Non-homogeneous — variation of parameters: Question bank

Vocabulary reused here, so it's never a surprise:

$y_{1}, y_{2}$ — two independent solutions of the homogeneous equation $y^{''} + p y^{'} + q y = 0$ .
$g (x)$ — the right-hand side of the standard-form equation (leading coefficient $1$ ).
$W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}$ — the Wronskian, the "are these two genuinely different?" scale.
$y_{p} = u_{1} y_{1} + u_{2} y_{2}$ — the particular solution built by letting the constants vary.

True or false — justify

VoP requires you to already know

y_{1}

and

y_{2}

True. The ansatz

y_{p} = u_{1} y_{1} + u_{2} y_{2}

literally builds on them; with no homogeneous solutions in hand there is nothing to "vary". Use Reduction of Order to find a second

y_{2}

if only one is known.

VoP only works when

g (x)

is a polynomial, exponential, sine or cosine.

False. That restriction belongs to Method of Undetermined Coefficients. VoP handles any continuous

g

—

sec x

ln x

1/ x

— because it integrates

g

rather than guessing its form.

W = 0

somewhere on the interval, the formula still applies safely at that point.

False.

W

sits in the denominator, so

W = 0

makes

u_{1}^{'}, u_{2}^{'}

blow up; it also signals

y_{1}, y_{2}

are linearly dependent, so they aren't a valid basis there.

The condition

u_{1}^{'} y_{1} + u_{2}^{'} y_{2} = 0

is a law of nature we must obey.

False. It is a free choice we impose to spend our one extra degree of freedom; we picked it precisely so second derivatives of

u_{1}, u_{2}

never appear.

Swapping the labels

y_{1} \leftrightarrow y_{2}

changes the final

y_{p}

False. Swapping flips the sign of

W

and swaps which solution carries the minus, so the two sign changes cancel and

y_{p}

is identical. You just must not mix conventions mid-problem.

You may drop the

+ C

when integrating

u_{1}^{'}

and

u_{2}^{'}

True. Any constant

C_{1}

from

u_{1}

contributes

C_{1} y_{1}

, and

C_{2}

from

u_{2}

contributes

C_{2} y_{2}

— both are already inside the complementary function

c_{1} y_{1} + c_{2} y_{2}

, so they are redundant.

For

a y^{''} + b y^{'} + cy = g

with constant

a \neq = 1

, you can plug

g

straight into the formula.

False. The derivation assumed leading coefficient

1

. You must divide by

a

first; the true right-hand side is

g / a

, not

g

Spot the error

W = y_{1}^{'} y_{2} - y_{1} y_{2}^{'}

, that's the Wronskian."

Signs are backwards. The correct order is

W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}

(top-left times bottom-right, minus the cross). The stated version is

- W

, which flips downstream signs.

u_{1}^{'} = + \frac{y _{2} g}{W}

and

u_{2}^{'} = + \frac{y _{1} g}{W}

— symmetric, so both plus."

The minus is missing on

u_{1}^{'}

. Cramer's rule gives

u_{1}^{'} = - y_{2} g / W

; forcing symmetry ignores that the

(0, g)

column enters the two determinants with opposite orientations.

"For

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

we have

g = x^{3}

Not in standard form. Divide by

x^{2}

first, giving

g = x

. Reading

g

off the un-normalized equation is the single most common VoP mistake.

"Since

u_{1}

and

u_{2}

are constants that we let vary, they're still constants."

Contradiction in terms — the whole method replaces constants with functions

u_{1} (x), u_{2} (x)

. If they stayed constant you'd just have the homogeneous solution and couldn't absorb

g

y_{p} = u_{1} y_{1} + u_{2} y_{2}

, so

y_{p}^{'} = u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'}

Wrong differentiation — this drops the product-rule terms

u_{1}^{'} y_{1} + u_{2}^{'} y_{2}

and the

u_{1} y_{1}^{'} + u_{2} y_{2}^{'}

terms. The correct first derivative keeps everything, then Condition 1 sets the extra bracket to zero, which is not the same as it never existing.

"The two homogeneous terms

u_{1} (y_{1}^{''} + p y_{1}^{'} + q y_{1})

vanish because

u_{1}

is small."

They vanish because the bracket is zero —

y_{1}

solves the homogeneous equation — regardless of the size of

u_{1}

. Nothing here depends on

u_{1}

being small.

W = e^{3 x}

for

y_{1} = e^{x}, y_{2} = e^{2 x}

, and it's never zero, so any

g

gives a nice closed form."

Non-vanishing

W

guarantees the setup is valid, but

u_{1} = \int - y_{2} g / W d x

may still be a non-elementary integral. Validity of the method

\neq =

existence of an elementary antiderivative.

Why questions

Why do we impose exactly one extra condition, not zero or two?

We have two unknown functions but only one equation (the ODE), so one degree of freedom is genuinely free — exactly one condition uses it up without over-constraining the system.

Why choose

u_{1}^{'} y_{1} + u_{2}^{'} y_{2} = 0

specifically, of all possible conditions?

It cancels precisely the terms whose next derivative would produce

u_{1}^{''}, u_{2}^{''}

, keeping the surviving system first-order and linear in

u_{1}^{'}, u_{2}^{'}

— algebraically the simplest possible outcome.

Why must the equation be in standard form before reading

g

The derivation grouped terms as

y^{''} + p y^{'} + q y = g

; the vanishing brackets

y_{i}^{''} + p y_{i}^{'} + q y_{i}

only equal zero when

p, q

are the normalized coefficients, which forces

g

to be the RHS after dividing by the leading coefficient.

Why does the Wronskian appear in the denominator rather than, say, the sum

y_{1} + y_{2}

It is the determinant of the coefficient matrix in the linear system for

u_{1}^{'}, u_{2}^{'}

, so solving by Cramer's Rule naturally divides by that determinant, which is exactly

W

Why is VoP called "variation of parameters"?

The parameters are the constants

c_{1}, c_{2}

of the general homogeneous solution; we let them vary — become functions of

x

— hence "variation of parameters".

Why does VoP always succeed in principle where undetermined coefficients can fail?

VoP produces

y_{p}

by integration of known quantities, and a continuous function always has an antiderivative; undetermined coefficients instead guesses a form, which only exists for a special catalogue of

g

Why is a non-zero Wronskian a guarantee that the method works?

W \neq = 0

certifies

y_{1}, y_{2}

are linearly independent, so they form a genuine basis and the

2 \times 2

system for

u_{1}^{'}, u_{2}^{'}

has a unique solution at every point.

Edge cases

What if

g (x) \equiv 0

Then

u_{1}^{'} = u_{2}^{'} = 0

, so

u_{1}, u_{2}

are constants and

y_{p}

collapses back into the complementary function — VoP correctly reports "no extra particular part needed" for a homogeneous equation.

What if

y_{1}

and

y_{2}

are accidentally the same solution (or proportional)?

Then

W \equiv 0

everywhere, the formula divides by zero, and the method breaks — a red flag that you don't actually have two independent solutions of the Second-order linear homogeneous ODE.

What if the repeated-root case gives

y_{1} = e^{r x}, y_{2} = x e^{r x}

Perfectly fine — these are independent (

W = e^{2 r x} \neq = 0

), so VoP applies unchanged; the repeated root only affected how you found

y_{1}, y_{2}

, not the method.

What happens at a point where

g (x)

has a jump or singularity (e.g.

g = 1/ x

x = 0

The integrals for

u_{1}, u_{2}

may diverge or the solution may only be valid on an interval avoiding that point; VoP is guaranteed only where

p, q, g

are continuous and

W \neq = 0

y_{p}

comes out containing a term proportional to

y_{1}

y_{2}

, is that a mistake?

No — such a term is a legitimate homogeneous piece and can be absorbed into

c_{1} y_{1} + c_{2} y_{2}

; the general solution is unaffected, as seen when

- e^{x}

merged into

c_{1} e^{x}

in Example 2.

Can VoP be extended to higher-order or non-constant-coefficient equations?

Yes — for order

n

you impose

n - 1

zero-conditions and solve an

n \times n

Cramer system with the higher-order Wronskian; non-constant coefficients are fine as long as you first divide to standard form. This viewpoint also underlies the Green's function.

Recall One-line summary of the traps

Standard form before $g$ ; minus only on $u_{1}^{'}$ ; $W$ in the denominator and never zero; drop the $+ C$ ; and remember Condition 1 is a choice, not a law.

Connections

Wronskian — the fairness scale; $W \neq = 0$ is the licence to proceed.
Cramer's Rule — why $W$ lands in the denominator with those signs.
Method of Undetermined Coefficients — the guess-based rival VoP outperforms on exotic $g$ .
Second-order linear homogeneous ODE — supplies $y_{1}, y_{2}$ .
Reduction of Order — how to get a second $y_{2}$ when only one is known.
Green's function — the integral-kernel generalization of this whole idea.