Linear Equations, Superposition, and Wronskians, differential equations lesson

The functions $e^t$ and $e^{2t}$ both solve $y'' - 3y' + 2y = 0$ , where $t$ is the independent variable and $y' = dy/dt$ . Is every combination $C_1e^t + C_2e^{2t}$ also a solution, and is every solution of this form? For linear equations both answers are yes, as long as the coefficients behave well on the interval in question. This lesson builds that structure from the operator form of the equation and gives a determinant test, the Wronskian, that decides when two solutions are enough.

Key terms

Linear differential operator

Definition Let $p$ and $q$ be functions on an open interval $I$ of the variable $t$ . The second-order linear differential operator with these coefficients assigns to each twice differentiable function $y$ on $I$ the function

L[y] = y'' + p(t)\,y' + q(t)\,y.

It is linear, which means that

L[au + bv] = aL[u] + bL[v]

for all twice differentiable functions $u$ and $v$ on $I$ and all constants $a$ and $b$ . This follows because differentiation and multiplication by a fixed coefficient both respect sums and constant multiples. A second-order linear equation is $L[y] = g$ for a given function $g$ on $I$ , and it is homogeneous when $g(t) = 0$ for every $t$ in $I$ . Here homogeneous means zero forcing, which is a different idea from the homogeneous first-order equations $y' = F(y/t)$ . An equation written as $P(t)\,y'' + Q(t)\,y' + R(t)\,y = G(t)$ takes this normalized form after division by $P(t)$ wherever $P(t) \ne 0$ , with $p = Q/P$ , $q = R/P$ and $g = G/P$ .

In plain words The operator takes a function and returns a new function built from it and its first two derivatives. Linearity says that it handles sums and constant multiples term by term.

Example Take $L[y] = y'' - 3y' + 2y$ on $\mathbb{R}$ . For a constant $r$ ,

L[e^{rt}] = (r^2 - 3r + 2)\,e^{rt},

so $L[e^t] = 0$ and $L[e^{2t}] = 0$ , while $L[1] = 2$ and $L[t] = 2t - 3$ . Linearity then gives $L[5e^t + 1] = 5 \cdot 0 + 2 = 2$ with no new differentiation. By contrast, the expression $N[y] = y'' + y^2$ is not linear, since $N[1] = 1$ but $N[2] = 4 \ne 2N[1]$ .

Regular interval

Definition For the equation $P(t)\,y'' + Q(t)\,y' + R(t)\,y = G(t)$ , a regular interval is an open interval $I$ on which $P(t) \ne 0$ at every point and the normalized functions $p = Q/P$ , $q = R/P$ and $g = G/P$ are continuous. On a regular interval the existence and uniqueness theorem for linear equations applies in the following form. For every $t_0$ in $I$ and all real numbers $k_0$ and $k_1$ , the initial-value problem

\begin{gathered} y'' + p(t)\,y' + q(t)\,y = g(t), \\[0.4em] y(t_0) = k_0, \qquad y'(t_0) = k_1 \end{gathered}

has exactly one solution, and that solution is defined on the whole of $I$ . Points where $P$ vanishes or where a normalized function is not continuous are excluded, and the theory is applied separately on each regular interval.

In plain words Inside a regular interval, a starting height and a starting slope pick out exactly one solution, and that solution lasts as long as the interval does. At the excluded points the theorem makes no promise at all.

Example The equation $t^2y'' - 2ty' + 2y = 0$ normalizes to

y'' - \frac{2}{t}\,y' + \frac{2}{t^2}\,y = 0,

whose coefficients are continuous for $t \ne 0$ . Its regular intervals are $(-\infty, 0)$ and $(0, \infty)$ , so data at $t_0 = 1$ select exactly one solution on all of $(0, \infty)$ . At $t = 0$ both conclusions of the theorem fail. Setting $t = 0$ in the equation leaves $2y(0) = 0$ , so no solution on an interval containing $0$ has $y(0) = 1$ . On the other hand, $y = t$ and $y = t + t^2$ both solve the equation on $\mathbb{R}$ with $y(0) = 0$ and $y'(0) = 1$ . A forcing term such as $G(t) = 1/(t - 1)$ would also exclude $t = 1$ , leaving the regular intervals $(-\infty, 0)$ , $(0, 1)$ and $(1, \infty)$ .

Superposition principle

Definition Let $L[y] = y'' + p(t)\,y' + q(t)\,y$ on an interval $I$ . If $u$ and $v$ solve the homogeneous equation $L[y] = 0$ on $I$ , then $au + bv$ solves it on $I$ for all constants $a$ and $b$ . For a forced equation $L[y] = g$ , superposition describes the whole solution set instead. If $y_p$ is one solution of $L[y] = g$ on $I$ , then the solutions of $L[y] = g$ on $I$ are exactly the functions

y = y_p + y_h, \quad L[y_h] = 0.

Indeed $L[y_p + y_h] = g + 0 = g$ , and any solution $y$ gives $L[y - y_p] = g - g = 0$ . More generally, $L[y_1] = g_1$ and $L[y_2] = g_2$ give $L[ay_1 + by_2] = ag_1 + bg_2$ . The multipliers must be constants. For a twice differentiable multiplier $c(t)$ the product rule leaves extra terms,

\begin{aligned} L[c\,u] &= c\,L[u] + c''u \\[0.4em] &\quad + 2c'u' + p\,c'u. \end{aligned}

In plain words Solutions of a homogeneous linear equation can be scaled and added freely. With forcing present, the sum of two solutions answers twice the forcing, but their difference answers zero forcing, so every forced solution is one fixed forced solution plus a homogeneous one.

Example For $L[y] = y'' - 3y' + 2y$ , both $e^t$ and $e^{2t}$ solve $L[y] = 0$ , so $3e^t + e^{2t}$ and $e^t - 7e^{2t}$ solve it too. The constant $y_p = 1$ satisfies $L[1] = 2$ , so every solution of $L[y] = 2$ has the form $1 + y_h$ with $L[y_h] = 0$ . A varying multiplier breaks superposition. With $c(t) = t$ and $u = e^t$ the formula above gives $L[te^t] = 0 + 0 + 2e^t - 3e^t = -e^t$ , which is not zero although $L[e^t] = 0$ .

Fundamental set of solutions

Definition Two functions $y_1$ and $y_2$ are linearly independent on an interval $I$ if the only constants with $C_1y_1(t) + C_2y_2(t) = 0$ for every $t$ in $I$ are $C_1 = C_2 = 0$ . For two functions this means that neither is a constant multiple of the other on $I$ . A fundamental set of solutions of $L[y] = 0$ on a regular interval $I$ is a pair of solutions $y_1$ and $y_2$ on $I$ that are linearly independent on $I$ . For such a pair, every solution of $L[y] = 0$ on $I$ can be written as

y = C_1y_1 + C_2y_2

for exactly one choice of the constants, and for each $t_0$ in $I$ every pair of initial values $y(t_0) = k_0$ , $y'(t_0) = k_1$ is met by exactly one such combination. A homogeneous linear equation of order $n$ has fundamental sets of $n$ solutions on a regular interval in the same way.

In plain words Two genuinely different solutions are enough to build every solution, and their two constants are exactly the two numbers needed to match a starting height and slope.

Example The equation $y'' - 3y' + 2y = 0$ has constant coefficients, so all of $\mathbb{R}$ is a regular interval. The solutions $e^t$ and $e^{2t}$ are independent there, because the ratio $e^{2t}/e^t = e^t$ is not constant. To meet $y(0) = 4$ and $y'(0) = 5$ , write $y = C_1e^t + C_2e^{2t}$ . The two conditions read

\begin{aligned} C_1 + C_2 &= 4, \\ C_1 + 2C_2 &= 5, \end{aligned}

and subtracting the first from the second gives $C_2 = 1$ , then $C_1 = 3$ . The solution is

y = 3e^t + e^{2t}.

As a check, $y' = 3e^t + 2e^{2t}$ and $y'' = 3e^t + 4e^{2t}$ . In $y'' - 3y' + 2y$ the coefficient of $e^t$ is $3 - 9 + 6 = 0$ and the coefficient of $e^{2t}$ is $4 - 6 + 2 = 0$ , and the values at $0$ are $y(0) = 4$ and $y'(0) = 5$ .

Wronskian

Definition The Wronskian of two differentiable functions $y_1$ and $y_2$ is

\begin{aligned} W(y_1, y_2)(t) &= \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} \\[0.4em] &= y_1y_2' - y_1'y_2, \end{aligned}

with every entry evaluated at $t$ . Its value at $t_0$ is the determinant of the system that matches initial data,

\begin{aligned} C_1y_1(t_0) + C_2y_2(t_0) &= k_0, \\ C_1y_1'(t_0) + C_2y_2'(t_0) &= k_1. \end{aligned}

Two facts follow.

For any differentiable functions on an interval $I$ , if $W(t_0) \ne 0$ at one point $t_0$ of $I$ , then $y_1$ and $y_2$ are linearly independent on $I$ . A combination that vanishes on $I$ has a vanishing derivative too, and the system above with $k_0 = k_1 = 0$ then forces $C_1 = C_2 = 0$ .
If $y_1$ and $y_2$ solve the same equation $L[y] = 0$ on a regular interval $I$ , they form a fundamental set on $I$ exactly when $W(t_0) \ne 0$ for some $t_0$ in $I$ . If instead $W(t_0) = 0$ , the system with zero data has a solution $(C_1, C_2) \ne (0, 0)$ , and the solution $C_1y_1 + C_2y_2$ has zero height and slope at $t_0$ , so by uniqueness it is the zero function.

When $W(t_0) \ne 0$ , any solution $y$ can be matched by constants with $k_0 = y(t_0)$ and $k_1 = y'(t_0)$ , and uniqueness then gives $y = C_1y_1 + C_2y_2$ on all of $I$ . That is why a fundamental set produces every solution. Without a common equation, even a Wronskian that is zero everywhere does not prove dependence. The functions $t^2$ and $t\lvert t \rvert$ are continuously differentiable on $\mathbb{R}$ , the derivative of $t\lvert t \rvert$ is $2\lvert t \rvert$ , and their Wronskian $t^2 \cdot 2\lvert t \rvert - 2t \cdot t\lvert t \rvert$ is $0$ for every $t$ . Yet they are independent on $\mathbb{R}$ , since the second function equals the first for $t > 0$ and its negative for $t < 0$ , so no single constant multiple works. They are not two solutions of one equation $L[y] = 0$ on a regular interval containing $0$ , and in fact $t\lvert t \rvert$ has no second derivative at $0$ .

In plain words The Wronskian asks whether the height and slope pairs $(y_1, y_1')$ and $(y_2, y_2')$ point in different directions. If they do, every starting height and slope can be built from them. For two solutions of one regular equation the answer is the same at every point.

Example For $y_1 = e^t$ and $y_2 = e^{2t}$ ,

\begin{aligned} W &= e^t \cdot 2e^{2t} - e^t \cdot e^{2t} \\[0.4em] &= e^{3t}, \end{aligned}

which is never zero, so the pair is a fundamental set on $\mathbb{R}$ , and the system for the data $y(0) = 4$ , $y'(0) = 5$ above has determinant $W(0) = 1$ and one solution. Listing the functions in the other order gives $W(e^{2t}, e^t) = -e^{3t}$ . The sign changes, but independence does not.

Abel’s identity

Definition Let $y_1$ and $y_2$ solve $y'' + p(t)\,y' + q(t)\,y = 0$ on a regular interval $I$ , and let $t_0$ be a point of $I$ . Then their Wronskian satisfies the first-order linear equation $W' = -p(t)\,W$ , and therefore

W(t) = W(t_0)\,e^{-\int_{t_0}^{t} p(s)\,ds}

for every $t$ in $I$ . To see this, differentiate $W = y_1y_2' - y_1'y_2$ . The two terms $y_1'y_2'$ cancel, which leaves $W' = y_1y_2'' - y_1''y_2$ . Substituting $y_j'' = -p\,y_j' - q\,y_j$ for $j = 1, 2$ makes the $q$ terms cancel and leaves $-p\,(y_1y_2' - y_1'y_2) = -pW$ . The coefficient $p$ is the one of the normalized equation. Because the exponential factor is never zero, the Wronskian is either zero at every point of $I$ or zero at no point of $I$ .

In plain words The Wronskian of two solutions follows its own first-order equation. It grows or shrinks by an exponential factor and can never reach zero inside a regular interval unless it is zero throughout. It also tells you the Wronskian, up to one constant, before you know any solution.

Example The equation $t^2y'' - 2ty' + 2y = 0$ has the solutions $y_1 = t$ and $y_2 = t^2$ , since $0 - 2t + 2t = 0$ and $2t^2 - 4t^2 + 2t^2 = 0$ . On $(0, \infty)$ the normalized coefficient is $p(t) = -2/t$ , and Abel’s identity from $t_0 = 1$ gives

\begin{aligned} W(t) &= W(1)\,e^{\int_1^t (2/s)\,ds} \\[0.4em] &= W(1)\,e^{2\ln t} \\[0.4em] &= W(1)\,t^2. \end{aligned}

The direct computation $W = t \cdot 2t - 1 \cdot t^2 = t^2$ agrees, with $W(1) = 1$ . On the whole line this Wronskian vanishes at $t = 0$ and nowhere else. That does not contradict Abel’s identity, because the leading coefficient $t^2$ vanishes at $t = 0$ , so no regular interval contains that point. For the constant-coefficient equation $y'' - 3y' + 2y = 0$ the identity reads $W(t) = W(0)\,e^{3t}$ , which matches $W(e^t, e^{2t}) = e^{3t}$ .

Check yourself

Show that $e^{-t}$ and $e^{3t}$ solve $y'' - 2y' - 3y = 0$ , compute their Wronskian, and find the solution with $y(0) = 0$ and $y'(0) = 4$ .

Show the answer

For $y = e^{rt}$ the left side is $(r^2 - 2r - 3)\,e^{rt} = (r - 3)(r + 1)\,e^{rt}$ , which is zero for $r = 3$ and $r = -1$ . The Wronskian is

\begin{aligned} W &= e^{-t} \cdot 3e^{3t} - (-e^{-t})\,e^{3t} \\[0.4em] &= 4e^{2t}, \end{aligned}

which is never zero, so the pair is a fundamental set on $\mathbb{R}$ . Abel’s identity with $p = -2$ agrees, since $W(0)\,e^{2t} = 4e^{2t}$ . Writing $y = C_1e^{-t} + C_2e^{3t}$ , the data give $C_1 + C_2 = 0$ and $-C_1 + 3C_2 = 4$ . Adding them gives $4C_2 = 4$ , so $C_2 = 1$ and $C_1 = -1$ , and

y = e^{3t} - e^{-t}.

Both $y = 1$ and $y = 1 + e^t$ solve $y'' - 3y' + 2y = 2$ . Does their sum solve the same equation? Describe every solution, and find the one with $y(0) = 0$ and $y'(0) = 0$ .

Show the answer

The sum $2 + e^t$ gives $L[2 + e^t] = 4 + 0 = 4$ , so it solves the equation with forcing $4$ , not $2$ . Free combinations of solutions are solutions only for homogeneous equations. The difference $e^t$ does solve the homogeneous equation, as it must. Since $e^t$ and $e^{2t}$ form a fundamental set of the homogeneous equation, every solution of the forced equation is

y = 1 + C_1e^t + C_2e^{2t}.

The data give $1 + C_1 + C_2 = 0$ and $C_1 + 2C_2 = 0$ , so $C_2 = 1$ and $C_1 = -2$ . The solution is $y = 1 - 2e^t + e^{2t}$ , which can also be written $(e^t - 1)^2$ .

Can $t^2$ and $t^3$ both solve an equation $y'' + p(t)\,y' + q(t)\,y = 0$ whose coefficients $p$ and $q$ are continuous on $(-1, 1)$ ?

Show the answer

No. Their Wronskian is

W = t^2 \cdot 3t^2 - 2t \cdot t^3 = t^4,

which is zero at $t = 0$ and positive at every other point. If both functions solved such an equation, then $(-1, 1)$ would be a regular interval, and Abel’s identity would make the Wronskian either zero everywhere on it or zero nowhere on it. The two functions do solve $t^2y'' - 4ty' + 6y = 0$ , as substitution shows, but the normalized coefficients $-4/t$ and $6/t^2$ of that equation are not defined at $t = 0$ .

Linear Equations, Superposition, and Wronskians

Key terms

Linear differential operator

Regular interval

Superposition principle

Fundamental set of solutions

Wronskian

Abel’s identity

Check yourself

How this lesson connects

Builds on

Used later in