Bernoulli equations — substitution — Hinglish notes

4.6.5 · Maths › Ordinary Differential Equations

One-line idea: Ek Bernoulli equation nonlinear lagti hai kyunki usme ek extra $y^{n}$ term hota hai, lekin ek smart substitution $v = y^{1 - n}$ use linear ODE mein badal deti hai jise tum pehle se solve karna jaante ho.

[!intuition] Yeh kaam karta hi kyun hai?

KYA hum face karte hain: ek equation jiska form hai $\frac{d y}{d x} + P (x) y = Q (x) y^{n}, n \neq = 0, 1.$ Right-hand side mein $y^{n}$ hai, isliye yeh $y$ mein linear nahi hai. Hum directly integrating-factor trick use nahi kar sakte.

KYUN yeh trick exist karti hai: Linearity ko sirf $y^{n}$ ki power rok rahi hai. Agar hum ek nayi variable $v$ define kar sakein jiska derivative naturally us power ko absorb kar le, toh equation $v$ mein ek linear equation ban jaayegi. $y^{1 - n}$ ko differentiate karne par exactly $y^{- n}$ factor milta hai — jo $y^{n}$ ki pareshani ka perfect "antidote" hai.

KAISE hum iska faayda uthate hain: pehle $y^{n}$ se divide karo (taaki $y^{1 - n}$ expose ho), phir $v = y^{1 - n}$ substitute karo. Algebra click karke ek linear ODE ban jaata hai.

Cases $n = 0$ aur $n = 1$ exclude hain kyunki tab equation pehle se hi linear hoti hai (koi substitution nahi chahiye): $n = 0$ deta hai $y^{'} + P y = Q$ ; $n = 1$ deta hai $y^{'} + (P - Q) y = 0$ .

[!definition] Bernoulli equation

Ek first-order ODE Bernoulli equation hai agar use is tarah likha ja sake $\frac{d y}{d x} + P (x) y = Q (x) y^{n}, n \in R, n \neq = 0, 1.$ Yahan $P, Q$ sirf $x$ ke functions hain, aur nonlinearity sirf power == $y^{n}$ == hai.

[!formula] Substitution — scratch se derive kiya hua

Step 1 — $y^{1 - n}$ expose karo. Har term ko $y^{n}$ se divide karo (assume karo $y \neq = 0$ ): $y^{- n} \frac{d y}{d x} + P (x) y^{1 - n} = Q (x) .$ Yeh step kyun? Hum chahte hain ki $y$ ki sirf ek power bache ( $y^{1 - n}$ ) aur ek clean $y^{- n} y^{'}$ jo chain rule absorb kar sake.

Step 2 — Nayi variable define karo. $v = y^{1 - n}$ Yeh choice kyun? Taaki iska derivative woh $y^{- n} y^{'}$ wapas de jo hamare paas hai.

Step 3 — $v$ differentiate karo (chain rule).

\quad\Longrightarrow\quad y^{-n}\frac{dy}{dx} = \frac{1}{1-n}\frac{dv}{dx}.$$ *Yeh step kyun?* Yahi engine hai: yeh awkward $y^{-n}y'$ ko ek plain $v'$ mein convert karta hai. **Step 4 — Step 1 mein substitute karo.** $$\frac{1}{1-n}\frac{dv}{dx} + P(x)\,v = Q(x).$$ $(1-n)$ se multiply karo: $$\boxed{\dfrac{dv}{dx} + (1-n)P(x)\,v = (1-n)Q(x)}$$ Yeh $v$ mein **linear hai** — integrating factor $\mu(x)=e^{\int (1-n)P\,dx}$ se solve karo, phir $v=y^{1-n}$ back-substitute karo. --- ![[4.6.05-Bernoulli-equations-—-substitution.png]] --- ## [!example] Worked Example 1 — ek classic Solve karo $\dfrac{dy}{dx} + \dfrac{y}{x} = x\,y^2.$ - **Identify karo:** $P=\frac1x,\ Q=x,\ n=2.$ *Kyun?* Standard form se match karo; $y^2$ Bernoulli flag karta hai. - **$y^2$ se divide karo:** $\;y^{-2}y' + \frac1x y^{-1} = x.$ *Kyun?* $y^{1-n}=y^{-1}$ expose karo. - **Substitute karo** $v=y^{-1}$, toh $v' = -y^{-2}y'\Rightarrow y^{-2}y'=-v'.$ *Kyun?* Step-3 relation plug karo. - **Linear ODE:** $-v' + \frac1x v = x \Rightarrow v' - \frac1x v = -x.$ *Kyun?* Sirf rearrange kiya. - **Integrating factor:** $\mu=e^{\int -1/x\,dx}=e^{-\ln x}=\frac1x.$ - **Solve karo:** $(\frac{v}{x})' = -1 \Rightarrow \frac{v}{x} = -x + C \Rightarrow v = -x^2 + Cx.$ - **Back-substitute karo:** $v=\frac1y$, toh $\;\boxed{\dfrac{1}{y} = Cx - x^2}.$ **Check (Forecast-then-Verify):** bade $|C|$ par curve ka behavior $y\approx 1/(Cx)$ jaisa hai, jo decay karta hai — $y/x$ damping term ke saath consistent hai. ✔ --- ## [!example] Worked Example 2 — fractional power Solve karo $\dfrac{dy}{dx} = y + y^{1/2}$ (likho $y' - y = y^{1/2}$ ki tarah). - **Identify karo:** $P=-1,\ Q=1,\ n=\tfrac12.$ - $1-n = \tfrac12$, toh $v=y^{1/2}$. - **Linear form (boxed result use karo):** $v' + (1-n)Pv = (1-n)Q$: $$v' + \tfrac12(-1)v = \tfrac12(1)\ \Rightarrow\ v' - \tfrac12 v = \tfrac12.$$ *Yeh step kyun?* Seedha derived formula mein plug karo — dobara derive karne ki zaroorat nahi. - **IF:** $\mu = e^{-x/2}$. Phir $(e^{-x/2}v)' = \tfrac12 e^{-x/2}$. - Integrate karo: $e^{-x/2}v = -e^{-x/2} + C \Rightarrow v = Ce^{x/2} - 1.$ - **Back karo:** $v=\sqrt{y}$, toh $\boxed{\sqrt{y} = Ce^{x/2} - 1}$, yaani $y=(Ce^{x/2}-1)^2.$ --- ## [!example] Worked Example 3 — $Q$ jo $x$ par depend karta hai Solve karo $x\,y' + y = -2x^6 y^4.$ - **Standard form:** $x$ se divide karo: $y' + \frac1x y = -2x^5 y^4.$ Toh $P=\frac1x,Q=-2x^5,n=4.$ - $1-n=-3$, $v=y^{-3}$. - **Linear:** $v' + (-3)\frac1x v = (-3)(-2x^5) \Rightarrow v' - \frac3x v = 6x^5.$ - **IF:** $\mu=e^{-3\ln x}=x^{-3}$. $(x^{-3}v)' = 6x^2.$ - Integrate karo: $x^{-3}v = 2x^3 + C \Rightarrow v = 2x^6 + Cx^3.$ - **Back karo:** $\boxed{y^{-3} = 2x^6 + Cx^3}.$ --- ## [!mistake] Common errors ko steel-man karte hain **Mistake A — $(1-n)$ factor bhool jaana.** *Kyun sahi lagta hai:* tum $v=y^{1-n}$ substitute karte ho aur expect karte ho clean $v'+Pv=Q$ milega. *Trap:* chain rule constant $(1-n)$ nikaalti hai, jo **dono** $P$ aur $Q$ ko multiply karta hai. **Fix:** Hamesha likho $v' + (1-n)P\,v = (1-n)Q$. **Mistake B — $y^n$ se divide karte waqt $y=0$ flag na karna.** *Kyun sahi lagta hai:* algebra theek kaam karta hai. *Trap:* agar $n>0$ hai, toh $y\equiv 0$ ek genuine (singular) solution hai jo division mein kho jaati hai. **Fix:** jab valid ho tab $y=0$ alag se note karo. **Mistake C — $n=1$ hone par substitution use karna.** *Kyun sahi lagta hai:* yeh Bernoulli *lagti hai*. *Trap:* $1-n=0$ se $v=y^0=1$ banta hai, jo useless hai. **Fix:** $n=0,1$ ko already-linear treat karo, seedha solve karo. **Mistake D — Back-substitute karna bhool jaana.** Answer ko $v$ mein chhod dena incomplete hai; hamesha $y$ mein wapas aao. --- ## [!recall]- Feynman: ek 12-saal ke bachche ko samjhao Socho ek equation hai jo *almost* easy hai, bas ek annoying piece hai jisme $y$ kisi weird power par hai, jaise $y^2$ ya $\sqrt{y}$. Woh power use bumpy aur mushkil banati hai. Toh hum uske ek chunk ka naam badal dete hain — $y^{1-n}$ ko ek brand-new letter $v$ bolte hain. Yeh bilkul waise hai jaise koi confusing nickname chhod kar clear naam rakh lo. Jab hum math carefully karte hain (chain rule use karke), toh saari bumpy $y$-power wali cheezein gayab ho jaati hain aur hum ek smooth, friendly "linear" equation par aa jaate hain jo hum pehle se solve karna jaante hain. $v$ ke liye solve karo, phir $v$ ko $y$ mein wapas translate karo. Ho gaya! --- ## [!mnemonic] **"Ise ONE minus n par lao."** - **B**ernoulli → $y^n$ se divide karo - **v = y^(1−n)** substitute karo - **(1−n)** dono $P$ aur $Q$ par **chhup ke aa jaata hai** - **L**inear → integrating factor → **back-substitute**. Yaad rakho: *"Divide, Define, Differentiate, Done (linear)."* --- ## Active Recall Flashcards #flashcards/maths Bernoulli equation ka standard form kya hai? ::: $y' + P(x)y = Q(x)y^n$, jahan $n\neq 0,1$. Kaun si substitution Bernoulli equation ko linearize karti hai? ::: $v = y^{1-n}$. Substitute karne ke baad $v$ mein kaun sa linear ODE milta hai? ::: $v' + (1-n)P\,v = (1-n)Q$. Pehle $y^n$ se divide kyun karna padta hai? ::: $y^{1-n}$ expose karne ke liye aur $y^{-n}y'$ create karne ke liye jo $v$ ki chain rule absorb karti hai. $n=0$ aur $n=1$ kyun exclude hain? ::: Un cases mein equation pehle se hi linear hai; substitution unnecessary/degenerate hai. Kaun sa constant factor aksar bhool jaata hai? ::: $(1-n)$ jo dono $P$ aur $Q$ ko multiply karta hai. $y^n$ se divide karne par kaun sa solution kho sakta hai ($n>0$)? ::: Trivial solution $y\equiv 0$. Transformed equation ka integrating factor kya hai? ::: $\mu(x)=e^{\int (1-n)P\,dx}$. $v$ ke liye solve karne ke baad aakhri step kya hai? ::: $v=y^{1-n}$ back-substitute karo taaki $y$ milega. --- ## Connections - [[Linear First-Order ODEs — Integrating Factor]] (woh engine jis par hum reduce karte hain) - [[Separable Equations]] (alternative jab $P$ ya $Q$ zero ho jaaye) - [[Exact Equations and Integrating Factors]] - [[Substitution Methods in ODEs]] (homogeneous, $v=y/x$ — same "simplify karne ke liye rename karo" spirit) - [[Riccati Equations]] (nonlinearity mein ek kadam *upar*; ek known solution milne par Bernoulli tak reduce hota hai) ## 🖼️ Concept Map ```mermaid flowchart TD B[Bernoulli ODE with y^n] -->|blocked by| NL[Not linear in y] B -->|excludes n=0,1| AL[Already linear cases] NL -->|divide by y^n| EXP[Expose y^1-n term] EXP -->|substitute| V["v = y^1-n"] V -->|chain rule| DV["dv/dx absorbs y^-n y'"] DV -->|convert| LIN[Linear ODE in v] LIN -->|solve with| IF["Integrating factor e^int of 1-n P dx"] IF -->|back-substitute| SOL[Solution for y] V -->|applied in| EX[Example y'+y/x=x y^2] EX -->|gives| RES["1/y = Cx - x^2"] ```