Visual walkthrough — Bernoulli equations — substitution

What this page is: we take the Bernoulli trick from the parent note and rebuild it one picture at a time, from absolute zero. By the last step you will see why renaming a chunk of $y$ turns a bumpy nonlinear equation into a smooth linear one you already know how to crack.

We assume only that you can read $\frac{dy}{dx}$ as "how fast $y$ changes as $x$ moves". Everything else we build.

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Step 1 — What we are staring at

WHAT. We meet a first-order differential equation shaped like this:

$\underset{\text{slope of the curve}}{\underbrace{\frac{dy}{dx}}}+\underset{\text{known }x\text{-only coefficient}}{\underbrace{P(x)}}\underset{\text{unknown function}}{\underbrace{y}}=\underset{\text{known }x\text{-only term}}{\underbrace{Q(x)}}\underset{\text{the troublemaker}}{\underbrace{y^n}}$

Every symbol earned:

• $\frac{dy}{dx}$ — the rate at which the unknown curve $y(x)$ rises or falls.
• $P(x)$ and $Q(x)$ — functions that depend only on $x$ (never on $y$). They are "known knobs".
• $y^n$ — the unknown $y$ raised to some fixed power $n$ (could be $2$, $\frac{1}{2}$, $-3$, ...).

WHY it matters. If the right side were just $Q(x)$ (no $y^n$), this would be a linear equation — the friendly kind we solve with an integrating factor. The lone $y^n$ is the only thing standing in our way.

PICTURE. On the left, the linear world: $y$ appears to the first power, a straight "ramp". On the right, the $y^n$ term bends that ramp into a curve — the source of all the trouble.

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Step 2 — Why $n=0$ and $n=1$ are boring (and excluded)

WHAT. Before doing anything clever, ask: are there values of $n$ where there is no trouble at all?

• If $n=0$: $y^0=1$, so the equation is already $y^{\prime }+Py=Q$. Already linear.
• If $n=1$: the right side is $Qy$, and moving it left gives $y^{\prime }+(P-Q)y=0$. Already linear.

WHY. No substitution is needed when the equation is already the friendly kind. That is why the definition says $n\ne 0,1$. (Foreshadow: our trick will divide by a factor $1-n$; at $n=1$ that factor is $0$, so the trick would literally divide by zero — the algebra itself refuses.)

PICTURE. A number line of $n$. The two dots $n=0$ and $n=1$ are marked "already easy — skip". Everything else is "Bernoulli territory".

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Step 3 — Expose the hidden power: divide by $y^n$

WHAT. Divide every term by $y^n$ (assume for now $y\ne 0$):

$\underset{\text{a }{y}^{-n}\text{ glued to the slope}}{\underbrace{y^{-n}\frac{dy}{dx}}}+P(x)\underset{\text{one clean leftover power}}{\underbrace{y^{1-n}}}=Q(x)$

Term-by-term, what division did:

• $\frac{dy}{dx}÷y^n=y^{-n}\frac{dy}{dx}$ — the slope now carries a factor $y^{-n}$.
• $Py÷y^n=P{y}^{1-n}$ — the linear term becomes a single power $y^{1-n}$.
• $Q{y}^n÷y^n=Q$ — the troublemaker is gone! It became a plain $Q(x)$.

WHY this exact move. We wanted two things: (1) kill the ugly $y^n$ on the right, and (2) leave behind exactly one new power, $y^{1-n}$, plus a $y^{-n}{y}^{\prime }$ chunk. As Step 4 shows, that chunk is precisely what a chain rule can swallow.

PICTURE. Think of dividing as "shifting every exponent down by $n$". The right-side $y^n$ slides down to $y^0=1$; the linear $y^1$ slides to $y^{1-n}$; the slope's hidden $y^0$ slides to $y^{-n}$.

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Step 4 — Rename the clean power: let $v=y^{1-n}$

WHAT. Define a brand-new variable:

$\boxed{v=y^{1-n}}$

WHY this specific name. Two matching leftovers appear in Step 3: the power $y^{1-n}$ and the chunk $y^{-n}{y}^{\prime }$. We hope they are two faces of one object — namely $v$ and its derivative $v^{\prime }$. Step 5 confirms it. Choosing $v=y^{1-n}$ is not luck: it is engineered so that differentiating it regurgitates a $y^{-n}$ factor.

PICTURE. A "relabel" box: the messy expression $y^{1-n}$ walks in, a single tidy letter $v$ walks out. Same object, cleaner name.

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Step 5 — The engine: differentiate $v$ with the chain rule

WHAT. Differentiate $v=y^{1-n}$ with respect to $x$. Because $y$ itself depends on $x$, we use the chain rule (bring the power down, subtract one, multiply by the inside's derivative):

$\frac{dv}{dx}=\underset{\text{power comes down}}{\underbrace{(1-n)}}\underset{\text{exponent drops by 1}}{\underbrace{y^{-n}}}\underset{\text{derivative of the inside }y}{\underbrace{\frac{dy}{dx}}}$

WHY the chain rule and nothing else. We are differentiating "a power of a function", $(y(x){)}^{1-n}$. That is the textbook trigger for the chain rule: outer operation = "raise to the power $1-n$", inner function = $y(x)$. No product rule (nothing is multiplied), no quotient rule (nothing is divided) — the chain rule is the exact right tool.

Now rearrange to isolate the chunk we saw in Step 3:

$y^{-n}\frac{dy}{dx}=\frac{1}{1-n}\frac{dv}{dx}$

Read it right-to-left: the awkward $y^{-n}{y}^{\prime }$ equals a plain $v^{\prime }$ scaled by $\frac{1}{1-n}$. This is the whole trick in one line.

PICTURE. The chain rule as a two-gear machine: turning the outer "power" gear and the inner "$y$" gear together produces the $(1-n)y^{-n}{y}^{\prime }$ output.

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Step 6 — Substitute and watch it collapse into a line

WHAT. Take Step 3 and replace its two matching pieces using Step 5:

• $y^{-n}{y}^{\prime }\to \frac{1}{1-n}\frac{dv}{dx}$
• $y^{1-n}\to v$

$\frac{1}{1-n}\frac{dv}{dx}+P(x)v=Q(x)$

Multiply through by $(1-n)$ to clear the fraction:

$\boxed{\frac{dv}{dx}+\underset{\text{new coefficient}}{\underbrace{(1-n)P(x)}}v=\underset{\text{new right side}}{\underbrace{(1-n)Q(x)}}}$

Term-by-term, what changed:

• The unknown is now $v$, appearing only to the first power — this is linear in $v$.
• The factor $(1-n)$ from the chain rule sticks to both $P$ and $Q$. This is the single most-forgotten detail: it is easy to write $v^{\prime }+Pv=Q$ and lose the $(1-n)$, because the substitution $v=y^{1-n}$ tempts you into expecting a clean copy of the original. Always carry the $(1-n)$ onto both sides.

WHY this is a victory. We have converted an unfamiliar nonlinear equation into the exact shape the integrating-factor method eats for breakfast.

PICTURE. The bumpy $y^n$ curve (Step 1) morphs into a straight ramp in $v$ — the nonlinearity has been "renamed away".

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Step 7 — Solve the line, then translate back

WHAT. Solve the linear equation with the integrating factor

$\mu (x)=e^{\int (1-n)P(x)dx},$

which makes the left side a perfect derivative $(\mu v{)}^{\prime }=\mu (1-n)Q$. Integrate once, then back-substitute $v=y^{1-n}$ to recover $y$.

WHY $\mu$. Multiplying by $\mu$ is the standard move that packages $v^{\prime }+(\text{coeff})v$ into $(\mu v{)}^{\prime }$, so a single integration finishes the job. (Full derivation lives in Linear First-Order ODEs — Integrating Factor.)

Concrete run (parent's Example 1): $y^{\prime }+\frac{y}{x}=x{y}^2$ has $P=\frac{1}{x},Q=x,n=2$, so $1-n=-1$ and $v=y^{-1}$:

$v^{\prime }-\frac{1}{x}v=-x\stackrel{\mu =1/x}{\to }(\frac{v}{x}{)}^{\prime }=-1\Rightarrow \frac{v}{x}=-x+C\Rightarrow \boxed{\frac{1}{y}=Cx-x^2}.$

PICTURE. Solution curves $\frac{1}{y}=Cx-x^2$ for several $C$, showing the family the method produces.

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Step 8 — The edge case: don't lose $y\equiv 0$

WHAT. Back in Step 3 we divided by $y^n$ and quietly assumed $y\ne 0$.

WHY it's dangerous. If $n>0$, the constant function $y\equiv 0$ makes both sides of the original equation zero — it is a solution. But dividing by $y^n$ erased it, because you cannot divide by zero. It never reappears through $v=y^{1-n}$.

Fix. After solving, add $y\equiv 0$ back by hand whenever $n>0$. It is a singular solution — not part of the $C$-family, but genuine. (And recall Step 3's stricter warning: if $n$ is a non-integer, we already restricted to $y>0$, so $y=0$ sits on the boundary of the allowed domain rather than inside it.)

PICTURE. The $C$-family of curves plus the flat red line $y=0$ hovering separately, labelled "lost in the division, restored by hand".

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The one-picture summary

Everything above, compressed into a single pipeline: nonlinear $y^n$ → divide by $y^n$ → rename $v=y^{1-n}$ → chain rule feeds the engine → linear ODE in $v$ → integrating factor → back-substitute → $y$ (plus the singular $y=0$ when $n>0$).

Recall Feynman retelling — the whole walkthrough in plain words

We started with an equation that was almost the easy kind, except one piece had $y$ raised to a weird power $n$. That power is the only villain. First we divided everything by $y^n$; this made the villain on the right vanish into a plain "$1$", and it left behind exactly two matching crumbs: a lonely power $y^{1-n}$ and a slope-chunk $y^{-n}{y}^{\prime }$. (If that power is a fraction, we quietly agree to stay on one side of zero — say $y>0$ — so $\sqrt{y}$ and friends make sense.) Then we gave the lonely power a nickname, $v=y^{1-n}$. When we differentiated that nickname using the chain rule, out popped the slope-chunk (times a constant $1-n$) — proving the two crumbs were really $v$ and $v^{\prime }$ in disguise. Swapping them in, the whole equation flattened into a straight, linear equation in $v$, which we solve with the integrating-factor trick we already know. Solve for $v$, translate the nickname back to $y$, and — if the power was positive — remember to hand back the flat solution $y=0$ that the division had swallowed. Done.

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Connections

• Bernoulli equations — substitution (the parent — full worked examples live there)
• Linear First-Order ODEs — Integrating Factor (the engine Step 7 hands off to)
• Separable Equations (what remains when $P$ or $Q$ vanishes)
• Exact Equations and Integrating Factors
• Substitution Methods in ODEs (same "rename to simplify" spirit)
• Riccati Equations (one step up in nonlinearity — reduces to Bernoulli)