First-order linear ODEs — integrating factor method (derivation): Question bank

True or false — justify

Every equation containing

\frac{d y}{d x}

and

y

is a linear ODE.

False. Linearity requires

y

and

\frac{d y}{d x}

to each appear only to the first power, never multiplied together or fed into a function; e.g.

y \frac{d y}{d x}

sin y

breaks it.

\frac{d y}{d x} + x y^{2} = x

can be solved by the integrating factor method as written.

False. The

y^{2}

term makes it nonlinear, so the product-rule collapse fails. It is a Bernoulli equation (

n = 2

) and needs the substitution

v = y^{1 - n}

first.

The integrating factor

μ = e^{\int P d x}

is always positive.

True. It's an exponential, and

e^{(anything real)} > 0

; this is why we can freely divide the final equation by

μ

without worrying about dividing by zero.

You must add a

+ C

every time you integrate, including when finding

μ

False. When finding

μ

we need only one working factor; an extra constant becomes a multiplicative

e^{C}

that cancels top and bottom. Keep

+ C

only at the final integration.

P (x) = 0

, the integrating factor method still applies.

True, and it degenerates gracefully:

μ = e^{\int 0 d x} = e^{0} = 1

, so the equation is already

\frac{d}{d x} (y) = Q

— just a plain integration, which is the separable baseline.

The method requires

Q (x)

to be integrable in closed form.

False. The general solution

y = \frac{1}{μ} [\int μ Q d x + C]

is always valid; if

\int μ Q d x

has no elementary form, we simply leave it as an integral. The method never fails, the antiderivative just may not simplify.

Multiplying by

μ

changes the solution set of the ODE.

False (as long as

μ \neq = 0

, which is guaranteed). Multiplying both sides by a never-zero function gives an equivalent equation, so no solutions are created or lost.

A first-order linear ODE always has exactly one solution.

False. The general solution carries one free constant

C

, giving a family of solutions; you get a single one only after imposing an initial condition like

y (x_{0}) = y_{0}

Spot the error

"For

x \frac{d y}{d x} + y = cos x

, I take

P = 1

μ = e^{x}

Error: the coefficient of

\frac{d y}{d x}

x

, not

1

. You must standardize first by dividing by

x

, giving

P = \frac{1}{x}

and

μ = x

, not

e^{x}

"After reaching

μ y = \int μ Q d x + C

, my answer is

y = \int μ Q d x + C

Error: you forgot to divide by $μ$ . The left side is

μ y

, so

y = \frac{1}{μ} [\int μ Q d x + C]

— and the

C

must be divided by

μ

too, so it is not a bare additive constant in

y

μ = e^{\int \frac{1}{x} d x} = e^{l n ∣ x ∣ + C}

, so I keep the

+ C

Error: dropping the constant is the whole point.

e^{l n ∣ x ∣ + C} = e^{C} ∣ x ∣

; the

e^{C}

factor cancels when it multiplies both sides, so we take the simplest

μ = ∣ x ∣

(usually

x

on the working domain).

μ = e^{\int 3 d x} = e^{3}

since

\int 3 d x = 3

Error:

\int 3 d x = 3 x

, not

3

. The integrating factor is

μ = e^{3 x}

; forgetting the

x

turns a function into a useless constant that won't collapse the left side.

"The left side becomes

\frac{d}{d x} (μ y)

, so I can also write it as

μ \frac{d y}{d x}

Error:

\frac{d}{d x} (μ y) = μ \frac{d y}{d x} + \frac{d μ}{d x} y

by the Product Rule — there are two terms. Dropping the

\frac{d μ}{d x} y

piece throws away exactly the term that made the collapse work.

"I applied the initial condition to

μ y = \int μ Q d x + C

before solving for

y

Not necessarily wrong, but risky and premature. It's cleanest to apply

y (x_{0}) = y_{0}

to the final

y (x)

; applying it early forces you to evaluate

μ (x_{0})

correctly and invites arithmetic slips.

Why questions

Why do we multiply by

μ

instead of adding or substituting something?

Because multiplication is what lets the Product Rule fire:

\frac{d}{d x} (μ y)

is a product's derivative, so only a multiplicative factor can reshape

μ \frac{d y}{d x} + μ P y

into a single derivative.

Why is

μ = e^{\int P d x}

and not some other function?

Because the collapse demands

\frac{d μ}{d x} = μ P

; that condition is a separable ODE whose solution is forced to be

μ = e^{\int P d x}

. It's not a guess — it's the unique factor the condition allows.

Why must the coefficient of

\frac{d y}{d x}

be exactly

1

before we start?

Because the term meant to equal

\frac{d μ}{d x} y

μ P y

, and

P

is read off assuming the

\frac{d y}{d x}

coefficient is

1

. If it's

a (x)

instead, the true

P

\frac{( coeff of y )}{a ( x )}

, so skipping normalization uses the wrong

P

Why does finding

μ

turn into a separable problem?

Because the collapse condition

\frac{d μ}{d x} = μ P (x)

has

μ

on one side and

x

on the other after rearranging to

\frac{d μ}{μ} = P d x

— the textbook shape of a separable equation.

Why can we ignore absolute-value subtleties in

e^{l n ∣ x ∣} = ∣ x ∣

and just use

x

Because we work on an interval where

x

keeps one sign (the domain of the solution); on that interval

∣ x ∣ = \pm x

, and the sign is absorbed by the multiplicative constant we already agreed to drop.

Why does the same

C

that appears once in the integration end up dividing by

μ

in the answer?

Because

C

is introduced inside the bracket at

μ y = \int μ Q d x + C

, and the very next step divides the entire right side by

μ

, so

C

rides along as

\frac{C}{μ ( x )}

— often a genuine function of

x

, not a constant.

Why is the constant-coefficient case (Linear Constant-Coefficient ODEs) just a special instance?

When

P

is a constant

k

\int P d x = k x

and

μ = e^{k x}

— the method runs identically; nothing about the derivation assumed

P

varied.

Edge cases

What happens to the method if

P (x)

has a singularity, say

P = \frac{1}{x}

, at

x = 0

The integrating factor

μ = x

vanishes at

x = 0

, so dividing by

μ

is invalid there. The solution is only guaranteed on an interval not containing the singularity (e.g.

x > 0

Q (x) \equiv 0

, does the method still work, and what do you get?

Yes. With

Q = 0

the equation is homogeneous;

\frac{d}{d x} (μ y) = 0

gives

μ y = C

, so

y = \frac{C}{μ} = C e^{- \int P d x}

— a pure decay/growth family, no particular part.

Can the integrating factor ever be zero and break the final division?

No. Since

μ = e^{(\dots)}

is an exponential, it is strictly positive wherever the exponent is finite; division by

μ

is always legal on the interval of definition.

\frac{d y}{d x} + P (x) y = Q (x)

with

P, Q

discontinuous still solvable in one piece?

Generally no. If

P

Q

jumps, the solution is guaranteed continuous only on each interval where they are continuous; you solve piece-by-piece and match values at the joins.

What if the equation is already an exact derivative before multiplying, i.e.

μ

turns out to be

1

Then

P = 0

and the left side is already

\frac{d}{d x} (y)

; the integrating factor is

μ = 1

and the method reduces to direct integration — consistent with the Exact ODEs viewpoint where the equation was exact from the start.

Does swapping the roles so

x

is the unknown,

y

the variable, ever help?

Yes — if an equation is nonlinear in

y

but linear when viewed as

\frac{d x}{d y} + P (y) x = Q (y)

, you apply the identical method with

x

and

y

interchanged; linearity is a property of how the unknown appears, not of its name.

Connections

Separable ODEs — the sub-problem that yields $μ$ .
Product Rule — the mechanism behind every collapse claim above.
Exact ODEs — the $μ = 1$ / already-exact edge case.
Bernoulli Equations — the " $y^{2}$ isn't linear" traps point here.
Linear Constant-Coefficient ODEs — the constant- $P$ special case.