Worked examples — Bernoulli equations — substitution

This page is the exhaustive practice sheet for Bernoulli equations. The parent taught the machine; here we drive it over every road: positive powers, negative powers, fractions, the degenerate cases, an initial-value problem, a word problem, and an exam trap. Nothing here contradicts the parent — we just go wider and slower.

Before anything, recall the single object we work with. A Bernoulli equation is a first-order ODE that can be forced into the shape $\frac{dy}{dx}+P(x)y=Q(x)y^n,n\ne 0,1,$ where $P,Q$ depend only on $x$, and the only thing that stops it being linear is the power $y^n$. The cure (derived in the parent) is:

Here "$y^{\prime }$" is shorthand for $\frac{dy}{dx}$ ("the rate at which $y$ changes as $x$ moves"), and $\int \dots dx$ means "the area-accumulating antiderivative". The letter $C$ is always the arbitrary constant that appears when we integrate.

The scenario matrix

Bernoulli problems differ along a few axes. If you can solve one example from each row, you can solve any Bernoulli equation an exam throws at you.

We now hit every cell.

[!example] Ex 1 — Cell C1 · positive integer power

Solve $\frac{dy}{dx}-\frac{y}{x}=x{y}^2.$ (Work on $x>0$.)

Forecast: Here $n=2$, so $v=y^{1-2}=y^{-1}=1/y$. Because $v$ is a reciprocal, guess the final answer will read "$1/y=(\text{something in }x)$", not "$y=\dots$" directly.

1. Identify. $P=-\frac{1}{x},Q=x,n=2.$ Why this step? The $y^2$ on the right flags Bernoulli; matching the standard form gives us the three ingredients the recipe needs.
2. Divide by $y^2$: $y^{-2}{y}^{\prime }-\frac{1}{x}{y}^{-1}=x.$ Why this step? This exposes $y^{1-n}=y^{-1}$ and creates the $y^{-2}{y}^{\prime }$ clump the chain rule will swallow.
3. Substitute $v=y^{-1}$, so $v^{\prime }=-y^{-2}{y}^{\prime }\Rightarrow y^{-2}{y}^{\prime }=-v^{\prime }.$ Why this step? This is the "engine" — it turns the awkward $y^{-2}{y}^{\prime }$ into a plain $-v^{\prime }$.
4. Linear ODE. Replace: $-v^{\prime }-\frac{1}{x}v=x\Rightarrow v^{\prime }+\frac{1}{x}v=-x.$ Why this step? Same as plugging into $v^{\prime }+(1-n)Pv=(1-n)Q$: $(1-2)(-1/x)=1/x$ and $(1-2)(x)=-x$. Both agree. ✔
5. Integrating factor. $\mu =e^{\int (1/x)dx}=e^{\ln |x|}=x$ (on $x>0$). Then $(xv{)}^{\prime }=x\cdot (-x)=-x^2.$ Why this step? Multiplying by $\mu$ makes the left side a perfect derivative $(\mu v{)}^{\prime }$, so we can just integrate. (On $x<0$ we'd use $\mu =-x$, giving the same ODE after cancelling.)
6. Integrate & back-substitute. $xv=-\frac{x^3}{3}+C\Rightarrow v=\frac{C}{x}-\frac{x^2}{3}.$ Since $v=1/y$: $\boxed{\frac{1}{y}=\frac{C}{x}-\frac{x^2}{3}}.$ Why this step? We integrate $(xv{)}^{\prime }=-x^2$ to recover $xv$, then undo the substitution $v=1/y$ because the recipe's Step 5 demands we translate back from the helper variable to the real unknown $y$.

Verify: Take $C=0$: $1/y=-x^2/3\Rightarrow y=-3/x^2$, so $y^{\prime }=6/x^3$. Plug into $y^{\prime }-y/x$: $6/x^3-(-3/x^2)/x=6/x^3+3/x^3=9/x^3$. And $x{y}^2=x\cdot 9/x^4=9/x^3$. Both sides match. ✔

[!example] Ex 2 — Cell C2 · fractional power $0<n<1$

Solve $y^{\prime }-y=y^{1/2}$ (the parent's Example 2, re-verified as our fraction case). We show all recipe steps this time.

Forecast: $n=\frac{1}{2}$ gives $1-n=\frac{1}{2}$, so $v=y^{1/2}=\sqrt{y}$. The new variable is a square root, so at the end we'll square to get $y$.

1. Identify. $P=-1,Q=1,n=\frac{1}{2}.$ Why this step? Written as $y^{\prime }+Py=Q{y}^n$, we read the constants directly.
2. Divide by $y^n=y^{1/2}$: $y^{-1/2}{y}^{\prime }-y^{1/2}=1.$ Why this step? Exactly as in Ex 1 — expose $y^{1-n}=y^{1/2}$ and create the $y^{-1/2}{y}^{\prime }$ clump the chain rule will absorb. We do not skip this step.
3. Substitute $v=y^{1/2}$, so $v^{\prime }=\frac{1}{2}{y}^{-1/2}{y}^{\prime }\Rightarrow y^{-1/2}{y}^{\prime }=2v^{\prime }.$ Why this step? This converts the awkward $y^{-1/2}{y}^{\prime }$ into a plain $2v^{\prime }$ (the "engine").
4. Linear ODE. Replace in Step 2: $2v^{\prime }-v=1\Rightarrow v^{\prime }-\frac{1}{2}v=\frac{1}{2}.$ Why this step? Divide by $2$. This matches the recipe formula $v^{\prime }+(1-n)Pv=(1-n)Q$ with $(1-n)=\frac{1}{2}$ multiplying both $P$ and $Q$ — the factor everyone forgets. ✔
5. Integrating factor. $\mu =e^{\int -\frac{1}{2}dx}=e^{-x/2}.$ Then $(e^{-x/2}v{)}^{\prime }=\frac{1}{2}{e}^{-x/2}.$ Why this step? Turns the left into a single derivative.
6. Integrate. $e^{-x/2}v=-e^{-x/2}+C\Rightarrow v=C{e}^{x/2}-1.$ Why this step? $\int \frac{1}{2}{e}^{-x/2}dx=-e^{-x/2}$; dividing by $\mu =e^{-x/2}$ isolates $v$.
7. Back-substitute $v=\sqrt{y}$: $\boxed{\sqrt{y}=C{e}^{x/2}-1,y=(C{e}^{x/2}-1{)}^2}.$ Why this step? We must translate the helper $v$ back to $y$; squaring is legitimate here only because $\sqrt{y}\ge 0$ — see the branch note.

Verify: With $C=1$, $y=(e^{x/2}-1{)}^2$. For $x>0$ we have $e^{x/2}-1>0$, so $\sqrt{y}=e^{x/2}-1$ and $y^{1/2}=e^{x/2}-1$. Then $y^{\prime }=2(e^{x/2}-1)\cdot \frac{1}{2}{e}^{x/2}=(e^{x/2}-1)e^{x/2}$. Also $y+y^{1/2}=(e^{x/2}-1{)}^2+(e^{x/2}-1)=(e^{x/2}-1)[(e^{x/2}-1)+1]=(e^{x/2}-1)e^{x/2}$. Equal on $x>0$. ✔

[!example] Ex 3 — Cell C3 · genuinely negative power $n<0$

Solve $y^{\prime }+y=y^{-2}$ (here $n=-2$, a truly negative index).

Forecast: $n=-2$ gives $1-n=1-(-2)=3$, so $v=y^3$ — a positive power this time. Since $v=y^3$, the final answer will read "$y^3=\dots$", and to get $y$ we take a cube root (odd root, so no $\pm$ ambiguity — every real number has exactly one real cube root).

1. Identify. In standard form $y^{\prime }+Py=Q{y}^n$ we read $P=1,Q=1,n=-2.$ Why this step? The $y^{-2}$ on the right is the Bernoulli power; $n<0$ is allowed since $n\ne 0,1$.
2. Divide by $y^n=y^{-2}$ (i.e. multiply by $y^2$): $y^2{y}^{\prime }+y^3=1.$ Substitute $v=y^3$, so $v^{\prime }=3y^2{y}^{\prime }\Rightarrow y^2{y}^{\prime }=\frac{1}{3}{v}^{\prime }.$ Why this step? Expose $y^{1-n}=y^3$ and turn $y^2{y}^{\prime }$ into $\frac{1}{3}{v}^{\prime }$.
3. Linear ODE. Replace: $\frac{1}{3}{v}^{\prime }+v=1\Rightarrow v^{\prime }+3v=3.$ (Same as recipe $v^{\prime }+(1-n)Pv=(1-n)Q$ with $(1-n)=3$ multiplying both $P$ and $Q$.) Why this step? Multiply by $3$; note $v=y^3$ is a positive power — the hallmark of the negative-$n$ case.
4. Integrating factor. $\mu =e^{\int 3dx}=e^{3x}.$ Then $(e^{3x}v{)}^{\prime }=3e^{3x}.$ Why this step? Multiplying by $\mu$ collapses the left side to a single derivative $(\mu v{)}^{\prime }$.
5. Integrate & isolate. $e^{3x}v=e^{3x}+C\Rightarrow v=1+C{e}^{-3x}.$ Why this step? We integrate $\int 3e^{3x}dx=e^{3x}$, then divide by $\mu =e^{3x}$ to free $v$.
6. Back-substitute $v=y^3$: $\boxed{y^3=1+C{e}^{-3x},y=(1+C{e}^{-3x}{)}^{1/3}.}$ Why this step? The recipe's Step 5: undo the helper. A cube root is single-valued over the reals, so no $\pm$ branch appears — unlike the square-root case of Ex 2.

Verify: With $C=0$, $y^3=1\Rightarrow y=1$, so $y^{\prime }=0$. Original: $y^{\prime }+y=0+1=1$ and $y^{-2}=1^{-2}=1$. Match. ✔ With general $C$: differentiate $y^3=1+C{e}^{-3x}$ to get $3y^2{y}^{\prime }=-3C{e}^{-3x}$, so $y^{\prime }=-\frac{C{e}^{-3x}}{y^2}$. Then $y^{\prime }+y=\frac{-C{e}^{-3x}+y^3}{y^2}=\frac{-C{e}^{-3x}+1+C{e}^{-3x}}{y^2}=\frac{1}{y^2}=y^{-2}$. ✔

[!example] Ex 4 — Cell C4 · initial-value problem

Solve $y^{\prime }+\frac{y}{x}=x{y}^2$ with $y(1)=\frac{1}{2}.$ (Work on $x>0$, which contains $x=1$.)

Forecast: This is Ex-1's cousin but with a plus $y/x$. We'll get a family $1/y=f(x)+Cx$, and the initial condition will pin one curve. At $x=1,y=\frac{1}{2}$ we have $1/y=2$, so $C$ should come out a small tidy number.

1. Identify & set up. $P=\frac{1}{x},Q=x,n=2,$ so $v=y^{-1},v^{\prime }=-y^{-2}{y}^{\prime }.$ Why this step? Same Bernoulli shape; we prepare the substitution.
2. Linear ODE. $v^{\prime }+(1-2)\frac{1}{x}v=(1-2)x\Rightarrow v^{\prime }-\frac{1}{x}v=-x.$ Why this step? Plug into the recipe formula.
3. Integrating factor. $\mu =e^{\int -1/xdx}=e^{-\ln |x|}=x^{-1}$ (on $x>0$). Then ${\left(\frac{v}{x}\right)}^{\prime }=-1.$ Why this step? Perfect-derivative trick; the $\ln$ carries an absolute value, resolved by working on $x>0$.
4. General solution. $\frac{v}{x}=-x+C\Rightarrow v=Cx-x^2\Rightarrow \frac{1}{y}=Cx-x^2.$ Why this step? We integrate $(v/x{)}^{\prime }=-1$ to get $v/x=-x+C$, multiply by $x$ to isolate $v$, then undo $v=1/y$ so the answer is stated in the original unknown $y$.
5. Apply $y(1)=\frac{1}{2}$. Then $1/y=2$: $2=C(1)-1\Rightarrow C=3.$ Why this step? The initial condition selects the one curve through the given point. $\boxed{\frac{1}{y}=3x-x^2,y=\frac{1}{3x-x^2}}.$

Verify: At $x=1$: $3(1)-1=2$, so $y=1/2$. ✔ Differentiate: $y^{\prime }=-\frac{3-2x}{(3x-x^2{)}^2}$. Check $y^{\prime }+y/x$ at $x=1$: $y^{\prime }=-(1)/(2{)}^2=-1/4$; $y/x=(1/2)/1=1/2$; sum $=1/4$. And $x{y}^2=1\cdot (1/2{)}^2=1/4$. Match. ✔

[!example] Ex 5 — Cell C5 · degenerate $n=0$ and $n=1$

Two mini-problems that look Bernoulli but aren't. (Work on $x>0$.)

(a) $n=0$: $y^{\prime }+\frac{2}{x}y=3x.$ Forecast: $n=0$ means $y^0=1$, so the "$y^n$" is invisible — this is already linear. No substitution.

1. $\mu =e^{\int 2/xdx}=e^{2\ln |x|}=x^2.$ Why? Standard integrating factor; $e^{2\ln |x|}=|x{|}^2=x^2$ for either sign, so no branch issue here.
2. $(x^{2}y{)}^{\prime }=3x^3\Rightarrow x^{2}y=\frac{3x^4}{4}+C\Rightarrow y=\frac{3x^2}{4}+\frac{C}{x^2}.$

Verify: $C=0$: $y=\frac{3}{4}{x}^2,y^{\prime }=\frac{3}{2}x$. Then $y^{\prime }+\frac{2}{x}y=\frac{3}{2}x+\frac{2}{x}\cdot \frac{3}{4}{x}^2=\frac{3}{2}x+\frac{3}{2}x=3x.$ ✔

(b) $n=1$: $y^{\prime }+\frac{1}{x}y=4y.$ Forecast: $n=1$ would give $v=y^0=1$ (useless!). Instead collect $y$: it's linear and separable.

1. Rewrite: $y^{\prime }=\left(4-\frac{1}{x}\right)y.$ Why? All $y$-terms together — the substitution $1-n=0$ is degenerate.
2. Separable (see Separable Equations): $\frac{dy}{y}=\left(4-\frac{1}{x}\right)dx.$
3. Integrate: $\ln |y|=4x-\ln |x|+C_1\Rightarrow \boxed{y=\frac{K{e}^{4x}}{x}}$ (with $K=e^{C_1}$, and note the $\ln |x|$ absolute value again).

Verify: $y=K{e}^{4x}/x$, $y^{\prime }=K\frac{4e^{4x}x-e^{4x}}{x^2}=K{e}^{4x}\frac{4x-1}{x^2}$. Then $y^{\prime }+\frac{1}{x}y=K{e}^{4x}\frac{4x-1}{x^2}+K{e}^{4x}\frac{1}{x^2}=K{e}^{4x}\frac{4x}{x^2}=4\cdot \frac{K{e}^{4x}}{x}=4y.$ ✔

[!example] Ex 6 — Cell C6 · the lost solution $y\equiv 0$

Solve $y^{\prime }=y^3$ and account for every solution.

Forecast: This is Bernoulli with $P=0,Q=1,n=3$. Dividing by $y^3$ silently assumes $y\ne 0$ — but $y\equiv 0$ does satisfy $y^{\prime }=y^3$ (both sides $0$). Watch for it.

1. Note the constant solution first. $y\equiv 0\Rightarrow y^{\prime }=0=0^3.$ Valid. Keep it aside. Why this step? Because Step 2 will divide by $y^3$ and erase it.
2. Divide by $y^3$: $y^{-3}{y}^{\prime }=1.$ Substitute $v=y^{-2}$ ($1-n=-2$), $v^{\prime }=-2y^{-3}{y}^{\prime }\Rightarrow y^{-3}{y}^{\prime }=-\frac{1}{2}{v}^{\prime }.$ Why this step? Standard exposure of $y^{1-n}$.
3. Linear ODE. $-\frac{1}{2}{v}^{\prime }=1\Rightarrow v^{\prime }=-2\Rightarrow v=-2x+C.$ Why this step? $P=0$ makes it trivially integrable (no $\ln$ needed).
4. Back-substitute $v=y^{-2}$: $\boxed{y^{-2}=C-2x,y=\pm \frac{1}{\sqrt{C-2x}}}\text{plus the singular solution }y\equiv 0.$ Why this step? Undo the helper $v=y^{-2}$. Since $y^{-2}=1/y^2$ loses the sign of $y$, solving for $y$ genuinely produces both signs $\pm$ — each is a legitimate branch, valid only where $C-2x>0$.

Verify: Take $C=0$: $y=1/\sqrt{-2x}$ (needs $x<0$), $y=(-2x{)}^{-1/2}$, $y^{\prime }=-\frac{1}{2}(-2x{)}^{-3/2}\cdot (-2)=(-2x{)}^{-3/2}$. And $y^3=(-2x{)}^{-3/2}$. Match. ✔ Also $y\equiv 0$ gives $0=0$. ✔

[!example] Ex 7 — Cell C7 · real-world word problem (logistic growth)

A fish population $y(t)$ (in thousands) grows, but crowding limits it. It obeys the logistic law $\frac{dy}{dt}=ry-\frac{r}{K}{y}^2,r=2{\text{yr}}^{-1},K=10\text{thousand}.$ Start with $y(0)=2$ thousand fish. Find $y(t)$ using the Bernoulli method, and predict the long-run population.

Forecast: The $y^2$ crowding term makes this Bernoulli with $n=2$. Because fish can't grow forever, expect $y\to K=10$ thousand as $t\to \infty$. So the answer should flatten toward $10$.

1. Standard form. $y^{\prime }-2y=-\frac{2}{10}{y}^2=-0.2y^2.$ So $P=-2,Q=-0.2,n=2.$ Why this step? Isolate the linear $y$ on the left; identify $P,Q,n$.
2. Substitute $v=y^{-1}$. Recipe: $v^{\prime }+(1-2)(-2)v=(1-2)(-0.2)\Rightarrow v^{\prime }+2v=\mathrm{0.2.}$ Why this step? $(1-n)=-1$ multiplies both $P$ and $Q$; the signs flip.
3. Integrating factor. $\mu =e^{\int 2dt}=e^{2t}.$ Then $(e^{2t}v{)}^{\prime }=0.2e^{2t}.$ Why this step? Collapse the left to a derivative (no $\ln$ here, $P$ constant).
4. Integrate. $e^{2t}v=0.1e^{2t}+C\Rightarrow v=0.1+C{e}^{-2t}.$ Why this step? $\int 0.2e^{2t}dt=0.1e^{2t}$; divide by $\mu$ to isolate $v$.
5. Back-substitute $v=1/y$: $\frac{1}{y}=0.1+C{e}^{-2t}.$ Apply $y(0)=2\Rightarrow \frac{1}{2}=0.1+C\Rightarrow C=\mathrm{0.4.}$ Why this step? Undo the helper (populations are $1/v$), then use the starting count to fix $C$. $\boxed{\frac{1}{y}=0.1+0.4e^{-2t},y(t)=\frac{1}{0.1+0.4e^{-2t}}.}$

The figure below plots this solution. Horizontal axis: time $t$ in years. Vertical axis: population $y$ in thousands of fish. The magenta S-curve climbs from the orange start point $y(0)=2$ and flattens onto the violet dashed line $y=K=10$ (the carrying capacity). Qualitatively: growth is fast when the pond is nearly empty (the $ry$ term dominates), then throttles as crowding — the $-\frac{r}{K}{y}^2$ term — bites. That flattening is the geometric meaning of the algebra we just did.

Verify (long-run & units): As $t\to \infty$, $e^{-2t}\to 0$, so $\frac{1}{y}\to 0.1\Rightarrow y\to \frac{1}{0.1}=10$ thousand $=K$. ✔ At $t=0$: $y=1/(0.1+0.4)=1/0.5=2$ thousand, matching the initial count. ✔ Units: $r$ has units ${\text{yr}}^{-1}$, so $ry$ has units (thousand)$\cdot {\text{yr}}^{-1}$, matching $\frac{dy}{dt}$; and $\frac{r}{K}{y}^2$ has units $\frac{{\text{yr}}^{-1}}{\text{thousand}}\cdot (\text{thousand}{)}^2=(\text{thousand})\cdot {\text{yr}}^{-1}$ too. Both terms carry the same units as $y^{\prime }$. ✔

[!example] Ex 8 — Cell C8 · exam twist (disguised, plus a Riccati sibling)

(a) Disguised Bernoulli. Solve $\frac{dy}{dx}=\frac{y}{x}+\frac{x^2}{y}.$ It doesn't look like the standard form at all. (Work on $x>0$.)

Forecast: The $\frac{x^2}{y}=x^2{y}^{-1}$ term means $n=-1$. Then $1-n=2$, so $v=y^2$ — a positive power. Expect the answer as "$y^2=\dots$", then a $\pm$ square root to recover $y$.

1. Rearrange to standard form. $y^{\prime }-\frac{1}{x}y=x^2{y}^{-1}.$ So $P=-\frac{1}{x},Q=x^2,n=-1.$ Why this step? Move the linear term left; recognise $y^{-1}$ as the Bernoulli power — the exam disguise is just the un-rearranged layout.
2. Divide by $y^n=y^{-1}$ (multiply by $y$): $y{y}^{\prime }-\frac{1}{x}{y}^2=x^2.$ Substitute $v=y^2$, so $v^{\prime }=2y{y}^{\prime }\Rightarrow y{y}^{\prime }=\frac{1}{2}{v}^{\prime }.$ Why this step? Expose $y^{1-n}=y^2$ and turn $y{y}^{\prime }$ into $\frac{1}{2}{v}^{\prime }$.
3. Linear ODE. Replace: $\frac{1}{2}{v}^{\prime }-\frac{1}{x}v=x^2\Rightarrow v^{\prime }-\frac{2}{x}v=2x^2.$ (Same as recipe with $(1-n)=2$ multiplying $P$ and $Q$.) Why this step? Multiply by $2$ to clear the $\frac{1}{2}$.
4. Integrating factor. $\mu =e^{\int -2/xdx}=e^{-2\ln |x|}=x^{-2}$ (on $x>0$). Then $(x^{-2}v{)}^{\prime }=x^{-2}\cdot 2x^2=2.$ Why this step? Perfect-derivative form; the $\ln$ carries $|x|$, resolved by $x>0$.
5. Integrate & back-substitute. $x^{-2}v=2x+C\Rightarrow v=2x^3+C{x}^2.$ With $v=y^2$: $\boxed{y^2=2x^3+C{x}^2,y=\pm \sqrt{2x^3+C{x}^2}.}$ Why this step? We integrate $(x^{-2}v{)}^{\prime }=2$ to get $x^{-2}v=2x+C$, multiply by $x^2$ to isolate $v$, then undo $v=y^2$. Because $y^2$ discards the sign of $y$, recovering $y$ needs both roots $\pm$ — each is a valid solution branch, defined only where $2x^3+C{x}^2\ge 0$.

Verify: $C=0$: $y^2=2x^3$, take $y=+\sqrt{2x^3}$ (the $+$ branch), so $2y{y}^{\prime }=6x^2\Rightarrow y^{\prime }=3x^2/y$. Compare $y/x+x^2/y$. Using $y^2=2x^3$: $y/x=\sqrt{2x^3}/x=\sqrt{2}{x}^{1/2}$ and $x^2/y=x^2/\sqrt{2x^3}=x^{1/2}/\sqrt{2}$. Sum $=\sqrt{2}{x}^{1/2}+x^{1/2}/\sqrt{2}=\frac{2+1}{\sqrt{2}}{x}^{1/2}=\frac{3}{\sqrt{2}}{x}^{1/2}$. And $3x^2/y=3x^2/\sqrt{2x^3}=\frac{3}{\sqrt{2}}{x}^{1/2}$. Match (the $-$ branch checks identically with $y\to -y$). ✔

(b) The Riccati sibling — why exams pair them. A Riccati equation $y^{\prime }=a(x)+b(x)y+c(x)y^2$ has an extra constant/known-free term $a(x)$, so it is not Bernoulli. But if you already know one particular solution $y_1(x)$, the substitution $y=y_1+1/z$ turns a Riccati into a linear ODE in $z$ — the exact same "rename to linearise" spirit. When $a(x)\equiv 0$, Riccati collapses back to a Bernoulli with $n=2$.

(No new numeric claim in (b) — recognition only.)

[!recall]- Quick self-test (reveal after guessing)

Connections

• Bernoulli equations — substitution (parent — the method these examples exercise)
• Linear First-Order ODEs — Integrating Factor (every example ends here)
• Separable Equations (Ex 5b solved this way)
• Exact Equations and Integrating Factors
• Substitution Methods in ODEs (same rename-to-simplify idea)
• Riccati Equations (Ex 8b — the step up in nonlinearity)

Scenario map