Exact equations — exactness condition, finding potential function

4.6.6 · Maths › Ordinary Differential Equations

1. Exact equation kya hoti hai?

Total differential kya hota hai? Multivariable calculus se, kisi bhi smooth $F (x, y)$ ke liye: $d F = \frac{\partial F}{\partial x} d x + \frac{\partial F}{\partial y} d y .$

Toh $d F = M d x + N d y$ ko term-by-term compare karne par do defining equations milti hain: $\frac{\partial F}{\partial x} = M, \frac{\partial F}{\partial y} = N$

Yahi sab kuch ka dil hai: $M$ aur $N$ ek potential function $F$ ki do partial derivatives hain.

2. Exactness condition derive karna (scratch se)

Hum ise kaise derive karte hain. Maan lo $F$ exist karta hai. Tab $M = F_{x}$ aur $N = F_{y}$ hain. Dono ko doosre variable mein differentiate karo:

\qquad \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}\!\left(\frac{\partial F}{\partial y}\right) = F_{yx}.$$ **Clairaut's / Schwarz's theorem** ke according, agar $F$ mein continuous second partials hain toh mixed partials equal hote hain: $F_{xy} = F_{yx}$. Isliye > [!formula] Exactness condition > $$\boxed{\;\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}\;}$$ > Yeh **necessary** hai. Ek simply-connected (no-holes) region par yeh **sufficient** bhi hai — toh yeh ek poora test hai. > [!recall]- Region assumption ke bina condition sirf "necessary" kyun hai? > Kyunki $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ guarantee karta hai ki field *curl-free* hai, lekin ise ek single-valued $F$ mein integrate karne ke liye region mein bhi koi holes nahi hone chahiye (simply connected). Pure plane par yeh automatic hai, toh exam ke liye yeh test necessary aur sufficient dono hai. --- ## 3. Potential function $F$ kaise dhundhein > [!intuition] "Integrate, phir patch karo" method > Hum $F$ ko uski do partial derivatives se rebuild karte hain. $F_x = M$ ko integrate karne se $F$ **ek $y$ ke function tak** milta hai (kyunki $x$-integration ke dauran $y$ constant ki tarah kaam karta hai). Phir hum $F_y = N$ use karte hain us bachi hui function ko pin down karne ke liye. **Step-by-step recipe:** 1. **Exactness check karo:** confirm karo ki $M_y = N_x$ hai. 2. **$M$ ko $x$ mein integrate karo:** $F = \displaystyle\int M\,dx + g(y)$ — unknown $g(y)$ $x$-integration ka "constant" hai. 3. **$y$ ke saath differentiate karo** aur $N$ ke barabar karo: $F_y = \frac{\partial}{\partial y}\!\int M\,dx + g'(y) = N$. 4. **$g'(y)$ solve karo**, integrate karke $g(y)$ nikalo. 5. **Solution likho** $F(x,y) = c$. (Symmetric alternative: $N$ ko $y$ mein integrate karo, $h(x)$ add karo, phir $F_x = M$ match karo.) --- ![[4.6.06-Exact-equations-—-exactness-condition,-finding-potential-function.png]] --- ## 4. Worked examples > [!example] Example 1 — clean exact equation > $(2xy + 3)\,dx + (x^2 - 1)\,dy = 0$ solve karo. > > **Identify karo:** $M = 2xy+3$, $N = x^2 - 1$. > > **Step 1 — Test.** $M_y = 2x$, $N_x = 2x$. Equal hain ⟹ exact hai. *Yeh step kyun? Jab tak $F$ exist nahi karta, dhundhhne ka koi faida nahi.* > > **Step 2 — $M$ ko $x$ mein integrate karo:** > $$F = \int (2xy+3)\,dx + g(y) = x^2 y + 3x + g(y).$$ > *$g(y)$ kyun? Jo bhi sirf $y$ par depend karta hai uska $x$-derivative zero hota hai, toh woh $M$ ke liye invisible hai.* > > **Step 3 — $y$ mein differentiate karo, $N$ se match karo:** > $$F_y = x^2 + g'(y) \stackrel{!}{=} x^2 - 1 \;\Rightarrow\; g'(y) = -1.$$ > *Kyun? Potential ki definition se $F_y$ ko $N$ ke equal hona chahiye.* > > **Step 4 — Integrate karo:** $g(y) = -y$. > > **Answer:** $\;x^2 y + 3x - y = c.$ > [!example] Example 2 — exponential/trig mix > $(e^{y} + \cos x)\,dx + (x e^{y} + 2y)\,dy = 0$ solve karo. > > **Test:** $M_y = e^{y}$, $N_x = e^{y}$ ⟹ exact hai. > > **$M$ integrate karo:** $F = \int(e^y + \cos x)\,dx + g(y) = x e^y + \sin x + g(y).$ > > **$N$ se match karo:** $F_y = x e^y + g'(y) = x e^y + 2y \Rightarrow g'(y)=2y \Rightarrow g(y)=y^2.$ > > **Answer:** $\;x e^y + \sin x + y^2 = c.$ > [!example] Example 3 — NON-exact equation verify karna > $y\,dx - x\,dy = 0$ test karo. Yahan $M=y$, $N=-x$ hai: $M_y = 1$, $N_x = -1$. **Equal nahi hain ⟹ exact nahi hai.** > *Yeh kyun important hai?* Potential method directly apply nahi karna chahiye. (Ise integrating factor $\mu = 1/x^2$ se exact banaya ja sakta hai, jo $d(y/x)=0$ deta hai — yeh alag topic hai.) --- ## 5. Common mistakes (Steel-man + fix) > [!mistake] "$M_y$ ka matlab $M$ ko $x$ ke saath differentiate karna hai." > **Yeh sahi kyun lagta hai:** $M$, $dx$ ko multiply karta hai, toh yeh "$x$ se tied" lagta hai. **Fix:** Exactness test deliberately *cross-differentiate* karta hai: $M$ ($dx$ coefficient) ko $y$ se differentiate karo, aur $N$ ($dy$ coefficient) ko $x$ se. Yeh $F_{xy}=F_{yx}$ ko mirror karta hai. > [!mistake] $g(y)$ bhool jaana (ya plain constant $+c$ likhna). > **Yeh sahi kyun lagta hai:** Indefinite integrals mein usually sirf "$+C$" add hota hai. **Fix:** Jab $x$ mein integrate karo, toh "constant" abhi bhi $y$ par depend kar sakta hai. Ise drop karna pura $-y$ ya $y^2$ term gawaata hai aur galat solution deta hai. > [!mistake] Non-exact equation par potential method apply karna. > **Yeh sahi kyun lagta hai:** Recipe mechanical hai aur tempting lagti hai. **Fix:** Hamesha PEHLE $M_y \stackrel{?}{=} N_x$ run karo. Agar fail ho, toh integrating factor dhundho. > [!mistake] Jo terms already account ho chuki hain unhe dobara integrate karna. > Jab tum $F_y$ ko $N$ se match karte ho, toh $N$ ka woh part jo $\partial_y(\int M\,dx)$ se aata hai woh **pehle se $F$ ke andar hai**. Sirf bacha hua hissa $g'(y)$ define karta hai. Carefully subtract karne se double-counting se bachte hain. --- ## 6. Active recall #flashcards/maths $M\,dx+N\,dy=0$ ko exact kab kehte hain? ::: Jab LHS kisi $F(x,y)$ ka total differential $dF$ ho, yaani $\exists F$ jiske liye $F_x=M,\ F_y=N$ ho. Exactness condition batao. ::: $\partial M/\partial y = \partial N/\partial x$ (simply-connected region par yeh sufficient bhi hai). Exactness condition mixed partials se kyun aati hai? ::: Kyunki $M=F_x,\ N=F_y\Rightarrow M_y=F_{xy},\ N_x=F_{yx}$, aur $F_{xy}=F_{yx}$ (Clairaut). $F$ dhundhhne ka pehla step kya hai? ::: $M$ ko $x$ ke saath integrate karo: $F=\int M\,dx + g(y)$. $g(y)$ kya hai aur ise kaise dhundhhte hain? ::: $x$-integration ka $y$-only "constant" hai; $F_y=N$ set karke aur $g'(y)$ solve karke, phir integrate karke milta hai. Jab $F$ mil jaaye toh general solution kya hoti hai? ::: $F(x,y)=c$. $y\,dx-x\,dy=0$ test karo: exact hai? ::: Nahi, kyunki $M_y=1\ne N_x=-1$. $(2xy+3)dx+(x^2-1)dy=0$ solve karo. ::: $x^2y+3x-y=c$. --- > [!recall]- Feynman: 12-saal ke bachche ko explain karo > Ek treasure map imagine karo jisme pahaad aur ghatiyan hain, aur har height par pahaad ke around lines khichi hain — bilkul asli map ki contour lines ki tarah. Equation $M\,dx+N\,dy=0$ ek rule hai jo kehta hai "aisa chalo ki teri height kabhi na badle — ek contour line par raho." Agar do numbers $M$ aur $N$ ek simple matching test pass karte hain ($M_y=N_x$), toh matlab hai ki ek asli pahaad $F$ exist karta hai. Phir hum **pahaad ko rebuild** karte hain uski slopes ko jod kar, aur answer sirf yeh hai ki "tum us line par ho jahan pahaad ki height $c$ ke barabar hai." > [!mnemonic] Test aur recipe yaad rakho > **"My Nx — cross your partials."** Test: *doosre* variable se differentiate karo ($M_y = N_x$). Recipe: **I-D-M-I** — **I**ntegrate $M$, **D**ifferentiate in $y$, **M**atch $N$, **I**ntegrate $g'$. > [!intuition] Connections > - [[Total Differential & Partial Derivatives]] — $dF=F_x dx+F_y dy$ ke peeche ka engine. > - [[Clairaut's Theorem (Equality of Mixed Partials)]] — exactness condition justify karta hai. > - [[Integrating Factors for ODEs]] — jab equation *exact nahi* ho tab kya karna hai. > - [[Separable Equations]] & [[Linear First-Order ODEs]] — special cases jo exact banaye ja sakte hain. > - [[Conservative Vector Fields & Potential Functions]] — exactness curl-free hai; $F$ potential hai. > - [[Level Curves & Contour Lines]] — $F(x,y)=c$ ki geometry. ## 🖼️ Concept Map ```mermaid flowchart TD ODE[M dx + N dy = 0] F[Potential function F x,y] TD[Total differential dF] FX[F_x = M] FY[F_y = N] CL[Clairaut theorem F_xy = F_yx] TEST[Exactness test M_y = N_x] SC[Simply-connected region] INT[Integrate M over x] PATCH[Patch using F_y = N] SOL[Solution F x,y = c] ODE -->|is exact if| TD TD -->|comes from| F TD -->|compare terms| FX TD -->|compare terms| FY FX -->|differentiate in y| CL FY -->|differentiate in x| CL CL -->|yields| TEST SC -->|makes sufficient| TEST TEST -->|then recover| F FX -->|integrate| INT INT -->|then| PATCH FY -->|pins down| PATCH PATCH -->|gives| SOL F -->|level curve| SOL ```