Exact equations — exactness condition, finding potential function: Worked examples

4.6.6 · D3 · Maths › Ordinary Differential Equations › Exact equations — exactness condition, finding potential fun

Intuition Yeh page kis liye hai

Parent note ne tumhe machinery di thi: $\partial M / \partial y = \partial N / \partial x$ test karo, phir hill $F (x, y)$ rebuild karo aur $F = c$ likho. Yahan hum us machinery ko har tarah ki equation pe stress-test karte hain jo tumhe mil sakti hai: seedhi-saadhi equations, sign-flipped wali, zero coefficient wali, test fail karne wali, sirf ek trick ke baad exact banne wali, aur ek real word problem.

Shuru karne se pehle, do reminders plain words mein taaki koi symbol unearned na ho:

$M$ woh coefficient hai jo $d x$ ke aage baitha hai; $N$ woh coefficient hai jo $d y$ ke aage hai. Inhe ek hill $F$ ki do slopes samjho: kitni steep hai jab tum east step karo ( $M = F_{x}$ ) aur jab north step karo ( $N = F_{y}$ ).
$\partial M / \partial y$ (padho "partial of $M$ with respect to $y$ ", likha jaata hai $M_{y}$ ) matlab hai: $x$ ko ek fixed number maan ke freeze karo, phir $M$ ko sirf $y$ ko variable treat karke differentiate karo. Yahi partial derivative hai.

Scenario matrix

Har exact-equation problem in cells mein se ek mein aati hai. Neeche ke examples mein cell label diya gaya hai jise woh cover karta hai, taaki milke koi gap na rahe.

#	Cell (scenario class)	Isme kya tricky hai	Example
A	Polynomial, cleanly exact	baseline mechanics	Ex 1
B	Mixed transcendental (exp, trig, log)	unusual functions ko integrate karna	Ex 2
C	Negative / sign-flipped coefficients	signs test mein survive karni chahiye	Ex 3
D	Ek coefficient sirf ek variable ka pure function hai	ek partial derivative zero hai	Ex 4
E	Sirf implicit solution (can't isolate $y$ )	answer $F = c$ ke roop mein rehta hai	Ex 5
F	Not exact — test fails	method refuse karna padega	Ex 6
G	Not exact but fixable by an [[Integrating Factors for ODEs	integrating factor]]	rescue karo phir solve karo
H	Real-world word problem + initial condition	constant $c$ pin karo	Ex 8

Hum degenerate limit bhi cover karte hain (kya hota hai jab $M$ ya $N$ poora vanish ho jaaye) Ex 4 ke andar, aur geometry (solution as level curves) Ex 1 aur Ex 8 ke figures mein.

Cell A — polynomial, cleanly exact

Worked example Example 1

Solve karo $(3 x^{2} y + 2 x y) d x + (x^{3} + x^{2} + 2 y) d y = 0.$

Forecast: Padhne se pehle guess karo — kya yeh test pass karega? Aur kya answer mein $x^{3} y$ term hogi? Apna guess note karo.

Identify karo: $M = 3 x^{2} y + 2 x y$ , $N = x^{3} + x^{2} + 2 y .$

Step 1 — Exactness test karo. Kyun yeh step? Ek aisi hill rebuild karne ka koi fayda nahi jo exist hi nahi karti. $x$ freeze karo, $M$ ko $y$ mein differentiate karo: har term se $y$ hat jaata hai, milta hai $M_{y} = 3 x^{2} + 2 x .$ $y$ freeze karo, $N$ ko $x$ mein differentiate karo: $N_{x} = 3 x^{2} + 2 x .$ Match karte hain ⟹ exact.

Step 2 — $M$ ko $x$ mein integrate karo. Kyun? $F_{x} = M$ , toh $x$ -derivative undo karne se $F$ milta hai ek $y$ -only piece tak. $F = \int (3 x^{2} y + 2 x y) d x + g (y) = x^{3} y + x^{2} y + g (y) .$

Step 3 — $y$ mein differentiate karo aur $N$ se match karo. Kyun? Bacha hua $g (y)$ $M$ ko invisible hai; sirf $F_{y} = N$ usse reveal kar sakta hai. $F_{y} = x^{3} + x^{2} + g^{'} (y) =! x^{3} + x^{2} + 2 y \Rightarrow g^{'} (y) = 2 y .$ Dhyan do ki $N$ ka $x^{3} + x^{2}$ part pehle se $F$ ke andar tha — isse dobara integrate mat karo (yahi double-counting trap hai).

Step 4 — $g^{'}$ integrate karo. $g (y) = y^{2} .$

Answer: $x^{3} y + x^{2} y + y^{2} = c .$

Verify karo: $F = x^{3} y + x^{2} y + y^{2}$ lo. Toh $F_{x} = 3 x^{2} y + 2 x y = M$ ✓ aur $F_{y} = x^{3} + x^{2} + 2 y = N$ ✓. Differential $d F$ original equation exactly rebuild karta hai.

Figure — Exact equations — exactness condition, finding potential function

Figure mein contour lines $F = c$ dikhaye gaye hain: ODE kehta hai "aisa move karo ki height kabhi na badle," toh har solution curve inhi level curves mein se ek hai.

Cell B — mixed transcendental

Cell C — negative / sign-flipped coefficients

Worked example Example 3

Solve karo $(2 x - y sin (x y)) d x - x sin (x y) d y = 0.$

Forecast: $N$ negative hai. Kya negative sign exactness tod deta hai? Guess karo.

Identify karo: $M = 2 x - y sin (x y)$ , $N = - x sin (x y) .$

Step 1 — Test karo. Kyun? Signs matter karte hain; carefully verify karo. $M_{y}$ : $- y sin (x y)$ ko $y$ mein product rule se differentiate karo ( $x$ freeze karo): $M_{y} = - [sin (x y) + y cos (x y) \cdot x] = - sin (x y) - x y cos (x y) .$ $N_{x}$ : $- x sin (x y)$ ko $x$ mein differentiate karo ( $y$ freeze karo): $N_{x} = - [sin (x y) + x cos (x y) \cdot y] = - sin (x y) - x y cos (x y) .$ Match karte hain ⟹ exact — minus sign dono sides pe survive kar gaya.

Step 2 — Is baar $N$ ko $y$ mein integrate karo (yeh simpler hai; symmetric route hamesha allowed hai). $x$ freeze karo: $F = \int - x sin (x y) d y + h (x) = cos (x y) + h (x),$ kyunki $\partial_{y} cos (x y) = - x sin (x y) .$

Step 3 — $M$ se match karo: Kyun? Ab $F_{x} = M$ use karo $h (x)$ find karne ke liye. $F_{x} = - y sin (x y) + h^{'} (x) =! 2 x - y sin (x y) \Rightarrow h^{'} (x) = 2 x .$

Step 4 — Integrate karo: $h (x) = x^{2} .$

Answer: $cos (x y) + x^{2} = c .$

Verify karo: $F_{x} = - y sin (x y) + 2 x = M$ ✓. $F_{y} = - x sin (x y) = N$ ✓.

Cell D — ek coefficient sirf ek variable pe depend karta hai (partial = 0)

Cell E — sirf implicit solution

Cell F — EXACT NAHI (method refuse karna zaroori)

Cell G — exact nahi, but fixable

Worked example Example 7

Solve karo $(y^{2} + 2 x y) d x - x^{2} d y = 0.$

Forecast: Integrating factor kis variable pe depend karega? $x$ ya $y$ guess karo.

Identify karo: $M = y^{2} + 2 x y, N = - x^{2} .$

Step 1 — Test karo. $M_{y} = 2 y + 2 x, N_{x} = - 2 x .$ Equal nahi ⟹ exact nahi.

Step 2 — $μ (x)$ try karo. Kyun? Check karo ki shortcut ratio sirf $x$ wala hai ya nahi. $\frac{M _{y} - N _{x}}{N} = \frac{( 2 y + 2 x ) - ( - 2 x )}{- x ^{2}} = \frac{2 y + 4 x}{- x ^{2}},$ jisme abhi bhi $y$ hai — sirf $x$ wala nahi. Toh $μ (x)$ fail; $μ (y)$ try karo.

Step 3 — $μ (y)$ try karo. Mirror shortcut: $\frac{N _{x} - M _{y}}{M}$ sirf $y$ wala hona chahiye. $\frac{N _{x} - M _{y}}{M} = \frac{- 2 x - ( 2 y + 2 x )}{y ^{2} + 2 x y} = \frac{- 4 x - 2 y}{y ( y + 2 x )} .$ Hmm, abhi bhi mixed hai. Iske bajaaye factor directly spot karo: $μ = 1/ y^{2}$ se multiply karo: $(1 + \frac{2 x}{y}) d x - \frac{x ^{2}}{y ^{2}} d y = 0.$ Re-test karo: naya $M^{*} = 1 + 2 x / y$ , naya $N^{*} = - x^{2} / y^{2}$ . $M_{y}^{*} = - 2 x / y^{2}$ , $N_{x}^{*} = - 2 x / y^{2}$ . Match ⟹ ab exact hai. $1/ y^{2}$ kyun? Kyunki $- x^{2} / y^{2}$ asal mein $x^{2} / y$ ka $\partial_{y}$ hai, jo potential hint karta hai.

Step 4 — $M^{*}$ ko $x$ mein integrate karo: $F = x + \frac{x ^{2}}{y} + g (y) .$

Step 5 — $N^{*}$ se match karo: $F_{y} = - \frac{x ^{2}}{y ^{2}} + g^{'} (y) =! - \frac{x ^{2}}{y ^{2}} \Rightarrow g^{'} (y) = 0 \Rightarrow g = const .$

Answer: $x + \frac{x ^{2}}{y} = c .$

Verify karo: $F_{x} = 1 + 2 x / y = M^{*}$ ✓, $F_{y} = - x^{2} / y^{2} = N^{*}$ ✓. $y^{2}$ se wapas multiply karne par original equation milti hai, toh solution curves unchanged hain.

Cell H — real-world word problem with initial condition

Worked example Example 8 — ek conservative force field

Ek particle plane mein move karta hai taaki uski motion hamesha ek hidden energy $F (x, y)$ ki level curve pe rahe. Measurements se differential relation milta hai $(2 x + y) d x + (x + 2 y) d y = 0,$ aur particle point $(x, y) = (1, 2)$ (units: metres) se guzarta hai. Woh energy contour find karo jis par woh ride karta hai.

Forecast: Yeh ek symmetric field hai ( $M$ aur $N$ $x \leftrightarrow y$ ke under roles swap karte hain). Guess karo kya $F$ mein ek $x y$ term hogi.

Identify karo: $M = 2 x + y, N = x + 2 y .$

Step 1 — Test karo. Kyun? Confirm karo ki ek potential/energy exist karti hai (ek conservative field). $M_{y} = 1, N_{x} = 1.$ Match ⟹ exact, toh ek energy $F$ exist karti hai.

Step 2 — $M$ ko $x$ mein integrate karo: $F = x^{2} + x y + g (y) .$

Step 3 — $N$ se match karo: $F_{y} = x + g^{'} (y) =! x + 2 y \Rightarrow g^{'} (y) = 2 y \Rightarrow g (y) = y^{2} .$

Step 4 — General energy: $F = x^{2} + x y + y^{2} .$ General solution $x^{2} + x y + y^{2} = c .$

Step 5 — Initial condition apply karo $(1, 2)$ : Kyun? Specific contour pin karne ke liye, poori family nahi (units: m²). $c = 1^{2} + (1) (2) + 2^{2} = 1 + 2 + 4 = 7.$

Answer: $x^{2} + x y + y^{2} = 7.$

Verify karo: $F_{x} = 2 x + y = M$ ✓, $F_{y} = x + 2 y = N$ ✓. $(1, 2)$ plug karo: $1 + 2 + 4 = 7$ ✓ (units m² ek energy contour ke liye consistent hain).

Red dot $(1, 2)$ hai; highlighted ellipse woh single contour $x^{2} + x y + y^{2} = 7$ hai jis par particle rehta hai. Paas ki patli curves aur energy levels $c$ hain.

Recall Kaun sa example kaun sa cell cover karta hai?

Ex1 = A (poly) ::: Ex2 = B (transcendental) Ex3 = C (signs) ::: Ex4 = D (degenerate / separable) Ex5 = E (implicit-only) ::: Ex6 = F (non-exact refusal) Ex7 = G (integrating factor rescue) ::: Ex8 = H (word problem + IC)