Exact equations — exactness condition, finding potential function: Exercises

4.6.6 · D4 · Maths › Ordinary Differential Equations › Exact equations — exactness condition, finding potential fun

Poore notes mein, equation ko $M (x, y) d x + N (x, y) d y = 0$ likha jaata hai, jahan:

$M$ wo ==coefficient hai jo $d x$ multiply karta hai== (chhupe hue hill ka " $x$ -slope"),
$N$ wo ==coefficient hai jo $d y$ multiply karta hai== (ya " $y$ -slope"),
$M_{y}$ ka matlab hai $\frac{\partial M}{\partial y}$ ( $M$ ko $y$ mein differentiate karo, $x$ ko frozen treat karo),
$N_{x}$ ka matlab hai $\frac{\partial N}{\partial x}$ ( $N$ ko $x$ mein differentiate karo, $y$ ko frozen treat karo).

Jo picture apne dimag mein rakhni hai:

Figure — Exact equations — exactness condition, finding potential function

Level 1 — Recognition

Goal: test ko sahi se run karo. Abhi solve nahi karna.

Exercise 1.1

Kya $(3 x^{2} + 2 y) d x + (2 x - 5 y^{4}) d y = 0$ exact hai?

Recall Solution 1.1

$M = 3 x^{2} + 2 y$ aur $N = 2 x - 5 y^{4}$ padho.

$M_{y} = \frac{\partial}{\partial y} (3 x^{2} + 2 y) = 2$ ( $3 x^{2}$ mein koi $y$ nahi, isliye woh vanish ho jaata hai).
$N_{x} = \frac{\partial}{\partial x} (2 x - 5 y^{4}) = 2$ ( $- 5 y^{4}$ mein koi $x$ nahi, isliye woh vanish ho jaata hai).

$M_{y} = N_{x} = 2$ ⟹ exact. ✓

Exercise 1.2

Kya $sin y d x + cos x d y = 0$ exact hai?

Recall Solution 1.2

$M = sin y$ , $N = cos x$ .

$M_{y} = cos y$ .
$N_{x} = - sin x$ .

$cos y \neq = - sin x$ generally ⟹ not exact. ✗

Exercise 1.3

Kya $(y e^{x y}) d x + (x e^{x y}) d y = 0$ exact hai?

Recall Solution 1.3

$M = y e^{x y}$ , $N = x e^{x y}$ . Inhe product rule ki zaroorat hai.

$M_{y} = \frac{\partial}{\partial y} (y e^{x y}) = e^{x y} + y \cdot (x e^{x y}) = e^{x y} (1 + x y)$ .
$N_{x} = \frac{\partial}{\partial x} (x e^{x y}) = e^{x y} + x \cdot (y e^{x y}) = e^{x y} (1 + x y)$ .

Equal hain ⟹ exact. ✓ (Yeh $d (e^{x y})$ hai.)

Level 2 — Application

Goal: clean equations par poora I-D-M-I recipe run karo.

Exercise 2.1

$(2 x + y) d x + (x + 2 y) d y = 0$ solve karo.

Recall Solution 2.1

$M = 2 x + y$ , $N = x + 2 y$ . Test: $M_{y} = 1$ , $N_{x} = 1$ ⟹ exact. $M$ ko $x$ mein integrate karo ( $y$ ko constant treat karo): $F = \int (2 x + y) d x + g (y) = x^{2} + x y + g (y) .$ $g (y)$ $x$ -integration ka "constant" hai — yeh $y$ par depend kar sakta hai kyunki $y$ frozen tha. $y$ mein differentiate karo, $N$ se match karo: $F_{y} = x + g^{'} (y) =! x + 2 y \Rightarrow g^{'} (y) = 2 y .$ Integrate karo: $g (y) = y^{2}$ . Answer: $x^{2} + x y + y^{2} = c .$

Exercise 2.2

$(3 x^{2} y + 2) d x + (x^{3} + 4 y) d y = 0$ solve karo.

Recall Solution 2.2

$M = 3 x^{2} y + 2$ , $N = x^{3} + 4 y$ . Test: $M_{y} = 3 x^{2}$ , $N_{x} = 3 x^{2}$ ⟹ exact. $M$ ko integrate karo: $F = \int (3 x^{2} y + 2) d x + g (y) = x^{3} y + 2 x + g (y) .$ $N$ se match karo: $F_{y} = x^{3} + g^{'} (y) = x^{3} + 4 y \Rightarrow g^{'} (y) = 4 y \Rightarrow g (y) = 2 y^{2} .$ Answer: $x^{3} y + 2 x + 2 y^{2} = c .$

Exercise 2.3

$(cos y) d x - (x sin y - y^{2}) d y = 0$ solve karo.

Recall Solution 2.3

Minus sign se careful raho: $N = - (x sin y - y^{2}) = - x sin y + y^{2}$ . $M = cos y$ . Test: $M_{y} = - sin y$ ; $N_{x} = - sin y$ ⟹ exact. $M$ ko integrate karo: $F = \int cos y d x + g (y) = x cos y + g (y)$ (yahan $cos y$ $x$ ke w.r.t. constant hai). $N$ se match karo: $F_{y} = - x sin y + g^{'} (y) = - x sin y + y^{2} \Rightarrow g^{'} (y) = y^{2} \Rightarrow g (y) = \frac{1}{3} y^{3} .$ Answer: $x cos y + \frac{1}{3} y^{3} = c .$

Level 3 — Analysis

Goal: decide karo ki exactness use karni hai ya nahi, aur failures diagnose karo.

Exercise 3.1

Dikhao ki $y d x - x d y = 0$ not exact hai, phir note karo ki kya fix karta hai.

Recall Solution 3.1

$M = y$ , $N = - x$ . $M_{y} = 1$ , $N_{x} = - 1$ . $1 \neq = - 1$ ⟹ not exact. Kya fix karta hai: integrating factor $μ = 1/ x^{2}$ se multiply karo (alag topic — Integrating Factors for ODEs dekho). Tab $\frac{y}{x ^{2}} d x - \frac{1}{x} d y = 0, M = \frac{y}{x ^{2}}, N = - \frac{1}{x} .$ Ab $M_{y} = 1/ x^{2}$ aur $N_{x} = 1/ x^{2}$ ⟹ exact, aur left side exactly $- d (y / x)$ hai. Solution: $y / x = c$ .

Exercise 3.2

Kis constant $a$ ke liye $(a x y + y^{2}) d x + (x^{2} + 2 x y) d y = 0$ exact hai? Phir ise solve karo.

Recall Solution 3.2

$M = a x y + y^{2}$ , $N = x^{2} + 2 x y$ . Test force karo: $M_{y} = a x + 2 y$ , $N_{x} = 2 x + 2 y$ . Inhe match karo: $a x + 2 y = 2 x + 2 y$ sab $x, y$ ke liye ⟹ $a = 2$ . $a = 2$ ke saath: $M = 2 x y + y^{2}$ . $M$ ko integrate karo: $F = \int (2 x y + y^{2}) d x + g (y) = x^{2} y + x y^{2} + g (y) .$ $N$ se match karo: $F_{y} = x^{2} + 2 x y + g^{'} (y) = x^{2} + 2 x y \Rightarrow g^{'} (y) = 0 \Rightarrow g (y) = 0.$ Answer: $a = 2$ aur $x^{2} y + x y^{2} = c .$

Exercise 3.3

Equation $(y cos x + 2 x e^{y}) d x + (sin x + x^{2} e^{y} - 1) d y = 0$ exact hone ka claim karti hai. Verify karo, phir $(0, 0)$ se guzarne wala particular solution nikalo.

Recall Solution 3.3

$M = y cos x + 2 x e^{y}$ , $N = sin x + x^{2} e^{y} - 1$ . Test: $M_{y} = cos x + 2 x e^{y}$ ; $N_{x} = cos x + 2 x e^{y}$ ⟹ exact. ✓ $M$ ko integrate karo: $F = \int (y cos x + 2 x e^{y}) d x + g (y) = y sin x + x^{2} e^{y} + g (y) .$ $N$ se match karo: $F_{y} = sin x + x^{2} e^{y} + g^{'} (y) = sin x + x^{2} e^{y} - 1 \Rightarrow g^{'} (y) = - 1 \Rightarrow g (y) = - y .$ General solution: $y sin x + x^{2} e^{y} - y = c .$ $(0, 0)$ apply karo: $0 \cdot sin 0 + 0 \cdot e^{0} - 0 = c \Rightarrow c = 0.$ Answer: $y sin x + x^{2} e^{y} - y = 0.$

Level 4 — Synthesis

Goal: exactness ko "integrate $N$ first" route ke saath, aur initial conditions ke saath combine karo.

Exercise 4.1

$(\frac{1}{y} + 2 x) d x + (- \frac{x}{y ^{2}} + 3 y^{2}) d y = 0$ ko pehle $N$ ko $y$ mein integrate karke solve karo (symmetric route). Domain restriction bhi batao.

Recall Solution 4.1

$M = \frac{1}{y} + 2 x$ , $N = - \frac{x}{y ^{2}} + 3 y^{2}$ . (Require karta hai $y \neq = 0$ .) Test: $M_{y} = - \frac{1}{y ^{2}}$ ; $N_{x} = - \frac{1}{y ^{2}}$ ⟹ exact. ✓ $N$ ko $y$ mein integrate karo ( $x$ ko constant treat karo), is baar $h (x)$ add karo: $F = \int (- \frac{x}{y ^{2}} + 3 y^{2}) d y + h (x) = \frac{x}{y} + y^{3} + h (x) .$ (Note: $\int - x / y^{2} d y = - x \cdot (- 1/ y) = x / y$ .) $F_{x} = M$ se match karo: $F_{x} = \frac{1}{y} + h^{'} (x) =! \frac{1}{y} + 2 x \Rightarrow h^{'} (x) = 2 x \Rightarrow h (x) = x^{2} .$ Answer: $\frac{x}{y} + y^{3} + x^{2} = c, y \neq = 0.$

Exercise 4.2

IVP $(2 x y - sec^{2} x) d x + (x^{2} + 2 y) d y = 0$ , $y (0) = 3$ solve karo.

Recall Solution 4.2

$M = 2 x y - sec^{2} x$ , $N = x^{2} + 2 y$ . Test: $M_{y} = 2 x$ ; $N_{x} = 2 x$ ⟹ exact. $M$ ko integrate karo: yaad karo $\int sec^{2} x d x = tan x$ , toh $F = \int (2 x y - sec^{2} x) d x + g (y) = x^{2} y - tan x + g (y) .$ $N$ se match karo: $F_{y} = x^{2} + g^{'} (y) = x^{2} + 2 y \Rightarrow g^{'} (y) = 2 y \Rightarrow g (y) = y^{2} .$ General: $x^{2} y - tan x + y^{2} = c .$ $y (0) = 3$ apply karo: $0 - tan 0 + 9 = c \Rightarrow c = 9.$ Answer: $x^{2} y - tan x + y^{2} = 9.$

Exercise 4.3

Ek student ne $M = 6 x^{2} + 4 x y$ integrate kiya aur $F = 2 x^{3} + 2 x^{2} y + g (y)$ paaya, phir $N = 2 x^{2} - 3 y^{2}$ se match kiya. Solution complete karo — aur picture use karke explain karo ki $N$ ka koi bhi term do baar integrate kyun nahi hota.

Recall Solution 4.3

Pehle test karo (student ne skip kiya tha): $M_{y} = 4 x$ , $N_{x} = 4 x$ ⟹ exact. ✓ $N$ se match karo: $F_{y} = 2 x^{2} + g^{'} (y) =! 2 x^{2} - 3 y^{2} \Rightarrow g^{'} (y) = - 3 y^{2} \Rightarrow g (y) = - y^{3} .$ Answer: $2 x^{3} + 2 x^{2} y - y^{3} = c .$ Double-counting kyun nahi hoti: $N$ ka woh part jo $2 x^{2}$ ke barabar hai, $\int M d x = 2 x^{3} + 2 x^{2} y$ ko $y$ mein differentiate karne se pehle se bana hua hai. Hill picture par, woh slope pehle se $F$ mein built-in hai. Sirf bachi hui slope $- 3 y^{2}$ — woh part jisme koi $x$ nahi — nayi information hai, aur wahi $g^{'} (y)$ banta hai. Ise subtract karna exactly I-D-M-I ka step M hai.

Level 5 — Mastery

Goal: machinery ko reverse-engineer karo aur degenerate / limiting cases handle karo.

Exercise 5.1

Ek potential $F (x, y) = x^{3} y^{2}$ design karo. Woh exact ODE likho jo yeh satisfy karta hai ( $M d x + N d y = 0$ form mein), phir $M$ aur $N$ se $F$ recover karo aur round trip confirm karo.

Recall Solution 5.1

Forward (ODE banao): $d F = F_{x} d x + F_{y} d y$ .

$M = F_{x} = 3 x^{2} y^{2}$ ,
$N = F_{y} = 2 x^{3} y$ .

ODE: $3 x^{2} y^{2} d x + 2 x^{3} y d y = 0.$ Sanity test: $M_{y} = 6 x^{2} y$ , $N_{x} = 6 x^{2} y$ ⟹ construction se exact. ✓ Reverse ( $F$ recover karo): $M$ ko $x$ mein integrate karo: $\int 3 x^{2} y^{2} d x + g (y) = x^{3} y^{2} + g (y)$ . $N$ se match karo: $F_{y} = 2 x^{3} y + g^{'} (y) = 2 x^{3} y \Rightarrow g^{'} (y) = 0 \Rightarrow g (y) = const$ . Recovered $F = x^{3} y^{2}$ ✓ — round trip close ho gayi. Solution family: $x^{3} y^{2} = c$ .

Exercise 5.2

$(y^{2} - 1) d x + (2 x y) d y = 0$ consider karo. Ise solve karo. Phir degenerate level $c = 0$ ko geometrically examine karo: $F (x, y) = 0$ kaunsa/kaunse curve describe karta hai?

Recall Solution 5.2

$M = y^{2} - 1$ , $N = 2 x y$ . Test: $M_{y} = 2 y$ , $N_{x} = 2 y$ ⟹ exact. $M$ ko integrate karo: $F = \int (y^{2} - 1) d x + g (y) = x (y^{2} - 1) + g (y) = x y^{2} - x + g (y) .$ $N$ se match karo: $F_{y} = 2 x y + g^{'} (y) = 2 x y \Rightarrow g^{'} (y) = 0 \Rightarrow g (y) = 0.$ General solution: $x y^{2} - x = c$ , yaani $x (y^{2} - 1) = c .$ Degenerate level $c = 0$ : $x (y^{2} - 1) = 0$ split hota hai $x = 0$ ( $y$ -axis) ya $y^{2} = 1$ , yaani $y = 1$ aur $y = - 1$ (do horizontal lines) mein. Toh single level curve $F = 0$ actually teen straight lines ka union hai jo plane ko cross karti hain — ek reminder ki contour $F = c$ ek smooth curve nahi bhi ho sakta; jahan level ek saddle/crossing se guzarta hai wahan yeh branch ya split ho sakta hai.

Exercise 5.3

Equation $M d x + N d y = 0$ jisme $M = 2 x y^{3} + y$ hai, ek suitable $N$ ke liye exact hai jisme $N (0, y) = y^{4}$ hai. $N$ reconstruct karo.

Recall Solution 5.3

Exactness demand karti hai $N_{x} = M_{y}$ . Compute karo $M_{y} = \frac{\partial}{\partial y} (2 x y^{3} + y) = 6 x y^{2} + 1.$ Toh $N_{x} = 6 x y^{2} + 1$ . $N$ recover karne ke liye $x$ mein integrate karo ("constant" $y$ ka function hai): $N = \int (6 x y^{2} + 1) d x + φ (y) = 3 x^{2} y^{2} + x + φ (y) .$ $N (0, y) = y^{4}$ use karke $φ$ pin karo: $x = 0$ set karo: $N (0, y) = 0 + 0 + φ (y) = y^{4} \Rightarrow φ (y) = y^{4} .$ Answer: $N = 3 x^{2} y^{2} + x + y^{4} .$ (Quick check ki real $F$ exist karta hai: $F = x^{2} y^{3} + x y + \frac{1}{5} y^{5}$ deta hai $F_{x} = 2 x y^{3} + y = M$ aur $F_{y} = 3 x^{2} y^{2} + x + y^{4} = N$ . ✓)

Exercise 5.4

Dikhao ki har separable equation $p (x) d x + q (y) d y = 0$ automatically exact hoti hai. Potential kahaan se aata hai?

Recall Solution 5.4

Yahan $M = p (x)$ sirf $x$ par depend karta hai; $N = q (y)$ sirf $y$ par depend karta hai. Test: $M_{y} = \frac{\partial}{\partial y} p (x) = 0$ (andar koi $y$ nahi), aur $N_{x} = \frac{\partial}{\partial x} q (y) = 0$ (andar koi $x$ nahi). $0 = 0$ ⟹ hamesha exact. ✓ The potential: $F = \int p (x) d x + \int q (y) d y$ , aur solution hai $\int p d x + \int q d y = c$ — exactly wahi answer jo separable method deta hai. Toh separable equations exact equations ka ek special, "already-split" case hain.

Recall Poori ladder ka one-line self-check

Agar tum (a) $M_{y} = ? N_{x}$ run kar sako, (b) kisi bhi coefficient ko integrate karo aur $g (y)$ ya $h (x)$ se patch karo, (c) $c$ fix karne ke liye initial condition apply karo, aur (d) process reverse karke ek chosen $F$ se exact equation build karo — toh yeh topic tumhara hai.