Exact equations — exactness condition, finding potential function: Question bank

4.6.6 · D5 · Maths › Ordinary Differential Equations › Exact equations — exactness condition, finding potential fun

True or false — justify karo

Recall Equation

$M d x + N d y = 0$ exact hai agar $M$ aur $N$ dono continuous hain. False ::: Sirf continuity kaafi nahi hai; exactness ke liye $M_{y} = N_{x}$ chahiye taaki $M, N$ ek potential $F$ ke do partials ho sakein. Continuity sirf yeh karne deti hai ki hum test state kar sakein.

Recall Agar

$M_{y} = N_{x}$ poore plane par har jagah hai, toh ek single-valued potential $F$ zaroor exist karta hai. True ::: Plane simply-connected hai (koi holes nahi), isliye curl-free condition $M_{y} = N_{x}$ sirf necessary nahi balki sufficient bhi hai — dekho Conservative Vector Fields & Potential Functions.

Recall

$M d x + N d y = 0$ ki exactness is baat par depend karti hai ki equation ko kaise scale kiya (jaise dono sides ko $x$ se multiply karna). True ::: Kisi function se multiply karne par $M$ aur $N$ change ho jaate hain, isliye $M_{y}$ aur $N_{x}$ bhi change ho jaate hain. Ek non-exact equation exact ban sakti hai (woh scaling factor ek integrating factor hota hai), aur ek exact equation kharaab bhi ho sakti hai.

Recall Har separable equation, jab

$M d x + N d y = 0$ ke form mein likhi jaaye, already exact hoti hai. True ::: Ek separable equation $f (x) d x + g (y) d y = 0$ hoti hai, isliye $M = f (x)$ (koi $y$ nahi) aur $N = g (y)$ (koi $x$ nahi). Tab $M_{y} = 0 = N_{x}$ automatically ho jaata hai. Dekho Separable Equations.

Recall Agar ek equation exact hai, toh uska potential

$F$ unique hota hai. False ::: $F$ sirf ek additive constant tak unique hota hai, kyunki $d (F + c) = d F$ . Isi liye solution $F (x, y) = c$ likha jaata hai na ki $F = 0$ .

Recall Solution curves

$F (x, y) = c$ ek doosre ko kabhi cross nahi karti. True ::: Yeh ek function ki level curves hain; ek point ke do alag heights nahi ho sakte, isliye alag- $c$ wale contours disjoint hote hain (singular points se door). Dekho Level Curves & Contour Lines.

Recall Agar

$M_{y} = N_{x}$ hai lekin sirf ek aisi region par jo origin ke aas-paas hole ke saath hai, toh exactness guaranteed hai. False ::: Test wahan sirf necessary hai. Ek hole ek single-valued $F$ ko exist karne se rok sakta hai even jab $M_{y} = N_{x}$ ho (classic example $\frac{- y d x + x d y}{x ^{2} + y ^{2}}$ ).

Error dhundho

Recall "

$M_{y}$ ka matlab hai $M$ ko $x$ ke respect mein differentiate karo, kyunki $M$ $d x$ ke paas baitha hai." Error ::: Subscript differentiation ke variable ka naam batata hai, aur woh variable $y$ hai, isliye $M_{y} = \partial M / \partial y$ . Test ka poora point yahi hai ki cross-differentiate karo: $M$ ko $y$ se, $N$ ko $x$ se, jo $F_{x y} = F_{y x}$ ko mirror karta hai.

Recall "

$M$ ko $x$ mein integrate karne ke baad, main $+ C$ add karta hoon, jaise kisi bhi indefinite integral mein." Error ::: $x$ -integration ka 'constant' abhi bhi $y$ par depend kar sakta hai, isliye woh $g (y)$ hona chahiye, plain $C$ nahi. $y$ -dependence drop karne se $F$ ke poore terms kho jaate hain.

Recall Student likhta hai

$F = \int M d x + g (y)$ , phir $g^{'} (y)$ ko poore $N$ ke barabar set kar deta hai. Error ::: $N$ ka kuch part pehle se $\partial_{y} \int M d x$ se aa jaata hai, jo $F$ mein already baked in hai. Tumhe pehle woh part subtract karna hoga; $g^{'} (y)$ sirf bacha hua part ke barabar hota hai.

Recall "Equation exact nahi hai, toh main

$M$ ko $x$ mein integrate karke phir bhi patch kar lunga." Error ::: Recipe assume karti hai ki ek potential exist karta hai. Agar $M_{y} \neq = N_{x}$ toh patch step ek aisa $g^{'}$ deta hai jo abhi bhi $x$ par depend karta hai — $g (y)$ ke liye ek impossible equation — jo failure signal karta hai. Pehle ek integrating factor dhundho.

Recall "Kyunki

$M = F_{x}$ aur $N = F_{y}$ hain, main bas dono integrals add kar sakta hoon: $F = \int M d x + \int N d y$ ." Error ::: Isse $x$ aur $y$ dono wale har term double-count ho jaata hai. Woh shared terms dono integrals mein aate hain, isliye unhe blindly add karne par woh do baar aa jaate hain. Integrate-then-patch method isi se bachne ke liye exist karta hai.

Recall "

$(2 x y) d x + (x^{2}) d y = 0$ ke liye mujhe $M_{y} = 2 x$ , $N_{x} = 2 x$ milta hai, toh yeh exact hai; answer hai $F_{x} = M \Rightarrow F = x^{2} y + 3 x$ ." Error ::: Koi $+ 3 x$ nahi hai — $M = 2 x y$ mein kuch bhi $3 x$ ko integrate karke nahi deta. Sahi: $F = \int 2 x y d x + g (y) = x^{2} y + g (y)$ , aur $F_{y} = x^{2} = N$ match karne par $g^{'} (y) = 0$ milta hai, isliye $F = x^{2} y = c$ .

Why questions

Recall Exactness condition mein

mixed second partials kyun involve hote hain, na ki pure ones jaise $F_{xx}$ ? Kyunki $M = F_{x}$ aur $N = F_{y}$ mein se har ek already ek derivative carry karta hai. $M$ ko $y$ se differentiate karne par $F_{x y}$ milta hai aur $N$ ko $x$ se differentiate karne par $F_{y x}$ — yeh dono paths in mixed partials par land karte hain, aur Clairaut unhe equal hone par majboor karta hai.

Recall Solution

$F (x, y) = c$ kyun likha jaata hai, $y$ ko explicitly solve karne ki jagah? Differential equation sirf yeh demand karti hai ki hum $F$ ki kisi level curve par rahein; woh curve implicitly defined hai. $y$ ke liye solve karna aksar impossible ya multivalued hota hai, isliye implicit form honest general solution hai.

Recall Ek integrating factor ek non-exact equation ko "rescue" kyun karta hai, solutions change kiye bina?

$M d x + N d y = 0$ ko $μ (x, y) \neq = 0$ se multiply karne par zero set unchanged rehta hai, isliye same curves ise solve karti hain. Lekin $μ M, μ N$ ab cross-partial test satisfy kar sakte hain, jisse potential method kaam kar sake.

Recall Geometric picture (ek contour par rehna) algebra

$d F = 0$ se kyun match karta hai? Ek contour $F = c$ ke along height kabhi nahi badlti, isliye uska total differential $d F = F_{x} d x + F_{y} d y$ zero hona chahiye. Yeh exactly $M d x + N d y = 0$ hai jab $M = F_{x}, N = F_{y}$ . Dekho Total Differential & Partial Derivatives.

Recall Exactness ek vector field ke conservative hone ke saath same idea kyun hai?

Field $(M, N)$ ka potential $F$ precisely tab hota hai jab $M = F_{x}, N = F_{y}$ — yahi conservative ki definition hai. Test $M_{y} = N_{x}$ 2D curl ka vanish hona hai. Dekho Conservative Vector Fields & Potential Functions.

Edge cases

Recall Kya

$0 d x + 0 d y = 0$ exact hai, aur uska solution kya hai? Trivially exact ( $M_{y} = 0 = N_{x}$ ), $F =$ koi bhi constant ke saath. Har point ise satisfy karta hai, isliye koi meaningful curve nahi hai — ek degenerate case jahan potential har jagah constant hai.

Recall Agar

$N \equiv 0$ hai, toh equation $M (x, y) d x = 0$ hai; yeh exact kab hai? Exactness ke liye $M_{y} = N_{x} = 0$ chahiye, isliye $M$ sirf $x$ par depend karna chahiye: $M = M (x)$ . Tab $F = \int M d x$ aur "curves" woh vertical lines hain jahan $M (x) = 0$ ya $F (x) = c$ hai.

Recall Ek

linear first-order ODE $y^{'} + P (x) y = Q (x)$ — kya yeh as written exact hai? Generally nahi. $(P (x) y - Q (x)) d x + d y = 0$ ke roop mein likha jaaye toh $M_{y} = P (x)$ , $N_{x} = 0$ milta hai, jo equal nahi hain jab tak $P \equiv 0$ na ho. Iska standard integrating factor $e^{\int P d x}$ exactly wahi hai jo ise exact banata hai.

Recall Recipe mein kya hota hai jab

$F_{y} = N$ match karne ke baad, bacha hua $g^{'}$ abhi bhi $x$ contain karta hai? Yeh failure ka signature hai: $g (y)$ $x$ par depend nahi kar sakta, isliye koi potential exist nahi karta — equation exact nahi thi. Yeh confirm karta hai ki method par trust karne se pehle $M_{y} = N_{x}$ check karna zaroori hai.

Recall Kya ek equation kuch points par exact ho sakti hai lekin kuch aur par nahi?

Haan — $M_{y} = N_{x}$ ek pointwise condition hai. Agar yeh plane ke sirf kuch hisse par hold karta hai, toh potential method sirf wahan kaam karta hai; full test ke liye yeh ek simply-connected region mein throughout required hai.