Bernoulli equations — substitution

One-line idea: A Bernoulli equation looks nonlinear because of a stray $y^n$ term, but a clever substitution $v = y^{1-n}$ turns it into a linear ODE you already know how to solve.

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[!intuition] Why does this work at all?

WHAT we face: an equation of the form $\frac{dy}{dx} + P(x)\,y = Q(x)\,y^n,\qquad n\neq 0,1.$ The right-hand side has $y^n$, so it's not linear in $y$. We can't use the integrating-factor trick directly.

WHY the trick exists: The only thing blocking linearity is the power $y^n$. If we could define a new variable $v$ whose derivative naturally absorbs that power, the equation would collapse into a linear one in $v$. Differentiating $y^{1-n}$ produces exactly a $y^{-n}$ factor — the perfect "antidote" to the $y^n$ nuisance.

HOW we exploit it: divide through by $y^n$ first (to expose $y^{1-n}$), then substitute $v = y^{1-n}$. The algebra clicks into a linear ODE.

Cases $n=0$ and $n=1$ are excluded because then the equation is already linear (no substitution needed): $n=0$ gives $y'+Py=Q$; $n=1$ gives $y'+(P-Q)y=0$.

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[!definition] Bernoulli equation

A first-order ODE is a Bernoulli equation if it can be written $\frac{dy}{dx} + P(x)\,y = Q(x)\,y^{n},\quad n\in\mathbb{R},\ n\neq 0,1.$ Here $P,Q$ are functions of $x$ only, and the nonlinearity is only the power ==$y^n$==.

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[!formula] The substitution — derived from scratch

Step 1 — Expose $y^{1-n}$. Divide every term by $y^n$ (assume $y\neq 0$): $y^{-n}\frac{dy}{dx} + P(x)\,y^{1-n} = Q(x).$ Why this step? We want a single power of $y$ left over ($y^{1-n}$) and a clean $y^{-n}y'$ that a chain rule can swallow.

Step 2 — Define the new variable. $\boxed{v = y^{1-n}}$ Why this choice? So that its derivative gives back the $y^{-n}y'$ we have.

Step 3 — Differentiate $v$ (chain rule).

\quad\Longrightarrow\quad y^{-n}\frac{dy}{dx} = \frac{1}{1-n}\frac{dv}{dx}.$$ *Why this step?* This is the engine: it converts the awkward $y^{-n}y'$ into a plain $v'$. **Step 4 — Substitute into Step 1.** $$\frac{1}{1-n}\frac{dv}{dx} + P(x)\,v = Q(x).$$ Multiply by $(1-n)$: $$\boxed{\dfrac{dv}{dx} + (1-n)P(x)\,v = (1-n)Q(x)}$$ This is **linear in $v$** — solve with integrating factor $\mu(x)=e^{\int (1-n)P\,dx}$, then back-substitute $v=y^{1-n}$. --- ![[4.6.05-Bernoulli-equations-—-substitution.png]] --- ## [!example] Worked Example 1 — a classic Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = x\,y^2.$ - **Identify:** $P=\frac1x,\ Q=x,\ n=2.$ *Why?* Match the standard form; the $y^2$ flags Bernoulli. - **Divide by $y^2$:** $\;y^{-2}y' + \frac1x y^{-1} = x.$ *Why?* Expose $y^{1-n}=y^{-1}$. - **Substitute** $v=y^{-1}$, so $v' = -y^{-2}y'\Rightarrow y^{-2}y'=-v'.$ *Why?* Plug Step-3 relation. - **Linear ODE:** $-v' + \frac1x v = x \Rightarrow v' - \frac1x v = -x.$ *Why?* Just rearranged. - **Integrating factor:** $\mu=e^{\int -1/x\,dx}=e^{-\ln x}=\frac1x.$ - **Solve:** $(\frac{v}{x})' = -1 \Rightarrow \frac{v}{x} = -x + C \Rightarrow v = -x^2 + Cx.$ - **Back-substitute:** $v=\frac1y$, so $\;\boxed{\dfrac{1}{y} = Cx - x^2}.$ **Check (Forecast-then-Verify):** at large $|C|$ the curve behaves like $y\approx 1/(Cx)$, decaying — consistent with a damping $y/x$ term. ✔ --- ## [!example] Worked Example 2 — fractional power Solve $\dfrac{dy}{dx} = y + y^{1/2}$ (write as $y' - y = y^{1/2}$). - **Identify:** $P=-1,\ Q=1,\ n=\tfrac12.$ - $1-n = \tfrac12$, so $v=y^{1/2}$. - **Linear form (use boxed result):** $v' + (1-n)Pv = (1-n)Q$: $$v' + \tfrac12(-1)v = \tfrac12(1)\ \Rightarrow\ v' - \tfrac12 v = \tfrac12.$$ *Why this step?* Plug straight into the derived formula — no re-deriving. - **IF:** $\mu = e^{-x/2}$. Then $(e^{-x/2}v)' = \tfrac12 e^{-x/2}$. - Integrate: $e^{-x/2}v = -e^{-x/2} + C \Rightarrow v = Ce^{x/2} - 1.$ - **Back:** $v=\sqrt{y}$, so $\boxed{\sqrt{y} = Ce^{x/2} - 1}$, i.e. $y=(Ce^{x/2}-1)^2.$ --- ## [!example] Worked Example 3 — $Q$ depending on $x$ Solve $x\,y' + y = -2x^6 y^4.$ - **Standard form:** divide by $x$: $y' + \frac1x y = -2x^5 y^4.$ So $P=\frac1x,Q=-2x^5,n=4.$ - $1-n=-3$, $v=y^{-3}$. - **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: $x^{-3}v = 2x^3 + C \Rightarrow v = 2x^6 + Cx^3.$ - **Back:** $\boxed{y^{-3} = 2x^6 + Cx^3}.$ --- ## [!mistake] Steel-manning the common errors **Mistake A — Forgetting the $(1-n)$ factor.** *Why it feels right:* you substitute $v=y^{1-n}$ and expect a clean $v'+Pv=Q$. *The trap:* the chain rule spits out the constant $(1-n)$, so it multiplies **both** $P$ and $Q$. **Fix:** Always write $v' + (1-n)P\,v = (1-n)Q$. **Mistake B — Dividing by $y^n$ without flagging $y=0$.** *Why it feels right:* algebra works fine. *The trap:* if $n>0$, $y\equiv 0$ is a genuine (singular) solution lost in the division. **Fix:** note $y=0$ separately when valid. **Mistake C — Using the substitution when $n=1$.** *Why it feels right:* it *looks* Bernoulli. *The trap:* $1-n=0$ makes $v=y^0=1$, useless. **Fix:** treat $n=0,1$ as already-linear, solve directly. **Mistake D — Forgetting to back-substitute.** Leaving the answer in $v$ is incomplete; always return to $y$. --- ## [!recall]- Feynman: explain to a 12-year-old Imagine an equation that's *almost* an easy one, except one annoying piece has $y$ raised to a weird power, like $y^2$ or $\sqrt{y}$. That power makes it bumpy and hard. So we rename a chunk of it — we call $y^{1-n}$ a brand-new letter $v$. It's like swapping a confusing nickname for a clear name. When we do the math carefully (using the chain rule), all the bumpy $y$-power stuff vanishes and we're left with a smooth, friendly "linear" equation that we already know how to crack. Solve for $v$, then translate $v$ back into $y$. Done! --- ## [!mnemonic] **"Bring it to ONE minus n."** - **B**ernoulli → divide by $y^n$ - substitute **v = y^(1−n)** - the **(1−n)** sneaks onto **both** $P$ and $Q$ - **L**inear → integrating factor → **back-substitute**. Remember: *"Divide, Define, Differentiate, Done (linear)."* --- ## Active Recall Flashcards #flashcards/maths What is the standard form of a Bernoulli equation? ::: $y' + P(x)y = Q(x)y^n$, with $n\neq 0,1$. Which substitution linearizes a Bernoulli equation? ::: $v = y^{1-n}$. After substituting, what linear ODE in $v$ do you get? ::: $v' + (1-n)P\,v = (1-n)Q$. Why must you first divide by $y^n$? ::: To expose $y^{1-n}$ and create the $y^{-n}y'$ that the chain rule of $v$ absorbs. Why are $n=0$ and $n=1$ excluded? ::: The equation is already linear in those cases; the substitution is unnecessary/degenerate. What constant factor is commonly forgotten? ::: The $(1-n)$ multiplying both $P$ and $Q$. What solution may be lost when dividing by $y^n$ (n>0)? ::: The trivial solution $y\equiv 0$. What is the integrating factor for the transformed equation? ::: $\mu(x)=e^{\int (1-n)P\,dx}$. Final step after solving for $v$? ::: Back-substitute $v=y^{1-n}$ to recover $y$. --- ## Connections - [[Linear First-Order ODEs — Integrating Factor]] (the engine we reduce to) - [[Separable Equations]] (alternative when $P$ or $Q$ vanish) - [[Exact Equations and Integrating Factors]] - [[Substitution Methods in ODEs]] (homogeneous, $v=y/x$ — same "rename to simplify" spirit) - [[Riccati Equations]] (a step *up* in nonlinearity; reduces to Bernoulli given one known solution) ## 🖼️ 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"] ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Bernoulli equation dikhne mein nonlinear lagti hai kyunki uske right side mein ek $y^n$ term ghoom raha hota hai: $y' + P(x)y = Q(x)y^n$. Direct integrating factor wala trick yahan kaam nahi karta, kyunki woh sirf linear equations pe chalta hai. Lekin ek smart move hai — hum naam badal dete hain. Pehle poore equation ko $y^n$ se divide karo, taaki $y^{1-n}$ saamne aa jaaye, phir substitution karo $v = y^{1-n}$. > > Iska kamaal yeh hai: jab tum $v$ ko differentiate karte ho (chain rule se), to nikalta hai $(1-n)y^{-n}y'$ — aur yahi cheez exactly woh $y^{-n}y'$ hai jo tumhare paas already pada hai. Sab nonlinear bumpy part cancel ho jaata hai, aur reh jaata hai ek seedha-saadha **linear** equation $v' + (1-n)Pv = (1-n)Q$. Yeh tum integrating factor se aaram se solve karte ho. > > Sabse common galti: woh $(1-n)$ wala factor bhool jaana — yaad rakho woh $P$ aur $Q$ dono pe lagta hai. Doosri baat: $n=0$ ya $n=1$ ho to substitution ki zaroorat hi nahi, equation already linear hai. Aur end mein $v$ ko wapas $y$ mein convert karna mat bhoolna, warna answer adhoora reh jaayega. > > Yeh topic isliye important hai kyunki bahut saari real-world growth/decay aur engineering problems Bernoulli form mein aati hain, aur yeh ek perfect example hai ki "variable rename karke problem easy banao" — yahi soch aage Riccati aur substitution methods mein bhi kaam aati hai. ![[audio/4.6.05-Bernoulli-equations-—-substitution.mp3]]