which factor? Slope ko step h se multiply karna; Δ y = h f , sirf f nahi.
Local error O of h squared