Bernoulli equations — substitution: Question bank

Before we start, one shared vocabulary reminder so no symbol is unearned:

True or false — justify

y^{'} + x y = x^{2} y^{3}

is a Bernoulli equation.

True. It matches

y^{'} + P y = Q y^{n}

with

P = x

Q = x^{2}

n = 3

, and

3 \neq = 0, 1

, so the substitution

v = y^{- 2}

applies.

y^{'} + x y^{2} = x^{2} y

is a Bernoulli equation in standard form as written.

False as written, but it is Bernoulli after rearranging: move to

y^{'} - x^{2} y = - x y^{2}

, giving

P = - x^{2}

Q = - x

n = 2

. The nonlinear power must sit alone on the right before you read off

n

The substitution

v = y^{1 - n}

always produces a linear ODE in

v

True, and that is the entire point: the chain rule turns the awkward

y^{- n} y^{'}

into

\frac{1}{1 - n} v^{'}

, leaving only first powers of

v

— no

v^{2}

, no

v \cdot v^{'}

For

n = 2

the linearizing substitution is

v = y^{2}

False. It is

v = y^{1 - n} = y^{- 1}

, i.e.

v = 1/ y

. Confusing

y^{n}

with

y^{1 - n}

is the single most common wrong start.

Q (x) \equiv 0

, the equation is no longer worth calling Bernoulli.

True in spirit: with

Q = 0

you get

y^{'} + P y = 0

, which is linear (and separable) already — no need for the

y^{- n}

division.

The trivial function

y \equiv 0

solves

y^{'} + P y = Q y^{n}

whenever

n > 0

True. Both sides become

0

, so

y \equiv 0

is a genuine solution — one you can lose if you carelessly divide by

y^{n}

The substitution works equally well for

n = 1

False. Then

1 - n = 0

and

v = y^{0} = 1

is a constant carrying no information; the equation was already linear, so no substitution is needed.

Once you find

v (x)

, the problem is finished.

False. You still must back-substitute

v = y^{1 - n}

to recover

y

; an answer left in

v

is incomplete.

The integrating factor for the transformed equation is

μ = e^{\int P d x}

False. It is

μ = e^{\int (1 - n) P d x}

— the factor

(1 - n)

rides along with

P

, so forgetting it gives the wrong

μ

Spot the error

Student writes: "

v = y^{- 1}

, so

v^{'} = y^{- 2} y^{'}

Error: differentiating

y^{- 1}

gives

v^{'} = - y^{- 2} y^{'}

(the minus sign from the power rule). Dropping that sign flips the sign of every later term.

Student divides

y^{'} + P y = Q y^{n}

y^{n}

and writes

y^{- n} y^{'} + P y^{1 - n} = Q y^{- n}

Error: the middle term

P y

becomes

P y^{1 - n}

(correct), but the right-hand side

Q y^{n}

divided by

y^{n}

is just

Q

, not

Q y^{- n}

. The whole reason we divide is to strip the power off the right side.

Student substitutes

v = y^{1 - n}

and writes the linear ODE as

v^{'} + P v = Q

Error: the

(1 - n)

from the chain rule multiplies both sides. The correct form is

v^{'} + (1 - n) P v = (1 - n) Q

Student solves a Bernoulli with

n = 3

and reports the general solution as

v = \dots

with a single arbitrary constant, then says "

y = v

Two errors: the constant is fine, but the final step is

v = y^{1 - n} = y^{- 2}

, so

y = \pm v^{- 1/2}

— you must invert the power and keep the

\pm

where relevant.

Student claims: "Dividing by

y^{n}

needs no comment since

y

is never zero."

Error:

y

can be zero, and if

n > 0

then

y \equiv 0

is a valid solution. You must set it aside before dividing and report it separately.

Student treats

P

as a function of

y

while building the integrating factor.

Error: in a Bernoulli equation

P

and

Q

are functions of

x

only. If

P

genuinely depended on

y

, the whole linear-ODE machinery would not apply.

Why questions

Why do we divide by

y^{n}

before substituting, rather than substituting directly?

Dividing exposes the exact pattern

y^{- n} y^{'}

that

\frac{d}{d x} y^{1 - n}

produces; without it the chain rule wouldn't line up and the

y

-powers wouldn't cancel.

Why is

v = y^{1 - n}

the "right" choice and not, say,

v = y^{n}

Because

\frac{d v}{d x} = (1 - n) y^{- n} y^{'}

manufactures precisely the

y^{- n} y^{'}

term left after dividing — it is the tailor-made antidote to the

y^{n}

nuisance.

Why does the factor

(1 - n)

appear on both

P

and

Q

The chain rule gives

y^{- n} y^{'} = \frac{1}{1 - n} v^{'}

; multiplying the whole equation by

(1 - n)

to clear that fraction scales every term, so both

P

and

Q

pick up the

(1 - n)

Why can a negative

n

(like

n = - 2

) still be handled the same way?

The derivation never assumed

n > 0

;

v = y^{1 - n} = y^{3}

works fine. Only the "lost

y \equiv 0

" caveat changes, since for

n < 0

the term

y^{n}

blows up at

y = 0

rather than vanishing.

Why does a Bernoulli equation reduce to a linear ODE rather than to a separable one?

The transformed equation still has an explicit

P (x) v

term coupling

v

and

x

, so its solution route is the integrating factor — separability only appears in the special case

P \equiv 0

Why is a Riccati equation harder than a Bernoulli, even though both are nonlinear?

A Riccati equation has a

y^{2}

term and a stand-alone term with no

y

; that extra term blocks the clean

y^{- n}

cancellation, so you first need one known particular solution to knock it down to Bernoulli form.

Edge cases

What happens to the substitution as

n \to 1

The factor

\frac{1}{1 - n}

blows up — a signal the method degenerates. At

n = 1

the equation

y^{'} + (P - Q) y = 0

is already linear, so no substitution is needed.

What happens at

n = 0

You get

y^{'} + P y = Q

, which is the standard linear form — Bernoulli's machinery would just multiply through by

1

, doing nothing useful.

For

n = \frac{1}{2}

, is

y \equiv 0

still a solution to worry about?

Yes:

n = \frac{1}{2} > 0

, so

y \equiv 0

satisfies

y^{'} - y = y^{1/2}

(both sides zero) and must be noted; it's easy to lose when dividing by

y^{1/2}

For

n = - 1

, does dividing by

y^{n}

risk losing

y \equiv 0

No: here

y^{n} = y^{- 1}

, and

y \equiv 0

does not solve the equation (the

Q y^{- 1}

term is undefined at

y = 0

), so there is no trivial solution to preserve.

If both

P \equiv 0

and

n = 2

, what does the equation become and how do you solve it?

It reduces to

y^{'} = Q y^{2}

, which is directly separable (

y^{- 2} d y = Q d x

) — the Bernoulli substitution still works but is overkill.

For a fractional

n

n = \frac{1}{3}

, why must you be careful stating the domain of

y

v = y^{2/3}

and recovering

y = v^{3/2}

can force sign or domain restrictions (roots of negatives), so the solution may only be valid where

y \geq 0

— always check the recovered

y

is real and consistent.

If the recovered relation is implicit, e.g.

y^{- 3} = 2 x^{6} + C x^{3}

, is that an acceptable final answer?

Yes: an implicit solution in

y

is complete and standard for ODEs; solving explicitly for

y

(here

y = (2 x^{6} + C x^{3})^{- 1/3}

) is optional unless the problem demands it.

Recall One-sentence summary

Bernoulli's method is fragile in exactly four spots — the $(1 - n)$ factor, the sign in $v^{'}$ , the lost $y \equiv 0$ , and the back-substitution — and every trap on this page is one of those four wearing a disguise.

Connections

Bernoulli equations — substitution (index 4.6.5) (parent)
Linear First-Order ODEs — Integrating Factor
Separable Equations
Substitution Methods in ODEs
Riccati Equations