7.3 The Wronskian, variation of parameters and undetermined coefficients

The Wronskian is a determinant of functions and their derivatives. A non-zero value proves independence; for solutions of one linear equation it is either zero everywhere or non-zero everywhere. The same determinant solves the inhomogeneous equation (variation of parameters), and for special right-hand sides a short guess does it faster (undetermined coefficients, annihilators).

Two questions were left open in 7.2. How do we know that $n$ solutions we found are independent, so that they are really a basis? And how do we find one particular solution of the inhomogeneous equation $Ly = R$ ? One determinant answers both. Its value tests independence; its inverse gives the particular solution by a formula. For the simplest right-hand sides there is a shorter route, a guess with unknown coefficients, and we prove why the guess works.

Before you start

You need determinants and Cramer’s rule, and the solution-space picture of 7.2. Three quick checks:

Quick checkWhat is

\det\begin{bmatrix} t & t^2\\ 1 & 2t\end{bmatrix}

Quick checkSolve

p + q = 0

p + 2q = 1

. What is

q

Quick checkWhat is the dimension of the solution space of

y''' + 5y' - 2y = 0

The picture

Take two functions $f_1, f_2$ and draw the curve $\gamma(t) = (f_1(t), f_2(t))$ in the plane. Its velocity is $\gamma'(t) = (f_1'(t), f_2'(t))$ . The Wronskian $W = f_1f_2' - f_1'f_2$ is the signed area of the parallelogram spanned by $\gamma$ and $\gamma'$ (Technau drew exactly this in the lecture of April 8). Choose a pair and slide $t$ .

Three things to see. First, for $\cos t, \sin t$ the area is the same everywhere: $1$ . Second, for $t, 2t$ the curve lies on a line through the origin and every parallelogram is flat: $W = 0$ , and $f_2 = 2f_1$ is exactly the dependence. Third, for $t^2, t|t|$ every parallelogram is also flat, yet the two functions are independent: the curve goes up the line $y = x$ for $t > 0$ and up the line $y = -x$ for $t < 0$ . A zero Wronskian says the curve is tangent to its radius at each point; dependence says it stays on a single line. For arbitrary functions these differ. For solutions of one linear equation they do not, and that is the main theorem of this chapter.

Where this came from· Wroński, Muir, and Lagrange's variation of parameters

The determinant was introduced by the Polish mathematician Józef Hoene-Wroński in 1812. Thomas Muir gave it the name “Wronskian” in his 1882 treatise on determinants. The method of variation of parameters grew out of celestial mechanics. Euler sketched an early version in 1748, studying how Jupiter and Saturn disturb each other’s orbits. Lagrange developed it in memoirs of the 1770s and 1780s (accounts differ on the first use, between 1766 and 1774), for the motion of planets and for finding a comet’s orbit from three observations; he gave it its final form in papers of 1808 to 1810. The idea was to treat the constants of a solution of the undisturbed problem (the orbital elements of an ellipse) as slowly varying functions when a small force is added. That is exactly the step $y = u_1b_1 + u_2b_2$ below, with the constants $c_i$ replaced by functions $u_i(t)$ .

The idea, in words

Why· Independence test, and why the same determinant solves the equation

To test whether $f_1, \dots, f_n$ are independent, suppose $c_1f_1 + \dots + c_nf_n = 0$ . Differentiating $n - 1$ times gives $n$ equations (the function and its derivatives) that the constants $c_i$ must satisfy at every $t$ . They form a linear system whose matrix is the Wronskian matrix. If its determinant is not zero at even one $t$ , the only solution is $c = 0$ . That is the test. For the inhomogeneous equation we do the reverse: we look for a particular solution $y = u_1b_1 + \dots + u_nb_n$ , where the $b_i$ are a basis of the homogeneous solutions and the “constants” $u_i$ are now functions. Demanding that $y', y'', \dots$ look as if the $u_i$ were constant (apart from the last step) gives a system with the SAME matrix for the unknowns $u_i'$ . Since the determinant is not zero, we can solve for the $u_i'$ and integrate.

Definitions and theorems

The Wronskian and independence

Definition T

Wronskian

Let $f_1, \dots, f_n$ be functions on an interval $J$ that are $n - 1$ times differentiable. Their Wronskian matrix is

\mathcal W(t) = \begin{bmatrix} f_1(t) & f_2(t) & \cdots & f_n(t)\\ f_1'(t) & f_2'(t) & \cdots & f_n'(t)\\ \vdots & & & \vdots\\ f_1^{(n-1)}(t) & f_2^{(n-1)}(t) & \cdots & f_n^{(n-1)}(t)\end{bmatrix}

and their Wronskian is the function $W(f_1, \dots, f_n)(t) = \det \mathcal W(t)$ . For two functions, $W = f_1f_2' - f_1'f_2$ . Column $i$ is the function $f_i$ and its derivatives, one per row. (Technau’s notes number the rows by derivative order, as here.)

In one lineFor $n$ functions with $n - 1$ derivatives, the determinant of the $n \times n$ matrix whose rows are the functions and their successive derivatives: $W = \det\big(f_i^{(j)}\big)$ .

Why it existsIt is the determinant of the linear system that the constants of a dependence relation must satisfy. A non-zero value proves the system has only the zero solution.

Worked example Technau, April 8 (Example 7.22); MATH 376 worksheets 21 #7, #8, 22 #2, #5; practice final 6

Compute the Wronskian of: t and t²; cos x and sin x; e^x and x e^x; cos 2x and sin 2x; e^x and e^(−x); t, t², t³ at t = 1.

Solution.

$t, t^2$ . $W = t \cdot 2t - 1 \cdot t^2 = t^2$ . Write the two rows $(f_1, f_2)$ and $(f_1', f_2')$ and take the determinant.
$\cos x, \sin x$ . $W = \cos x\cdot\cos x - (-\sin x)\sin x = \cos^2x + \sin^2x = 1$ . The area picture: a unit square.
$e^x, xe^x$ . $W = e^x(e^x + xe^x) - e^x\cdot xe^x = e^{2x}$ . The $xe^{2x}$ terms cancel; this is the double-root pair.
$\cos2x, \sin2x$ . $W = \cos2x\cdot2\cos2x - (-2\sin2x)\sin2x = 2$ . The factor $2$ comes from the chain rule.
$e^x, e^{-x}$ . $W = e^x(-e^{-x}) - e^x\,e^{-x} = -2$ . Be careful with the sign: the answer is negative. This sign matters in variation of parameters.
$t, t^2, t^3$ at $t = 1$ . $W(t) = \det\begin{bmatrix} t & t^2 & t^3\\ 1 & 2t & 3t^2\\ 0 & 2 & 6t\end{bmatrix}$ , so $W(1) = \det\begin{bmatrix}1 & 1 & 1\\1 & 2 & 3\\0 & 2 & 6\end{bmatrix} = 1(12 - 6) - 1(6 - 0) + 1(2 - 0) = 2$ . Expand along the first row (2.7). In general $W = 2t^3$ .

Answer. $t^2$ ; $1$ ; $e^{2x}$ ; $2$ ; $-2$ ; $W(1) = 2$ .

Your answerWhat is

W(e^{-t}, e^{-2t})

t = 1

? (Three decimals.)

Theorem T MATH 376 homework 8 #1; Apostol 6.12

A non-zero Wronskian proves independence

Let $f_1, \dots, f_n$ be $n - 1$ times differentiable on $J$ . If $\det\mathcal W(t_0) \ne 0$ for some $t_0 \in J$ , then $c_1f_1 + \dots + c_nf_n = 0$ on $J$ forces $c_1 = \dots = c_n = 0$ .

In one lineIf $W(f_1, \dots, f_n)(t_0) \ne 0$ at even one point $t_0$ of $J$ , then $f_1, \dots, f_n$ are linearly independent on $J$ .

Why it is trueA dependence relation, differentiated $n - 1$ times, says that the Wronskian matrix at every point sends the coefficient vector to zero. An invertible matrix sends only the zero vector there.

ProofMATH 376 homework 8 #1

The idea: Suppose a combination vanishes everywhere. Differentiate it n-1 times. At t0 you get the matrix equation W(t0)c = 0, and the matrix is invertible.

Suppose

c_1f_1 + \dots + c_nf_n = 0

J

, for some numbers

c_i

Linear independence is the statement that a vanishing combination has all coefficients zero (2.2).

Differentiating

k

times:

c_1f_1^{(k)} + \dots + c_nf_n^{(k)} = 0

J

, for

k = 0, 1, \dots, n - 1

The derivative of the zero function is zero, and the derivative of a combination is the combination of derivatives, for each order

k \le n - 1

t = t_0

these

n

equations are the single matrix equation

\mathcal W(t_0)\,c = O

Row

k

\mathcal W(t)

times the column

c

\sum_ic_if_i^{(k)}(t)

Since

\det\mathcal W(t_0) \ne 0

c = O

\blacksquare

A square matrix with non-zero determinant is invertible (2.7); multiply by the inverse.

Look at the picture again. A non-zero area at any one time means the curve and its velocity are not parallel there, so the curve cannot lie on one line through the origin. The converse is false for arbitrary functions.

Worked example MATH 376 homework 8 #1(2), homework 9 #3, Piazza post 18

Find two functions with Wronskian zero everywhere that are linearly independent; show a case where the Wronskian is zero at a point only.

Solution.

Choose the pair. $f_1(t) = t^2$ and $f_2(t) = t|t|$ on $\mathbb R$ . For $t > 0$ , $f_2 = t^2$ , and for $t < 0$ , $f_2 = -t^2$ . Piazza post 18: when the problem asks for an example, you must show the example has the property.
They are differentiable. $f_2' = 2|t|$ everywhere (at $t = 0$ the derivative is $0$ from both sides). The Wronskian needs one derivative of each.
Compute the Wronskian. $W = f_1f_2' - f_1'f_2 = t^2\cdot2|t| - 2t\cdot t|t| = 2t^2|t| - 2t^2|t| = 0$ for every $t$ . Directly from the definition.
Show they are independent. Suppose $at^2 + bt|t| = 0$ for all $t$ . For $t > 0$ this reads $(a + b)t^2 = 0$ , so $a + b = 0$ . For $t < 0$ it reads $(a - b)t^2 = 0$ , so $a - b = 0$ . Hence $a = b = 0$ . Use two values of $t$ , one on each side of $0$ .
A zero at a single point. $W(t, t^2) = t^2$ vanishes at $t = 0$ only, yet $t, t^2$ are independent (take $t_0 = 1$ in the theorem). The theorem needs one non-zero value, not all values.
A larger family. $W(t, t^2, \dots, t^r) = Ct^r$ with $C \ne 0$ , which is zero only at $t = 0$ (homework 9 #3, proved in the problems). Independent functions can have a Wronskian that vanishes at isolated points.

Answer. $t^2$ and $t|t|$ : $W \equiv 0$ but independent. The converse of the theorem fails for general functions.

Quick check

f(t) = t^2

and

g(t) = t|t|

have Wronskian identically

0

. What does that tell you?

Where does the converse hold? When the functions are solutions of one linear equation, or when they are analytic. Here is the first case.

Theorem T Technau, April 8 (Lemma 7.23); Apostol 6.12 (Theorem 6.12)

Wronskian of solutions of a linear equation

Let $b_1, \dots, b_n$ be solutions of the homogeneous linear equation $Ly = 0$ of order $n$ on $J$ (continuous coefficients). Then: (i) $b_1, \dots, b_n$ are linearly independent; (ii) $W(b_1, \dots, b_n)(t) \ne 0$ for every $t \in J$ ; (iii) $W(b_1, \dots, b_n)(t_0) \ne 0$ for some $t_0 \in J$ . These are equivalent. In particular a Wronskian of solutions has no isolated zeros.

In one lineFor $n$ solutions of $Ly = 0$ the following are equivalent: they are linearly independent; $W(t) \ne 0$ for every $t$ ; $W(t_0) \ne 0$ for some $t_0$ . So $W$ is either zero everywhere or non-zero everywhere.

Why it is trueThe map from a solution to its starting numbers at $t$ is an isomorphism for each $t$ , and the columns of the Wronskian matrix at $t$ are exactly the starting numbers of the $n$ solutions.

Why· The picture behind the proof

In the picture, a pair of solutions of one equation such as $y'' + y = 0$ gives a curve $\gamma(t) = (b_1, b_2)$ that is never tangent to its own radius. The reason is the starting-number isomorphism of 7.2: the column $(b_i(t), b_i'(t), \dots)$ is the starting vector of $b_i$ at time $t$ , and two independent solutions have independent starting vectors at EVERY time, not only at one. The pair $t, t^2$ cannot be solutions of a linear equation with continuous coefficients near $0$ : its Wronskian $t^2$ has an isolated zero. (They solve $t^2y'' - 2ty' + 2y = 0$ , whose leading coefficient $t^2$ vanishes at $0$ , so the equation is not of the form the theorem needs.)

ProofTechnau, April 8; Apostol 6.12

The idea: The columns of the Wronskian matrix at t are the starting numbers of the b_i at time t. The starting-number map is an isomorphism for each fixed t, so independent solutions have independent columns, at every t.

For every

t \in J

\mu_t

is an isomorphism.

Fix

t \in J

and define

\mu_t : \ker L \to \mathbb R^n

\mu_t(y) = (y(t), y'(t), \dots, y^{(n-1)}(t))

. The proof of the dimension theorem in 7.2 works for every choice of base point.

\mathcal W(t) = \big[\mu_t(b_1)\ \cdots\ \mu_t(b_n)\big]

The

i

-th column of

\mathcal W(t)

(b_i(t), b_i'(t), \dots, b_i^{(n-1)}(t)) = \mu_t(b_i)

(i)

\Rightarrow

(ii): if the

b_i

are independent, the columns of

\mathcal W(t)

are independent for every

t

, so

\det\mathcal W(t) \ne 0

An isomorphism sends independent vectors to independent vectors, and a square matrix has non-zero determinant exactly when its columns are independent.

(ii)

\Rightarrow

(iii).

Trivial: if it holds for all

t

it holds for one.

(iii)

\Rightarrow

(i).

\blacksquare

The previous theorem (non-zero Wronskian at a point proves independence).

Theorem A Apostol 6.12 (Theorem 6.12); Abel

Abel's formula

Let $b_1, \dots, b_n$ solve $Ly = 0$ with $L = D^n + a_{n-1}D^{n-1} + \dots + a_0$ on $J$ . Then $W' + a_{n-1}W = 0$ on $J$ , and for $t_0 \in J$ : $W(t) = W(t_0)\exp\!\Big(-\displaystyle\int_{t_0}^{t}a_{n-1}(s)\,ds\Big)$ . If $a_{n-1} \equiv 0$ , $W$ is constant.

In one lineIf $b_1, \dots, b_n$ solve $y^{(n)} + a_{n-1}y^{(n-1)} + \dots = 0$ , their Wronskian satisfies $W' = -a_{n-1}W$ , so $W(t) = W(t_0)\exp\!\big(-\int_{t_0}^{t}a_{n-1}\big)$ .

Why it is trueOnly the last row of the Wronskian matrix is not already a derivative of the row above it. Differentiating the determinant differentiates that row, and the equation turns the new row into $-a_{n-1}$ times the old last row plus lower rows.

ProofApostol 6.12

The idea: Differentiate the determinant row by row: all terms but one have two equal rows. Replace the last row's derivative by the equation, and subtract multiples of the lower rows.

W' = \det

of the matrix with the last row replaced by

(b_1^{(n)}, \dots, b_n^{(n)})

; the other terms vanish.

The derivative of a determinant is the sum of determinants in which one row at a time is differentiated (product rule for the permutation formula, 2.7). Differentiating row

k < n - 1

gives the row

b^{(k+1)}

, which is already the next row, so the determinant has two equal rows and is

0

So the new last row is

-a_{n-1}\cdot(\text{old last row}) - a_{n-2}\cdot(\text{row } n-1) - \dots - a_0\cdot(\text{row }1)

Each

b_i

satisfies

b_i^{(n)} = -a_{n-1}b_i^{(n-1)} - \dots - a_0b_i

W' = -a_{n-1}W

Adding a multiple of one row to another does not change a determinant. The terms with the rows

1, \dots, n - 1

are removed, leaving

-a_{n-1}

times the last row.

W(t) = W(t_0)\exp\big(-\int_{t_0}^{t}a_{n-1}\big)

\blacksquare

The linear equation

W' = -a_{n-1}W

is solved by the integrating factor (7.2):

\big(We^{\int a_{n-1}}\big)' = 0

Abel’s formula gives the equivalence theorem a second proof, and a number. For $y'' + 3y' + 2y = 0$ the solutions $e^{-t}$ and $e^{-2t}$ have $W = -e^{-3t}$ , and Abel’s formula with $a_1 = 3$ says $W(t) = W(0)e^{-3t}$ with $W(0) = -1$ . The two agree.

Fact T MATH 376 homework 8 #1(3)

Wronskian of exponentials

Let $r_1, \dots, r_n$ be distinct. Then $W(e^{r_1t}, \dots, e^{r_nt})(t) = e^{(r_1 + \dots + r_n)t}\det\big(r_i^{\,j}\big)_{i = 1..n,\ j = 0..n-1}$ , and the determinant is non-zero.

In one lineFor distinct numbers $r_1, \dots, r_n$ , $W(e^{r_1t}, \dots, e^{r_nt}) = e^{(r_1 + \dots + r_n)t}\cdot V$ with $V = \det(r_i^{\,j})$ non-zero, so the exponentials are independent.

Why it is trueFactor $e^{r_it}$ out of each column. What remains is a Vandermonde matrix, which is invertible when the $r_i$ are distinct.

ProofMATH 376 homework 8 #1(3)

The idea: Differentiating e^{rt} j times gives r^j e^{rt}, so column i of the Wronskian matrix is e^{r_i t} times (1, r_i, r_i^2, ...). Then show the Vandermonde matrix has independent rows by a polynomial with too many roots.

Entry

(j, i)

\mathcal W

r_i^je^{r_it}

for

j = 0, \dots, n - 1

(e^{r_it})^{(j)} = r_i^je^{r_it}

W = e^{(r_1 + \dots + r_n)t}\det\big(r_i^{\,j}\big)

A common factor of a whole column comes out of a determinant (2.7).

Suppose

\det V = 0

Suppose the Vandermonde matrix

V

is singular. Then some non-zero

c = (c_0, \dots, c_{n-1})

has

\sum_jc_jr_i^j = 0

for every

i

, that is,

p(x) = \sum_jc_jx^j

has the

n

distinct roots

r_1, \dots, r_n

That is impossible, so

\det V \ne 0

\blacksquare

A non-zero polynomial of degree at most

n - 1

has at most

n - 1

roots (0.6). So

p = 0

and

c = O

, contradicting

c \ne O

Theorem T MATH 376 final 7

Wronskian test for power series

Let $f, g : (-1, 1) \to \mathbb R$ be given by power series with radius of convergence $1$ (1.6). Then $f, g$ are linearly independent if and only if their Wronskian $W = fg' - f'g$ is not identically zero on $(-1, 1)$ . You may use the fact: a non-zero power series $h$ has only finitely many zeros in $(-1 + \varepsilon, 1 - \varepsilon)$ , for each $\varepsilon > 0$ .

In one lineIf $f$ and $g$ are power series with radius of convergence $1$ , then $f, g$ are linearly independent on $(-1, 1)$ if and only if $W(f, g)$ is not identically zero.

Why it is trueDependent functions always have $W \equiv 0$ (easy direction). For power series the converse holds because a power series that vanishes on an interval vanishes everywhere.

ProofMATH 376 final 7

The idea: Prove the contrapositive of each direction: dependent implies W is zero (differentiate the relation), and W zero implies dependent (divide by g where g is not zero, get f over g constant on an interval, then the identity theorem spreads the relation to all of (-1, 1)).

(

\Rightarrow

, contrapositive) Suppose

f, g

are dependent:

af + bg = 0

(-1, 1)

with

(a, b) \ne (0, 0)

f, g

are dependent there are

a, b

, not both

0

, with

af + bg = 0

. This is the definition of linear dependence.

Then also

af' + bg' = 0

Power series can be differentiated term by term inside the radius (1.6), so

f, g

are differentiable; differentiate the relation.

W(t) = fg' - f'g = 0

for all

t

These two equations say

\begin{bmatrix}f & g\\f' & g'\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} = O

at every

t

, with a non-zero vector. So the matrix is singular at every

t

If one of them is the zero function, they are dependent. So assume neither is.

(

\Leftarrow

, contrapositive) Suppose

W \equiv 0

. If

f \equiv 0

then

1\cdot f + 0\cdot g = 0

is a non-trivial relation, and the same if

g \equiv 0

Fix

\varepsilon > 0

. There is a non-empty open interval

I \subseteq (-1 + \varepsilon, 1 - \varepsilon)

on which

g \ne 0

g \not\equiv 0

is a non-zero power series, so it has finitely many zeros in

(-1 + \varepsilon, 1 - \varepsilon)

(the fact you may use). A finite set has gaps: some open interval

I

inside has no zero of

g

I

\left(\dfrac fg\right)' = \dfrac{f'g - fg'}{g^2} = -\dfrac{W}{g^2} = 0

Quotient rule, then

W = fg' - f'g = 0

f/g = c

I

for some constant

c

, that is,

f - cg = 0

I

A function with zero derivative on an interval is constant (1.4).

h = f - cg

is a power series vanishing on

I

, hence

h \equiv 0

(-1, 1)

The difference of two power series centred at

0

is a power series with radius at least

1

. It has infinitely many zeros in

(-1 + \varepsilon, 1 - \varepsilon)

(all points of

I

), so by the fact it is the zero series.

f, g

are dependent.

\blacksquare

f = cg

on all of

(-1, 1)

is the non-trivial relation

1\cdot f - c\cdot g = 0

Proof workshopRebuild the proof of the Wronskian test for power seriesMATH 376 final 7

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Watch out· What the grader wanted on the final's Problem 7

On the 376 final (Problem 7, 5 points) you wrote the definition of independence, “Let’s say otherwise,” and stopped. The rubric gave $1$ of $5$ . It would have given $2$ more points for the easy direction alone (dependent implies $W \equiv 0$ ) and $3$ for the hard direction. The lesson for any “if and only if” proof: write the easy direction completely first, in the first five minutes. Then the structure of the hard direction is the three sentences above (divide where $g \ne 0$ , the quotient is constant, spread by the identity theorem).

Your answerPractice final 6: what is the Wronskian of

t, t^2, t^3

t = 1

Variation of parameters

Now the particular solution. Take a second-order equation $y'' + a_1(t)y' + a_0(t)y = R(t)$ whose homogeneous solutions $b_1, b_2$ we know. The general solution of the homogeneous equation is $c_1b_1 + c_2b_2$ with constants. Lagrange’s idea: let the constants vary.

Definition T

Variation of parameters

Variation of parameters finds a particular solution of $Ly = R$ in the form $y_p = \sum_iu_ib_i$ with $u_i$ functions of $t$ , subject to the conditions $\sum_iu_i'b_i^{(j)} = 0$ for $j = 0, \dots, n - 2$ (and the equation itself gives the last condition $\sum_iu_i'b_i^{(n-1)} = R$ ).

In one lineLook for a particular solution $y = u_1(t)b_1(t) + \dots + u_n(t)b_n(t)$ , where the $b_i$ are a basis of the homogeneous solutions and the coefficients $u_i$ are unknown functions; impose $n - 1$ extra conditions that make the derivatives clean.

Why it existsThe constants of the homogeneous solution become functions. We get $n$ unknown functions and only one equation, so we are free to choose $n - 1$ more equations, and we choose them to simplify the derivatives.

Here is the derivation for $n = 2$ , which Technau did on April 8 and worksheet 22 #3 asks for.

Worked example Technau, April 8; MATH 376 worksheet 22 #3

Derive the formulas u1' = −b2 R / W and u2' = b1 R / W for y'' + a1 y' + a0 y = R.

Solution.

Set the form. $y = u_1b_1 + u_2b_2$ where $b_1, b_2$ are independent solutions of $y'' + a_1y' + a_0y = 0$ . If $u_1, u_2$ were constants this would be a homogeneous solution. We let them vary.
Differentiate once. $y' = u_1'b_1 + u_1b_1' + u_2'b_2 + u_2b_2'$ . Product rule on each term.
Impose the first condition. We choose $u_1'b_1 + u_2'b_2 = 0$ , so $y' = u_1b_1' + u_2b_2'$ . We have two unknown functions and one equation, so we may impose one more condition. This one makes $y'$ look like the constants case, and avoids $u''$ terms later.
Differentiate again. $y'' = u_1'b_1' + u_1b_1'' + u_2'b_2' + u_2b_2''$ . Product rule once more.
Substitute into the equation. $y'' + a_1y' + a_0y = u_1'b_1' + u_2'b_2' + u_1(b_1'' + a_1b_1' + a_0b_1) + u_2(b_2'' + a_1b_2' + a_0b_2)$ . Collect the terms with $u_1$ and with $u_2$ separately.
Use that $b_1, b_2$ are homogeneous solutions. Both brackets are $0$ , so the equation becomes $u_1'b_1' + u_2'b_2' = R$ . This is the second condition: the only place the right-hand side enters.
Solve the system. $\begin{bmatrix}b_1 & b_2\\b_1' & b_2'\end{bmatrix}\begin{bmatrix}u_1'\\u_2'\end{bmatrix} = \begin{bmatrix}0\\R\end{bmatrix}$ . The matrix is the Wronskian matrix, with determinant $W = b_1b_2' - b_1'b_2 \ne 0$ . By Cramer’s rule, $u_1' = \dfrac{\det\begin{bmatrix}0 & b_2\\R & b_2'\end{bmatrix}}{W} = -\dfrac{b_2R}{W}$ and $u_2' = \dfrac{\det\begin{bmatrix}b_1 & 0\\b_1' & R\end{bmatrix}}{W} = \dfrac{b_1R}{W}$ . $W \ne 0$ because $b_1, b_2$ are independent solutions, by the equivalence theorem above.

Answer. $u_1' = -\dfrac{b_2R}{W}$ , $u_2' = \dfrac{b_1R}{W}$ . Integrate to get $u_1, u_2$ and set $y_p = u_1b_1 + u_2b_2$ .

The same idea works for any order, and there is a clean proof.

Theorem T Technau, April 8 and 10 (Proposition 7.24); Apostol 6.11

Variation of parameters formula

Let $L = D^n + a_{n-1}D^{n-1} + \dots + a_0$ with continuous coefficients and $R$ continuous on $J$ . Let $b_1, \dots, b_n$ be a basis of $\ker L$ and $\mathcal W(t)$ its Wronskian matrix (invertible for all $t$ ). Let $t_0 \in J$ and define $u(t) = \int_{t_0}^{t}\mathcal W(s)^{-1}\,(0, \dots, 0, R(s))^{\mathsf T}\,ds$ (coordinate by coordinate). Then $y_p(t) = \sum_{i}u_i(t)\,b_i(t)$ satisfies $Ly_p = R$ .

In one lineIf $b_1, \dots, b_n$ is a basis of $\ker L$ with Wronskian matrix $\mathcal W$ , and $u' = \mathcal W^{-1}(0, \dots, 0, R)^{\mathsf T}$ , then $y_p = u_1b_1 + \dots + u_nb_n$ solves $Ly = R$ .

Why it is trueThe equation $\mathcal Wu' = (0, \dots, 0, R)$ says the first $n - 1$ derivatives of $y_p$ look like those of a homogeneous solution, and the $n$ -th differs from it by exactly $R$ .

ProofTechnau, April 10

The idea: Differentiating y_p, the extra terms u'·b^(j) vanish for j up to n-2, so y_p^(j) = u·b^(j) for j below n. The n-th derivative picks up u'·b^(n-1) = R. Put it into L and use that each b_i solves Ly = 0.

u'(t) = \mathcal W(t)^{-1}(0, \dots, 0, R)^{\mathsf T}

, that is,

\mathcal W u' = (0, \dots, 0, R)^{\mathsf T}

Fundamental theorem of calculus, coordinate by coordinate. Multiply by

\mathcal W

\sum_iu_i'b_i^{(j)} = 0

for

j = 0, \dots, n - 2

and

\sum_iu_i'b_i^{(n-1)} = R

Row

j

\mathcal Wu'

\sum_iu_i'b_i^{(j)}

. The vector

(0, \dots, 0, R)

has

0

in rows

0, \dots, n-2

and

R

in row

n - 1

Hence

y_p^{(j)} = u \cdot b^{(j)}

for

j = 0, 1, \dots, n - 1

, by induction on

j

Product rule. Write

u \cdot b^{(j)} = \sum_iu_ib_i^{(j)}

. Then

(u \cdot b^{(j)})' = u'\cdot b^{(j)} + u \cdot b^{(j+1)}

, and the first term is

0

for

j \le n - 2

y_p^{(n)} = u' \cdot b^{(n-1)} + u\cdot b^{(n)} = R + u\cdot b^{(n)}

One more derivative: the first term is now

u'\cdot b^{(n-1)} = R

Substitute into

L

Linearity:

Ly_p = y_p^{(n)} + \sum_ja_jy_p^{(j)} = R + u\cdot\big(b^{(n)} + \sum_ja_jb^{(j)}\big)

Ly_p = R

\blacksquare

Each

b_i

solves

Lb_i = 0

, so the bracket is the zero vector.

Proof workshopRebuild: variation of parameters gives a particular solutionTechnau, April 10

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

Let $u' = \mathcal W^{-1}(0, \dots, 0, R)^{\mathsf T}$ , so that $\sum_iu_i'b_i^{(j)} = 0$ for $j \le n - 2$ and $= R$ for $j = n - 1$ .
Then $y_p^{(j)} = \sum_iu_ib_i^{(j)}$ for $j \le n - 1$ .
Then $y_p^{(n)} = R + \sum_iu_ib_i^{(n)}$ .
So $Ly_p = R + \sum_iu_i\,(Lb_i) = R$ .

Watch out· Normalise first: the leading coefficient must be 1

The formulas $u_1' = -b_2R/W$ and $u_2' = b_1R/W$ and the vector $(0, \dots, 0, R)$ assume the equation is written $y^{(n)} + \dots = R$ , with coefficient $1$ in front of $y^{(n)}$ . For $x^2y'' + xy' - y = x^3$ you divide first: the $R$ to use is $x$ .

Worked example MATH 376 worksheets 22 #4, #5, #6 and 23 #3, #4

Variation of parameters, order 2: y'' + y = tan x; y'' − y = e^x; y'' − 2y' + y = e^x/(1 + x²); y'' + 4y = sec 2x.

Solution.

First equation: basis and Wronskian. $y'' + y = 0$ has $b_1 = \cos x$ , $b_2 = \sin x$ , $W = 1$ . The roots are $\pm i$ .
The $u'$ . $u_1' = -\sin x\tan x = -\dfrac{\sin^2x}{\cos x} = -(\sec x - \cos x)$ and $u_2' = \cos x\tan x = \sin x$ . The formulas with $R = \tan x$ , $W = 1$ . And $\sin^2x = 1 - \cos^2x$ .
Integrate. $u_1 = -\ln|\sec x + \tan x| + \sin x$ and $u_2 = -\cos x$ . $\int\sec x\,dx = \ln|\sec x + \tan x|$ ; constants of integration are dropped since they only add homogeneous solutions.
Assemble. $y_p = u_1\cos x + u_2\sin x = -\cos x\ln|\sec x + \tan x| + \sin x\cos x - \cos x\sin x = -\cos x\ln|\sec x + \tan x|$ . The two $\sin x\cos x$ terms cancel.
Second equation. $y'' - y = e^x$ : $b_1 = e^x$ , $b_2 = e^{-x}$ , $W = -2$ . $u_1' = -\dfrac{e^{-x}e^x}{-2} = \dfrac12$ and $u_2' = \dfrac{e^xe^x}{-2} = -\dfrac{e^{2x}}{2}$ . So $u_1 = \frac x2$ , $u_2 = -\frac{e^{2x}}{4}$ and $y_p = \frac x2e^x - \frac14e^x$ . The term $-\frac14e^x$ is a homogeneous solution and may be dropped: $y_p = \frac x2e^x$ . Remember the sign of $W$ , and integrate $u_1'$ : it is a constant $\frac12$ , not a final answer.
Third equation. $b_1 = e^x$ , $b_2 = xe^x$ , $W = e^{2x}$ . $u_1' = -\dfrac{xe^x\cdot e^x/(1 + x^2)}{e^{2x}} = -\dfrac{x}{1 + x^2}$ and $u_2' = \dfrac{e^x\cdot e^x/(1 + x^2)}{e^{2x}} = \dfrac{1}{1 + x^2}$ . So $u_1 = -\frac12\ln(1 + x^2)$ , $u_2 = \arctan x$ , and $y_p = -\frac12e^x\ln(1 + x^2) + xe^x\arctan x$ . Exponentials cancel against $W$ here.
Fourth equation. $b_1 = \cos2x$ , $b_2 = \sin2x$ , $W = 2$ . $u_1' = -\dfrac{\sin2x\sec2x}{2} = -\dfrac{\tan2x}{2}$ and $u_2' = \dfrac{\cos2x\sec2x}{2} = \dfrac12$ . So $u_1 = \frac14\ln|\cos2x|$ , $u_2 = \frac x2$ , and $y_p = \frac14\cos2x\ln|\cos2x| + \frac x2\sin2x$ . $\int\tan2x\,dx = -\frac12\ln|\cos2x|$ .

Answer. The general solutions are $c_1\cos x + c_2\sin x - \cos x\ln|\sec x + \tan x|$ ; $c_1e^x + c_2e^{-x} + \frac x2e^x$ ; $c_1e^x + c_2xe^x - \frac12e^x\ln(1 + x^2) + xe^x\arctan x$ ; $c_1\cos2x + c_2\sin2x + \frac14\cos2x\ln|\cos2x| + \frac x2\sin2x$ .

Worked example MATH 376 final 4(b)

Finish what the final started: find the general solution of y'' + 2y' + y = t by variation of parameters.

Solution.

Solve the homogeneous equation. $t^2 + 2t + 1 = (t + 1)^2$ , a double root $-1$ . So $b_1 = e^{-t}$ , $b_2 = te^{-t}$ . The double-root basis of 7.2.
Wronskian. $W = e^{-t}(e^{-t} - te^{-t}) - (-e^{-t})(te^{-t}) = e^{-2t}$ . The $te^{-2t}$ terms cancel.
The $u'$ with $R = t$ . $u_1' = -\dfrac{te^{-t}\cdot t}{e^{-2t}} = -t^2e^t$ and $u_2' = \dfrac{e^{-t}\cdot t}{e^{-2t}} = te^t$ . This is where your final stopped: the system for $u_1', u_2'$ was never solved.
Integrate by parts. $\int t^2e^t\,dt = e^t(t^2 - 2t + 2)$ and $\int te^t\,dt = e^t(t - 1)$ . So $u_1 = -e^t(t^2 - 2t + 2)$ and $u_2 = e^t(t - 1)$ . Differentiate to check: $\big(e^t(t - 1)\big)' = te^t$ .
Assemble. $y_p = u_1b_1 + u_2b_2 = -(t^2 - 2t + 2) + t(t - 1) = t - 2$ . Everything with $e^t$ cancels against $e^{-t}$ , as it should for a polynomial right-hand side.
Write the general solution. $y = (C_1 + C_2t)e^{-t} + t - 2$ . Homogeneous plus particular.
Check. $y_p = t - 2$ has $y_p' = 1$ and $y_p'' = 0$ , so $y_p'' + 2y_p' + y_p = 0 + 2 + t - 2 = t$ . Substituting the particular solution back is the fastest check.

Answer. $y = (C_1 + C_2t)e^{-t} + t - 2$ . (The next section does this in two lines.)

Worked example MATH 376 worksheet 23 #5; Technau, April 13 (Example 7.26)

Order 3: y''' = e^x with the basis 1, x, x²; and the 376 lecture's example y''' + 3y'' + 2y' = e^(−2t).

Solution.

Set up for $y''' = e^x$ . The homogeneous solutions are $b = (1, x, x^2)$ . We need $u_1', u_2', u_3'$ with $u_1' + xu_2' + x^2u_3' = 0$ , $u_2' + 2xu_3' = 0$ , $2u_3' = e^x$ . The conditions are the Wronskian system $\mathcal Wu' = (0, 0, R)$ written out row by row.
Solve from the bottom. $u_3' = \frac{e^x}{2}$ ; $u_2' = -2xu_3' = -xe^x$ ; $u_1' = -xu_2' - x^2u_3' = x^2e^x - \frac{x^2}{2}e^x = \frac{x^2e^x}{2}$ . The system is triangular, so back-substitution works.
Integrate and assemble. $u_3 = \frac{e^x}{2}$ , $u_2 = (1 - x)e^x$ , $u_1 = \frac12e^x(x^2 - 2x + 2)$ . So $y_p = u_1 + xu_2 + x^2u_3 = e^x\big[\frac{x^2}2 - x + 1 + x - x^2 + \frac{x^2}2\big] = e^x$ . $(e^x)''' = e^x$ , so the answer could have been guessed. The method finds it without guessing.
The lecture’s example. For $y''' + 3y'' + 2y' = e^{-2t}$ the roots of $t^3 + 3t^2 + 2t = t(t + 1)(t + 2)$ are $0, -1, -2$ , so $b = (1, e^{-t}, e^{-2t})$ . The Wronskian matrix is $\begin{bmatrix}1 & e^{-t} & e^{-2t}\\0 & -e^{-t} & -2e^{-2t}\\0 & e^{-t} & 4e^{-2t}\end{bmatrix}$ with determinant $-2e^{-3t}$ . Expand along the first column; the minor is $-4e^{-3t} + 2e^{-3t}$ .
Cramer’s rule with the last-row cofactors. The cofactors of the entries $(3, 1), (3, 2), (3, 3)$ are $-e^{-3t}$ , $2e^{-2t}$ , $-e^{-t}$ , so $u' = \dfrac{R}{-2e^{-3t}}\big(-e^{-3t}, 2e^{-2t}, -e^{-t}\big) = R\big(\tfrac12, -e^{t}, \tfrac12e^{2t}\big)$ . With $R = e^{-2t}$ : $u' = \big(\frac12e^{-2t}, -e^{-t}, \frac12\big)$ . The solution of $\mathcal Wu' = (0, 0, R)$ uses the third column of $\mathcal W^{-1}$ , which is the cofactors of the third row divided by $\det\mathcal W$ .
Integrate and assemble. $u = \big(-\frac14e^{-2t}, e^{-t}, \frac t2\big)$ , so $y_p = -\frac14e^{-2t} + e^{-2t} + \frac t2e^{-2t} = \big(\frac34 + \frac t2\big)e^{-2t}$ . The part $\frac34e^{-2t}$ is a homogeneous solution; the new information is $\frac t2e^{-2t}$ .
The question of the lecture: is there a solution with $\lvert y'''\rvert \le 2$ on $t \ge 0$ ? The general solution is $y = \frac t2e^{-2t} + c_1 + c_2e^{-t} + c_3e^{-2t}$ , with $y''' = (6 - 4t)e^{-2t} - c_2e^{-t} - 8c_3e^{-2t}$ . Take $c_2 = 0$ , $c_3 = \frac34$ : $y''' = -4te^{-2t}$ , whose largest size on $t \ge 0$ is $\frac2e = 0.74$ at $t = \frac12$ . The function $\frac t2e^{-2t}$ alone has $y''' = 6$ at $t = 0$ , so the choice of the constants matters. The notes’ claim ‘max is 3/2 at t = 0’ is a computational slip; the conclusion that a solution exists is still true.

Answer. $y = c_1 + c_2x + c_3x^2 + e^x$ ; and $y_p = \big(\frac34 + \frac t2\big)e^{-2t}$ , with $\lvert y'''\rvert \le 2$ achieved for $y = \big(\frac34 + \frac t2\big)e^{-2t}$ .

Why· The Green's function: variation of parameters as a response to a kick

For $n = 2$ the formula gives $y_p(t) = \int_{t_0}^{t}\frac{b_1(s)b_2(t) - b_1(t)b_2(s)}{W(s)}\,R(s)\,ds$ . The kernel $G(t, s) = \frac{b_1(s)b_2(t) - b_1(t)b_2(s)}{W(s)}$ is the solution in $t$ that starts at time $s$ with value $0$ and speed $1$ . The picture: the forcing $R$ is a sequence of tiny kicks, each kick at time $s$ of size $R(s)\,ds$ launches its own homogeneous solution $G(t, s)$ , and $y_p$ is the sum of all of them. This is the Duhamel principle, and the same formula in 7.4 will read $e^{(t-s)A}$ .

Undetermined coefficients and the annihilator method

When the right-hand side is a polynomial, an exponential, a sine or a cosine (or sums and products of these), a guess with unknown coefficients is faster than any integral. The theory is the annihilator method of Apostol 6.14, which we prove, so that we know when the guess works.

Definition A

Annihilating operator

A constant-coefficient operator $B$ is an annihilating operator (an annihilator) of $f$ if $Bf = 0$ . The functions that have annihilators are exactly the solutions of constant-coefficient homogeneous equations: sums of $t^je^{\alpha t}\cos\beta t$ and $t^je^{\alpha t}\sin\beta t$ (7.2).

In one lineA constant-coefficient operator $B$ with $Bf = 0$ : it sends the function $f$ to zero. Example: $D - r$ annihilates $e^{rt}$ , $(D - r)^m$ annihilates $t^{m-1}e^{rt}$ .

Why it existsIf $B$ kills the right-hand side of $Ly = R$ , then applying $B$ to the equation gives a homogeneous equation $BLy = 0$ that we can solve.

Function	Annihilator
$t^{m-1}$	$D^m$
$e^{at}$	$D - a$
$t^{m-1}e^{at}$	$(D - a)^m$
$\cos\beta t$ or $\sin\beta t$	$D^2 + \beta^2$
$t^{m-1}\cos\beta t$ or $t^{m-1}\sin\beta t$	$(D^2 + \beta^2)^m$
$e^{\alpha t}\cos\beta t$ or $e^{\alpha t}\sin\beta t$	$D^2 - 2\alpha D + \alpha^2 + \beta^2$
$t^{m-1}e^{\alpha t}\cos\beta t$ or $t^{m-1}e^{\alpha t}\sin\beta t$	$\big(D^2 - 2\alpha D + \alpha^2 + \beta^2\big)^m$

(Apostol’s Table 6.1. The last operator has roots $\alpha \pm i\beta$ .)

Definition T

Undetermined coefficients

Undetermined coefficients is the method: for $Ly = R$ with $L$ of constant coefficients and $R$ annihilated by a constant-coefficient operator $B$ , every solution satisfies $BLy = 0$ . So $y_p$ lies in a space we can write down (the kernel of $BL$ , from 7.2), and we find the constants by substituting into $Ly = R$ .

In one lineGuess a particular solution of the form dictated by the right-hand side, with unknown constants, substitute, and match coefficients.

Why it existsIt replaces integration by algebra, and the annihilator method tells us exactly which form to guess.

Theorem A Apostol 6.14; standard

Shape of a particular solution

Let $L = P_L(D)$ have constant coefficients and let $R(t) = q(t)e^{rt}$ with $q$ a polynomial of degree $d$ (real or complex coefficients, $r$ complex). Let $s$ be the multiplicity of $r$ as a root of $P_L$ . Then there is a polynomial $h$ of degree at most $d$ such that $y_p = t^sh(t)e^{rt}$ satisfies $Ly_p = R$ .

In one lineIf $R = q(t)e^{rt}$ with $q$ a polynomial of degree $d$ and $r$ is a root of $P_L$ of multiplicity $s$ ( $s = 0$ if $r$ is not a root), then $Ly = R$ has a particular solution $y_p = t^s\,h(t)\,e^{rt}$ with $h$ a polynomial of degree at most $d$ .

Why it is trueMultiplying a solution by $e^{rt}$ shifts $D$ to $D + r$ , so the equation becomes a polynomial in $D$ acting on $h$ . The root of multiplicity $s$ means the lowest power of $D$ present is $D^s$ , which needs $s$ extra integrations: that is the factor $t^s$ .

ProofApostol 6.14; standard

The idea: Shift: L(e^{rt}h) = e^{rt}P_L(D + r)h. The polynomial P_L(D + r) has a zero of order s at D = 0, so it is D^s times an invertible polynomial in D on polynomials. Solve the invertible part, then integrate s times.

The identity

D(e^{rt}h) = e^{rt}(D + r)h

holds for any differentiable

h

, and by induction

D^k(e^{rt}h) = e^{rt}(D + r)^kh

Product rule:

D(e^{rt}h) = e^{rt}(Dh + rh) = e^{rt}(D + r)h

L(e^{rt}h) = e^{rt}\,P_L(D + r)h

Add up the powers with the coefficients of

P_L

Hence

P_L(D + r) = D^sQ(D)

with

Q(0) \ne 0

P_L(t + r)

is a polynomial in

t

whose root at

t = 0

has multiplicity

s

(the multiplicity of

r

P_L

). So

P_L(t + r) = t^sQ(t)

with

Q(0) \ne 0

There is a polynomial

w

of degree at most

d

with

Q(D)w = q

On polynomials of degree at most

d

Q(D) = Q(0) + Q_1D + \dots

acts by a triangular matrix with the non-zero number

Q(0)

on the diagonal (the operator

D

lowers the degree). So it is invertible there.

Choose a polynomial

H

with

D^sH = w

and no terms below

t^s

Integrate

w

s

times, choosing the constants of integration to be

0

; the result is a polynomial

H

of degree

d + s

with lowest power

t^s

, so

H = t^sh

with

\deg h \le d

y_p = e^{rt}H = t^sh(t)e^{rt}

solves

Ly = R

\blacksquare

Combine:

L(e^{rt}H) = e^{rt}D^sQ(D)H = e^{rt}Q(D)D^sH = e^{rt}Q(D)w = e^{rt}q = R

. Constant-coefficient operators commute.

When $R$ is a sine or cosine, write it as the real or imaginary part of $e^{i\omega t}$ and use the theorem with $r = i\omega$ (the equation has real coefficients, so the real part of a complex solution is a solution). In real form: if $R = \cos\omega t$ or $\sin\omega t$ and $i\omega$ is not a root, try $A\cos\omega t + B\sin\omega t$ , with both terms. If $i\omega$ is a root of multiplicity $s$ (resonance), multiply by $t^s$ .

Worked example MATH 376 final 4(b); homework 9 #2

Finish final 4(b) in two lines: y'' + 2y' + y = t; and HW9 #2: q'' + 5q' + 4q = 10 sin 2t.

Solution.

First equation: choose the guess. The right-hand side is $t = te^{0t}$ , with $r = 0$ and $d = 1$ . The number $0$ is not a root of $t^2 + 2t + 1$ , so $s = 0$ and we try $y_p = at + b$ . The theorem: a polynomial of the same degree.
Substitute. $y_p' = a$ , $y_p'' = 0$ , so $0 + 2a + at + b = t$ . Matching coefficients: $a = 1$ and $2a + b = 0$ , so $b = -2$ . Equate the coefficient of $t$ and the constant term.
Write the general solution. $y = (C_1 + C_2t)e^{-t} + t - 2$ . The same answer as variation of parameters, without a single integral.
Second equation: homogeneous part. $t^2 + 5t + 4 = (t + 1)(t + 4)$ , so $q_h = C_1e^{-t} + C_2e^{-4t}$ . Two distinct real roots.
Guess. The right-hand side is $10\sin2t$ ; $2i$ is not a root. Try $q_p = A\sin2t + B\cos2t$ . The derivative of a sine is a cosine, so both appear. Trying only $A\sin2t$ fails.
Substitute. $q_p' = 2A\cos2t - 2B\sin2t$ and $q_p'' = -4A\sin2t - 4B\cos2t$ . The $\sin2t$ coefficient of $q_p'' + 5q_p' + 4q_p$ is $-4A - 10B + 4A = -10B$ and the $\cos2t$ coefficient is $-4B + 10A + 4B = 10A$ . Match with $10\sin2t$ : $-10B = 10$ , $10A = 0$ . Collect separately the terms in $\sin2t$ and in $\cos2t$ .
Solve. $B = -1$ , $A = 0$ , so $q_p = -\cos2t$ and $q = C_1e^{-t} + C_2e^{-4t} - \cos2t$ . Check: $q_p'' + 5q_p' + 4q_p = 4\cos2t + 10\sin2t - 4\cos2t = 10\sin2t$ .

Answer. $y = (C_1 + C_2t)e^{-t} + t - 2$ and $q = C_1e^{-t} + C_2e^{-4t} - \cos2t$ .

Worked example MATH 376 final 6(b); Apostol 6.14 style

Resonance: y'' + y = cos t (final 6(b)); and y'' − 3y' + 2y = t e^t.

Solution.

Test the obvious guess. The forcing $\cos t$ has $r = i$ , and $t^2 + 1$ has the root $i$ with multiplicity $s = 1$ . So the guess $A\cos t + B\sin t$ is itself a homogeneous solution, and cannot work (it gives $0$ on the left). When the forcing matches a natural frequency, the response is not at that frequency. This is resonance.
Multiply by $t^s = t$ . Try $y_p = t(A\cos t + B\sin t)$ . Then $y_p' = A\cos t + B\sin t + t(-A\sin t + B\cos t)$ and $y_p'' = -2A\sin t + 2B\cos t - t(A\cos t + B\sin t)$ . Product rule twice.
Substitute. $y_p'' + y_p = -2A\sin t + 2B\cos t$ . Matching with $\cos t$ : $A = 0$ , $B = \frac12$ . The terms in $t$ cancel because $t(A\cos t + B\sin t)$ already solves the homogeneous part.
Conclude. $y_p = \frac t2\sin t$ , and the solutions are $\frac t2\sin t + a\cos t + b\sin t$ . The amplitude of $\frac t2\sin t$ grows linearly without bound. This is what makes a swing go high when pushed at its natural period.
Second equation. $t^2 - 3t + 2 = (t - 1)(t - 2)$ , so $r = 1$ is a simple root ( $s = 1$ ), and $R = te^t$ has $d = 1$ . Try $y_p = t(At + B)e^t$ . The theorem: degree $d = 1$ times $t^s = t$ .
Solve. Substituting and matching coefficients of $t^2e^t$ , $te^t$ gives $A = -\frac12$ , $B = -1$ , so $y_p = -\big(\frac{t^2}2 + t\big)e^t$ . The computation is routine; check by substituting numerically at one point.

Answer. $\frac t2\sin t$ ; and $-\big(\frac{t^2}{2} + t\big)e^t$ .

Worked example Apostol 6.14, Examples 1 and 2

Annihilator method (Apostol): find a particular solution of (D⁴ − 16)y = t⁴ + t + 1, and of y'' − 5y' + 6y = t e^t.

Solution.

Annihilate the right side. $t^4 + t + 1$ is annihilated by $D^5$ (table: $t^{m-1}$ with $m = 5$ ). Five derivatives of a polynomial of degree $4$ give $0$ .
Apply $D^5$ to the equation. Every solution satisfies $D^5(D^4 - 16)y = 0$ . The roots of $t^5(t^4 - 16)$ are $0$ (five times), $2, -2, 2i, -2i$ . The kernel of $BL$ is known from 7.2.
Keep only what is not homogeneous. The terms $e^{2t}, e^{-2t}, \cos2t, \sin2t$ solve $(D^4 - 16)y = 0$ , so we try the polynomial part $y_p = a t^4 + bt^3 + ct^2 + dt + e$ . The other four functions are killed by $L$ and contribute nothing to the right-hand side.
Substitute. $y_p^{(4)} = 24a$ , so $(D^4 - 16)y_p = 24a - 16(at^4 + bt^3 + ct^2 + dt + e)$ . Matching with $t^4 + t + 1$ : $-16a = 1$ , $b = c = 0$ , $-16d = 1$ , $24a - 16e = 1$ . So $a = -\frac1{16}$ , $d = -\frac1{16}$ , $e = -\frac{5}{32}$ . Compare coefficients of $t^4, t^3, t^2, t, 1$ .
Second problem. $te^t$ is annihilated by $(D - 1)^2$ . The roots of $(t - 1)^2(t - 2)(t - 3)$ give the form $ae^t + bte^t + (\text{terms } e^{2t}, e^{3t}\text{ that solve the homogeneous part})$ . Substituting $y_p = ae^t + bte^t$ : $(2a - 3b)e^t + 2bte^t = te^t$ , so $b = \frac12$ , $a = \frac34$ . Matching the coefficients of $e^t$ and $te^t$ .

Answer. $y_p = -\frac{t^4}{16} - \frac t{16} - \frac5{32}$ and $y_p = \frac34e^t + \frac12te^t$ .

Worked example MATH 376 worksheet 28 #3; Apostol 6.13

Another route (Apostol 6.13): factor the operator and solve two first-order equations. Do y'' − 3y' + 2y = e^t.

Solution.

Factor. $y'' - 3y' + 2y = (D - 1)(D - 2)y$ . Roots $1, 2$ .
Name the inner part. Let $v = (D - 2)y$ . The equation is $(D - 1)v = e^t$ , that is, $v' - v = e^t$ . A first-order linear equation for $v$ .
Solve for $v$ . $(e^{-t}v)' = 1$ , so $v = te^t + C_1e^t$ . Integrating factor $e^{-t}$ .
Solve $(D - 2)y = v$ . $y' - 2y = te^t + C_1e^t$ , so $(e^{-2t}y)' = (t + C_1)e^{-t}$ , and $e^{-2t}y = -(t + 1)e^{-t} - C_1e^{-t} + C_2$ . $\int te^{-t}\,dt = -(t + 1)e^{-t}$ .
Read off. $y = -(t + 1)e^t - C_1e^t + C_2e^{2t}$ . Absorbing $-C_1e^t - e^t$ into one constant: $y = c_1e^t + c_2e^{2t} - te^t$ . Compare worksheet 28 #3 (variation of parameters gives $-te^t - e^t$ and the second term is a homogeneous solution).

Answer. $y = c_1e^t + c_2e^{2t} - te^t$ .

Quick checkWhich method fits

y'' + y = \tan x

Quick checkFor

y'' - 4y = e^{2t}

, what is the right guess?

Your answerThe particular solution of

y'' + 2y' + y = t

found above is

y_p = t - 2

. A student's general solution has

y(0) = 3

and

y'(0) = 0

. What is

y(1)

? (Three decimals.)

Where it is used

Used inside maths· Inside maths: Liouville's formula, Green's functions and the geometry of volume

Abel’s formula is the scalar case of Liouville’s formula for systems: if $\Phi(t)$ is a matrix whose columns solve $x' = A(t)x$ , then $\det\Phi(t) = \det\Phi(t_0)\exp\int_{t_0}^{t}\operatorname{tr}A(s)\,ds$ (for the companion matrix the trace is $-a_{n-1}$ ). Read geometrically: $\det\Phi$ is the volume of the parallelepiped spanned by $n$ solutions, and the trace says the flow expands or shrinks volume at the rate $\operatorname{tr}A$ . In 7.4 it becomes $\det e^{tA} = e^{t\operatorname{tr}A}$ . Variation of parameters is also where Green’s functions come from, and from there the solution formulas of Poisson’s equation and the wave equation.

Used in the real world· Electronics: tuning a radio is resonance

A series RLC circuit driven by a source $V\cos\omega t$ has the charge equation $Lq'' + Rq' + q/C = V\cos\omega t$ . The natural frequency is $\omega_0 = 1/\sqrt{LC}$ . Take a coil $L = 250\ \mu$ H and a variable capacitor set to $C = 100$ pF: $\omega_0 = 1/\sqrt{2.5\times10^{-14}} = 6.32\times10^{6}$ rad/s, that is $f_0 = 1.0066$ MHz. With no resistance the charge response to a forcing at exactly $\omega_0$ is $\frac{V}{2L\omega_0}\,t\sin\omega_0t$ : the amplitude grows linearly, the resonance case of undetermined coefficients above. With resistance $R = 10\ \Omega$ the growth stops and the amplitude of the capacitor voltage reaches $Q = \omega_0L/R = 6.32\times10^6\times2.5\times10^{-4}/10 = 158$ times the source voltage. A station at $1.0066$ MHz is amplified $158$ -fold, while a station at $1.1$ MHz is far off resonance and barely affects the circuit. Turning the dial changes $C$ and so moves $\omega_0$ .

Traps

Watch out· Do not stop halfway through variation of parameters

On the 376 final (4(b)) you set up variation of parameters correctly: the homogeneous basis $e^{-t}, te^{-t}$ , the substitution $y = v_1e^{-t} + v_2te^{-t}$ , and the extra condition $v_1'e^{-t} + v_2'te^{-t} = 0$ . The page ends there, with no equations for $v_1', v_2'$ , no particular solution, no general solution. For a polynomial right-hand side use undetermined coefficients ( $y_p = at + b$ , two lines), and keep variation of parameters for forcing such as $\tan x$ or $\frac{1}{1 + x^2}$ . If time runs out, write the plan: ‘the Wronskian is $e^{-2t}$ , $u_1' = -t^2e^t$ , $u_2' = te^t$ , integrate, $y_p = u_1b_1 + u_2b_2$ ’. The exam cover sheet says a plan earns partial credit.

Watch out· Zero Wronskian does not mean dependent, and a zero at one point does not either

$t^2$ and $t|t|$ have $W \equiv 0$ and are independent. And $t, t^2$ have $W(0) = 0$ but are independent. What IS true: one non-zero value proves independence; for solutions of one linear equation $W$ is either identically zero (dependent) or never zero (independent); for analytic functions $W \equiv 0$ implies dependence (final 7).

Watch out· Sign and normalisation in the u' formulas

$u_1' = -b_2R/W$ (with the minus) and $u_2' = +b_1R/W$ . For $e^x, e^{-x}$ the Wronskian is $-2$ , not $2$ . The equation must have coefficient $1$ in front of $y^{(n)}$ . And $u_1', u_2'$ must be integrated: in worksheet 22 #5 the answer $u_1' = \frac12$ gives $u_1 = \frac x2$ .

Watch out· Terms of a particular solution that solve the homogeneous equation can be dropped

In $y_p = \frac x2e^x - \frac14e^x$ the term $-\frac14e^x$ is a homogeneous solution; it is absorbed into the constant $c_1$ of the general solution. You do not need it. But the term $\frac x2e^x$ is not a homogeneous solution; it must stay.

Watch out· Sine forcing needs both sine and cosine in the guess

For $q'' + 5q' + 4q = 10\sin2t$ the guess $A\sin2t$ alone cannot work: the first-derivative term produces $\cos2t$ . Use $A\sin2t + B\cos2t$ (homework 9 #2). And when the guess is already a homogeneous solution (resonance), multiply by $t$ .

Watch out· Operators on a product: the first derivative of u1 b1 + u2 b2 has extra terms

Without the condition $u_1'b_1 + u_2'b_2 = 0$ the formula for $y'$ has four terms, and $y''$ has $u''$ terms. The condition is not an assumption about the solution; it is a free choice that we are allowed to make because we have two unknown functions and only one equation.

Problems

checkMATH 376 worksheet 21 #7 to #9, worksheet 22 #2

Compute the Wronskian: (a) $W(\cos x, \sin x)$ ; (b) $W(e^x, xe^x)$ ; (c) $W(\cos2x, \sin2x)$ ; (d) $W(e^x, e^{-x})$ . Which pairs are bases of solutions of which equation?

Solution

(a) $1$ : basis of $y'' + y = 0$ . (b) $e^{2x}$ : basis of $(D - 1)^2y = 0$ . (c) $2$ : basis of $y'' + 4y = 0$ . (d) $-2$ : basis of $y'' - y = 0$ . All are non-zero, so each pair is independent, and two independent solutions of a second-order equation form a basis.

checkMATH 376 practice final 6

Compute $W(t, t^2, t^3)$ as a function of $t$ , and say where it vanishes. Can $t, t^2, t^3$ be solutions of a third-order linear homogeneous equation with continuous coefficients on $(-1, 1)$ ?

Solution

$W(t) = 2t^3$ . It vanishes at $t = 0$ only. On an interval containing $0$ the answer is no: the Wronskian of three independent solutions of such an equation never vanishes. (They are independent, but they would need coefficients with a singularity at $0$ .)

check

True or false: (a) If $W(f, g)(1) \ne 0$ then $f, g$ are independent. (b) If $f, g$ are independent then $W(f, g)(1) \ne 0$ . (c) If $f, g$ solve $y'' + 3y' + y = 0$ and $W(f, g)(1) = 0$ then $f, g$ are dependent.

Solution

(a) True (non-zero Wronskian at a point). (b) False: $t, t^2$ are independent and $W(0) = 0$ , so independence does not give a non-zero Wronskian at every point; and for $t^2, t|t|$ the Wronskian is zero at every point. (c) True: for solutions of one linear equation $W$ is zero everywhere or nowhere.

checkApostol 6.14

Write an annihilating operator for each: (a) $5t^2 + 1$ ; (b) $e^{-3t}$ ; (c) $t\cos2t$ ; (d) $e^t\sin3t$ .

Solution

(a) $D^3$ . (b) $D + 3$ . (c) $(D^2 + 4)^2$ . (d) $D^2 - 2D + 10$ (roots $1 \pm 3i$ ).

useMATH 376 worksheet 22 #4 and worksheet 23 #4

By variation of parameters, find a particular solution of $y'' + y = \tan x$ on $(-\frac\pi2, \frac\pi2)$ .

Solution

$b_1 = \cos x$ , $b_2 = \sin x$ , $W = 1$ . $u_1' = -\sin x\tan x = -(\sec x - \cos x)$ , $u_2' = \sin x$ . $u_1 = -\ln|\sec x + \tan x| + \sin x$ , $u_2 = -\cos x$ . $y_p = -\cos x\ln|\sec x + \tan x|$ . Check by differentiating twice: $y_p'' + y_p = \tan x$ .

useMATH 376 worksheet 22 #6

Find the general solution of $y'' + 4y = \sec2x$ on $(-\frac\pi4, \frac\pi4)$ .

Solution

$b_1 = \cos2x$ , $b_2 = \sin2x$ , $W = 2$ . $u_1' = -\frac12\tan2x$ , $u_2' = \frac12$ . $u_1 = \frac14\ln|\cos2x|$ , $u_2 = \frac x2$ . $y = c_1\cos2x + c_2\sin2x + \frac14\cos2x\ln|\cos2x| + \frac x2\sin2x$ .

useMATH 376 worksheet 22 #7

For $y'' + 2y' + y = \dfrac{1}{1 + x^2}$ , find the homogeneous solutions, the Wronskian, and $u_1, u_2$ as integrals (you need not evaluate them).

Solution

$t^2 + 2t + 1 = (t+1)^2$ , so $b_1 = e^{-x}$ , $b_2 = xe^{-x}$ . $W = e^{-x}(e^{-x} - xe^{-x}) - (-e^{-x})(xe^{-x}) = e^{-2x}$ . With $R = \frac{1}{1 + x^2}$ : $u_1' = -\dfrac{b_2R}{W} = -\dfrac{xe^{-x}}{(1 + x^2)e^{-2x}} = -\dfrac{xe^{x}}{1 + x^2}$ and $u_2' = \dfrac{b_1R}{W} = \dfrac{e^{x}}{1 + x^2}$ . So $u_1 = -\int\frac{xe^x}{1 + x^2}\,dx$ and $u_2 = \int\frac{e^x}{1 + x^2}\,dx$ , and $y_p = u_1e^{-x} + u_2xe^{-x}$ . These integrals have no elementary closed form, which is why the worksheet stops here.

useMATH 376 worksheet 23 #6

Solve $y'''' = e^x$ by variation of parameters with the basis $1, x, x^2, x^3$ .

Solution

The Wronskian system $\mathcal Wu' = (0, 0, 0, e^x)^{\mathsf T}$ reads $u_1' + xu_2' + x^2u_3' + x^3u_4' = 0$ , $u_2' + 2xu_3' + 3x^2u_4' = 0$ , $2u_3' + 6xu_4' = 0$ , $6u_4' = e^x$ . Solve from the bottom: $u_4' = \frac{e^x}{6}$ , $u_3' = -3xu_4' = -\frac{xe^x}{2}$ , $u_2' = -2xu_3' - 3x^2u_4' = x^2e^x - \frac{x^2e^x}{2} = \frac{x^2e^x}{2}$ , $u_1' = -xu_2' - x^2u_3' - x^3u_4' = -\frac{x^3e^x}{2} + \frac{x^3e^x}{2} - \frac{x^3e^x}{6} = -\frac{x^3e^x}{6}$ . Integrating (constants dropped): $u_4 = \frac{e^x}{6}$ , $u_3 = -\frac{(x - 1)e^x}{2}$ , $u_2 = \frac{(x^2 - 2x + 2)e^x}{2}$ , $u_1 = -\frac{(x^3 - 3x^2 + 6x - 6)e^x}{6}$ . Then $y_p = u_1 + xu_2 + x^2u_3 + x^3u_4 = e^x\Big[-\frac{x^3 - 3x^2 + 6x - 6}{6} + \frac{x^3 - 2x^2 + 2x}{2} - \frac{x^3 - x^2}{2} + \frac{x^3}{6}\Big] = e^x$ . The general solution is $y = c_1 + c_2x + c_3x^2 + c_4x^3 + e^x$ , as one sees directly since $(e^x)'''' = e^x$ .

useMATH 376 homework 9 #2; new

Find particular solutions of: (a) $y'' + 5y' + 4y = 10\sin2t$ ; (b) $y'' - y = e^t$ ; (c) $y'' + 4y = \cos2t$ ; (d) $y'' + 2y' + 5y = 5t + 7$ .

Solution

(a) $-\cos2t$ . (b) $r = 1$ is a root, so try $Ate^t$ : $2Ae^t = e^t$ , $y_p = \frac t2e^t$ . (c) $2i$ is a root of $t^2 + 4$ : try $t(A\cos2t + B\sin2t)$ ; $y_p'' + 4y_p = -4A\sin2t + 4B\cos2t$ , so $A = 0$ , $B = \frac14$ : $y_p = \frac t4\sin2t$ . (d) Try $at + b$ : $2a + 5at + 5b = 5t + 7$ , so $a = 1$ , $b = 1$ : $y_p = t + 1$ .

useMATH 376 worksheet 23 #5; worksheet 26 style

Solve $y''' - y' = 2$ by undetermined coefficients, and give the general solution.

Solution

Homogeneous: $t^3 - t = t(t - 1)(t + 1)$ , so $y_h = c_1 + c_2e^t + c_3e^{-t}$ . The right-hand side $2$ has $r = 0$ , a simple root of $t^3 - t$ , so $s = 1$ and we try $y_p = at$ . Then $y_p''' - y_p' = -a = 2$ , so $a = -2$ . General solution: $y = c_1 + c_2e^t + c_3e^{-t} - 2t$ .

proveMATH 376 homework 8 #1(1)

Prove that if $W(f_1, \dots, f_n)(t_0) \ne 0$ for some $t_0$ , the functions are linearly independent.

Solution

Plan. Suppose a combination vanishes, differentiate, and evaluate at $t_0$ . See the proof of the theorem in this chapter: $\sum c_if_i^{(k)}(t_0) = 0$ for $k = 0, \dots, n - 1$ says $\mathcal W(t_0)c = O$ with $\mathcal W(t_0)$ invertible, so $c = O$ . $\blacksquare$

proveMATH 376 homework 9 #3

Prove that $W(t, t^2, \dots, t^r) = C\,t^r$ with a constant $C \ne 0$ .

Solution

Plan. Pull the powers of $t$ out of the rows and columns, and show the remaining constant matrix is invertible.

The entry in row $j + 1$ and column $i$ of the Wronskian matrix is $\frac{d^j}{dt^j}t^i = i(i - 1)\cdots(i - j + 1)\,t^{i - j}$ . Factor $t^i$ from column $i$ and $t^{-j}$ from row $j + 1$ : the total power is $\sum_{i=1}^{r}i - \sum_{j=0}^{r-1}j = \frac{r(r+1)}2 - \frac{r(r-1)}2 = r$ . The remaining matrix has entries $(i)_j := i(i - 1)\cdots(i - j + 1)$ (a falling factorial), which do not depend on $t$ . Its determinant $C$ is non-zero: if $\sum_{j=0}^{r-1}c_j(x)_j$ (a polynomial in $x$ of degree at most $r - 1$ , because $(x)_j$ has degree $j$ ) vanishes at $x = 1, \dots, r$ , it has $r$ roots, so it is the zero polynomial; the $(x)_j$ have distinct degrees, so all $c_j = 0$ . Hence the rows of the constant matrix are independent and $C \ne 0$ . (Indeed $C = 0!\,1!\cdots(r - 1)!$ : $1, 2, 12$ for $r = 2, 3, 4$ .) $\blacksquare$

proveMATH 376 worksheet 22 #3; new for order 3

Derive the system for $u_1', u_2', u_3'$ in variation of parameters for $y''' + a_2y'' + a_1y' + a_0y = R$ with homogeneous solutions $b_1, b_2, b_3$ .

Solution

Plan. Differentiate $y = \sum u_ib_i$ and impose zero extra terms twice.

$y' = \sum u_i'b_i + \sum u_ib_i'$ . Impose $\sum u_i'b_i = 0$ : $y' = \sum u_ib_i'$ . Then $y'' = \sum u_i'b_i' + \sum u_ib_i''$ . Impose $\sum u_i'b_i' = 0$ : $y'' = \sum u_ib_i''$ . Then $y''' = \sum u_i'b_i'' + \sum u_ib_i'''$ . Substituting in the equation and using $b_i''' + a_2b_i'' + a_1b_i' + a_0b_i = 0$ , we obtain $\sum u_i'b_i'' = R$ . The system is $\sum u_i'b_i = 0$ , $\sum u_i'b_i' = 0$ , $\sum u_i'b_i'' = R$ , whose matrix is the Wronskian matrix. (Worksheet 23 #5 with $b = (1, x, x^2)$ and $R = e^x$ is a case.) $\blacksquare$

proveMATH 376 final 7

Let $f(t) = \sum_k a_kt^k$ and $g(t) = \sum_kb_kt^k$ have radius of convergence $1$ . Show: if $f$ and $g$ are linearly dependent then $W(f, g) \equiv 0$ on $(-1, 1)$ ; and prove the converse using the fact that a non-zero power series has finitely many zeros in $(-1 + \varepsilon, 1 - \varepsilon)$ .

Solution

Plan. Easy direction by differentiating the relation; hard direction by dividing where $g \ne 0$ , then the identity theorem.

See the proof of the power-series Wronskian test above: ( $\Rightarrow$ ) $af + bg = 0$ and $af' + bg' = 0$ with $(a, b) \ne O$ give $\det\begin{bmatrix}f & g\\f' & g'\end{bmatrix} = 0$ . ( $\Leftarrow$ ) If neither is the zero function, choose an interval $I$ where $g \ne 0$ ; $(f/g)' = -W/g^2 = 0$ so $f = cg$ on $I$ ; $f - cg$ is a power series with infinitely many zeros in $I$ , hence zero. $\blacksquare$ Note where analyticity is used: the contrast $t^2$ versus $t|t|$ shows the converse fails for $C^1$ functions.

stretchApostol 6.12

Use Abel’s formula to find $W(b_1, b_2)$ for solutions of $y'' + \frac1ty' + \left(1 - \frac1{t^2}\right)y = 0$ (Bessel’s equation of order $1$ ) on $(0, \infty)$ , up to a constant. Then, given $b_1 = J_1$ , show how the second solution can be obtained.

Solution

Here $a_1 = \frac1t$ , so Abel gives $W(t) = W(t_0)\exp\left(-\int_{t_0}^{t}\frac1s\,ds\right) = \frac{C}{t}$ . If $b_1$ is known and non-zero on an interval, then $W = b_1b_2' - b_1'b_2 = b_1^2\left(\frac{b_2}{b_1}\right)'$ , so $\left(\frac{b_2}{b_1}\right)' = \frac{C}{t\,b_1(t)^2}$ and $b_2 = b_1\int\frac{C}{t\,b_1^2}\,dt$ . This is reduction of order: Abel’s formula turns the second-order equation into a quadrature. $\blacksquare$

stretch

Find the response of $y'' + y = f(t)$ , $y(0) = y'(0) = 0$ in the form $y(t) = \int_0^tG(t - s)f(s)\,ds$ and compute it for $f(t) = t$ .

Solution

Here $b_1 = \cos t$ , $b_2 = \sin t$ , $W = 1$ , so $G(t, s) = \cos s\sin t - \cos t\sin s = \sin(t - s)$ , which depends only on $t - s$ . So $y(t) = \int_0^t\sin(t - s)f(s)\,ds$ . For $f(s) = s$ : integrate by parts, $\int_0^ts\sin(t - s)\,ds = \left[s\cos(t - s)\right]_0^t - \int_0^t\cos(t - s)\,ds = t - \sin t$ . Check: $y = t - \sin t$ has $y'' + y = \sin t + t - \sin t = t$ , $y(0) = 0$ , $y'(0) = 1 - 1 = 0$ . $\blacksquare$

Summary

The Wronskian $W(f_1, \dots, f_n) = \det\big(f_i^{(j)}\big)$ is the determinant of the functions and their first $n - 1$ derivatives. For two functions it is the area between the curve $(f_1, f_2)$ and its tangent vector.
Test. $W(t_0) \ne 0$ at one point proves independence. The converse fails for general functions ( $t^2$ and $t|t|$ : $W \equiv 0$ , independent), and $W$ can vanish at isolated points of independent functions ( $t, t^2$ ).
For solutions of one linear equation the Wronskian is zero everywhere or nowhere, because the starting-number map is an isomorphism at every time. Abel’s formula: $W(t) = W(t_0)\exp\left(-\int a_{n-1}\right)$ . For power series, $W \not\equiv 0$ is equivalent to independence (final 7).
Variation of parameters: write $y_p = \sum u_ib_i$ , impose $\sum u_i'b_i^{(j)} = 0$ for $j < n - 1$ , and the equation gives $\sum u_i'b_i^{(n-1)} = R$ . So $\mathcal Wu' = (0, \dots, 0, R)$ . For $n = 2$ : $u_1' = -b_2R/W$ , $u_2' = b_1R/W$ . The equation must have leading coefficient $1$ ; constants of integration and homogeneous terms in $y_p$ can be dropped.
Undetermined coefficients: for $R = q(t)e^{rt}$ try $y_p = t^sh(t)e^{rt}$ with $\deg h = \deg q$ and $s$ the multiplicity of $r$ as a characteristic root (resonance when $s > 0$ ). For sines and cosines use both. It fails for $\tan x$ , $\log x$ , $e^{x^2}$ , where variation of parameters is needed.
Your 376 misses: finish variation of parameters or choose undetermined coefficients for a polynomial right side (final 4(b)); write the easy direction of an iff first (final 7); a zero Wronskian is not a proof of dependence.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

7.3 The Wronskian, variation of parameters and undetermined coefficients

The picture

The idea, in words

Definitions and theorems

The Wronskian and independence

Wronskian

A non-zero Wronskian proves independence

Wronskian of solutions of a linear equation

Abel's formula

Wronskian of exponentials

Wronskian test for power series

Variation of parameters

Variation of parameters

Variation of parameters formula

Undetermined coefficients and the annihilator method

Annihilating operator

Undetermined coefficients

Shape of a particular solution

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere