Variation of Parameters, differential equations lesson

The equation $y'' + y = \sec t$ looks like an ordinary oscillator with an applied force, yet undetermined coefficients cannot solve it, because $\sec t$ and its derivatives do not stay inside the span of finitely many functions. Variation of parameters needs no trial family. It starts from two independent solutions of the homogeneous equation, lets their constant multipliers vary with $t$ , and writes a particular solution as integrals that make sense for every forcing that is continuous on the interval. Throughout, $t$ is the independent variable and a prime means $d/dt$ .

Key terms

Fundamental pair

Definition Variation of parameters works with the normalized equation

y'' + p(t)\,y' + q(t)\,y = g(t)

on a regular interval $I$ , an open interval on which $p$ , $q$ and $g$ are continuous. An equation $P(t)\,y'' + Q(t)\,y' + R(t)\,y = G(t)$ takes this form after division by $P(t)$ on an interval where $P(t) \ne 0$ , with $p = Q/P$ , $q = R/P$ and $g = G/P$ . A fundamental pair on $I$ is a pair of solutions $y_1$ and $y_2$ of the homogeneous equation $y'' + p\,y' + q\,y = 0$ on $I$ whose Wronskian

W = y_1y_2' - y_1'y_2

is nonzero at one point of $I$ . By Abel’s identity the Wronskian is then nonzero at every point of $I$ , and every solution of the homogeneous equation on $I$ is $C_1y_1 + C_2y_2$ for exactly one choice of the constants.

In plain words A fundamental pair is all the information about the homogeneous equation that the method needs. The Wronskian condition says that the two solutions are genuinely different at every instant, not multiples of each other.

Example For $y'' + y = g$ , the functions $y_1 = \cos t$ and $y_2 = \sin t$ solve $y'' + y = 0$ , and their Wronskian is $W = \cos^2 t + \sin^2 t = 1$ . For $y'' - 2y' + y = g$ , the pair $e^t$ and $te^t$ has

\begin{aligned} W &= e^t(e^t + te^t) - e^t \cdot te^t \\[0.4em] &= e^{2t}. \end{aligned}

Both Wronskians are nonzero on all of $\mathbb{R}$ . By contrast, $\cos t$ and $3\cos t$ have $W = 0$ and do not form a fundamental pair.

Variation of parameters

Definition Let $y_1$ and $y_2$ be a fundamental pair on a regular interval $I$ . Variation of parameters looks for a particular solution of $y'' + p\,y' + q\,y = g$ of the form

y_p = u_1(t)\,y_1 + u_2(t)\,y_2,

where the constants of the complementary solution $C_1y_1 + C_2y_2$ are replaced by differentiable functions $u_1$ and $u_2$ , called the parameters. Two unknown functions for one equation leave room for one extra condition, and the method imposes

u_1'y_1 + u_2'y_2 = 0

at every point of $I$ . With this auxiliary condition the first derivative contains no derivatives of the parameters,

y_p' = u_1y_1' + u_2y_2',

so second derivatives of $u_1$ and $u_2$ never appear when $y_p$ is substituted into the equation.

In plain words The constants that describe the homogeneous solutions are allowed to drift with time, and the drift is chosen so that the equation produces the forcing. The auxiliary condition is a convenient choice, not a consequence of the equation, and it keeps the first derivative as simple as if the parameters were constant.

Example For $y'' + y = g$ with the pair $\cos t$ and $\sin t$ , the candidate is $y_p = u_1\cos t + u_2\sin t$ . The product rule gives

\begin{aligned} y_p' &= u_1'\cos t + u_2'\sin t \\[0.4em] &\quad - u_1\sin t + u_2\cos t, \end{aligned}

and the auxiliary condition $u_1'\cos t + u_2'\sin t = 0$ removes the first two terms. What is left, $y_p' = -u_1\sin t + u_2\cos t$ , has the same shape as the derivative of $C_1\cos t + C_2\sin t$ .

Equations for the parameter derivatives

Definition Differentiating $y_p' = u_1y_1' + u_2y_2'$ once more gives

\begin{aligned} y_p'' &= u_1y_1'' + u_2y_2'' \\[0.4em] &\quad + u_1'y_1' + u_2'y_2'. \end{aligned}

In $y_p'' + p\,y_p' + q\,y_p$ the terms containing $u_1$ and $u_2$ without derivatives add up to $u_1(y_1'' + p\,y_1' + q\,y_1) + u_2(y_2'' + p\,y_2' + q\,y_2)$ , which is zero because $y_1$ and $y_2$ solve the homogeneous equation. The equation therefore reduces to $u_1'y_1' + u_2'y_2' = g$ . Together with the auxiliary condition, the parameter derivatives satisfy the linear system

\begin{aligned} y_1\,u_1' + y_2\,u_2' &= 0, \\[0.4em] y_1'\,u_1' + y_2'\,u_2' &= g. \end{aligned}

At each $t$ in $I$ the determinant of this system is $W(t) \ne 0$ , so it has exactly one solution $(u_1'(t), u_2'(t))$ .

In plain words The two unknowns are the rates at which the parameters change. One equation is the auxiliary choice and the other is the differential equation itself, and once the homogeneous terms cancel only these rates remain.

Example For $y'' + y = \sec t$ on $(-\pi/2, \pi/2)$ , with $y_1 = \cos t$ and $y_2 = \sin t$ , the system reads

\begin{aligned} \cos t\,u_1' + \sin t\,u_2' &= 0, \\[0.4em] -\sin t\,u_1' + \cos t\,u_2' &= \sec t. \end{aligned}

Multiplying the first equation by $\cos t$ and the second by $-\sin t$ and adding gives $u_1' = -\sin t\sec t = -\tan t$ . Multiplying the first by $\sin t$ and the second by $\cos t$ and adding gives $u_2' = \cos t\sec t = 1$ .

Wronskian denominator

Definition Solving the system by Cramer’s rule gives

\begin{gathered} u_1' = -\frac{y_2\,g}{W}, \\[0.4em] u_2' = \frac{y_1\,g}{W}, \end{gathered}

where $W = y_1y_2' - y_1'y_2$ is the Wronskian of the pair, which is nonzero on $I$ . Here $g$ is the forcing of the normalized equation, whose coefficient of $y''$ is $1$ . For $P(t)\,y'' + Q(t)\,y' + R(t)\,y = G(t)$ the forcing in these formulas is $g = G/P$ , not $G$ . The signs depend on the order of the pair. Swapping $y_1$ and $y_2$ changes the sign of $W$ and exchanges the roles of the two formulas, and it leads to the same particular solutions.

In plain words The Wronskian measures how independent the two homogeneous solutions are at each instant, and dividing by it turns the forcing into rates of change for the two parameters. The formulas assume that the equation was divided by its leading coefficient first, and skipping that step feeds in the wrong forcing.

Example Solve $y'' - 2y' + y = e^t/t$ on $(0, \infty)$ , where the forcing is continuous. With $y_1 = e^t$ , $y_2 = te^t$ and $W = e^{2t}$ ,

\begin{gathered} u_1' = -\frac{te^t \cdot e^t/t}{e^{2t}} = -1, \\[0.4em] u_2' = \frac{e^t \cdot e^t/t}{e^{2t}} = \frac{1}{t}. \end{gathered}

Taking $u_1 = -t$ and $u_2 = \ln t$ gives $y_p = -te^t + te^t\ln t$ . The first term solves the homogeneous equation, so a simpler particular solution is

y_p = te^t\ln t.

Undetermined coefficients could not reach this forcing, since no finite trial family contains $e^t/t$ and all its derivatives.

Particular solution by definite integrals

Definition Fix a point $t_0$ in the regular interval $I$ and integrate the parameter derivatives from $t_0$ ,

\begin{gathered} u_1(t) = -\int_{t_0}^{t} \frac{y_2(s)\,g(s)}{W(s)}\,ds, \\[0.6em] u_2(t) = \int_{t_0}^{t} \frac{y_1(s)\,g(s)}{W(s)}\,ds. \end{gathered}

The integrands are continuous on $I$ , so both integrals exist for every $t$ in $I$ , and $y_p = u_1y_1 + u_2y_2$ solves $y'' + p\,y' + q\,y = g$ on all of $I$ . In one formula,

y_p(t) = \int_{t_0}^{t} K(t, s)\,g(s)\,ds,

with the kernel

\begin{aligned} K(t, s) = \frac{1}{W(s)}\big[&y_1(s)\,y_2(t) \\[0.4em] &- y_1(t)\,y_2(s)\big]. \end{aligned}

This particular solution satisfies $y_p(t_0) = 0$ and $y_p'(t_0) = 0$ . Both integrals vanish at $t_0$ , which gives the height, and by the auxiliary condition the slope is $u_1y_1' + u_2y_2'$ , which also vanishes at $t_0$ . The solution with $y(t_0) = k_0$ and $y'(t_0) = k_1$ is therefore $y_p$ plus the homogeneous solution $C_1y_1 + C_2y_2$ with those same data. A different lower limit, or constants added to $u_1$ and $u_2$ , changes $y_p$ only by a homogeneous solution. The integral formula is an exact answer even when the integrals have no elementary antiderivative.

In plain words Starting both integrals at the same time $t_0$ picks out the response that starts at rest, with zero height and zero slope at $t_0$ . Any other starting height and slope is reached by adding a homogeneous solution, and the forcing only has to be continuous for the integrals to make sense.

Example For $y'' + y = \sec t$ on $(-\pi/2, \pi/2)$ take $t_0 = 0$ . The earlier cards gave $u_1' = -\tan t$ and $u_2' = 1$ . Since $\cos t > 0$ on this interval,

u_1(t) = -\int_0^t \tan s\,ds = \ln(\cos t),

and $u_2(t) = t$ , so

y_p = \cos t\,\ln(\cos t) + t\sin t.

Differentiating, the terms $-\sin t$ and $+\sin t$ cancel, and $y_p' = -\sin t\,\ln(\cos t) + t\cos t$ . One more derivative gives

\begin{aligned} y_p'' + y_p &= \frac{\sin^2 t}{\cos t} + \cos t \\[0.4em] &= \frac{1}{\cos t} = \sec t, \end{aligned}

and $y_p(0) = y_p'(0) = 0$ as promised. The solution with $y(0) = 1$ and $y'(0) = 2$ is $y = \cos t + 2\sin t + y_p$ . These solutions live on $(-\pi/2, \pi/2)$ only. As $t$ approaches either endpoint the height $y_p(t)$ tends to $\pi/2$ , but the slope $y_p'(t)$ is unbounded and $\sec t$ is not defined at the endpoints, so the formula gives no solution through them.

Check yourself

Use variation of parameters to find a particular solution of $y'' - y = 2e^t$ , and compare the result with undetermined coefficients.

Show the answer

The pair $y_1 = e^t$ and $y_2 = e^{-t}$ has $W = e^t(-e^{-t}) - e^te^{-t} = -2$ . The formulas give

\begin{gathered} u_1' = -\frac{e^{-t} \cdot 2e^t}{-2} = 1, \\[0.4em] u_2' = \frac{e^t \cdot 2e^t}{-2} = -e^{2t}. \end{gathered}

Taking $u_1 = t$ and $u_2 = -\tfrac12e^{2t}$ gives $y_p = te^t - \tfrac12e^t$ . The term $-\tfrac12e^t$ solves the homogeneous equation and can be dropped, which leaves $y_p = te^t$ . Undetermined coefficients gives the same function, since $1$ is a simple root of $r^2 - 1$ , the trial is $Ate^t$ , and $L[Ate^t] = 2Ae^t$ forces $A = 1$ .

For $y'' + y = g(t)$ with $g$ continuous on an interval containing $0$ , show that the definite-integral formula with $t_0 = 0$ gives $y_p(t) = \int_0^t \sin(t - s)\,g(s)\,ds$ . Evaluate it for $g(t) = t$ , and explain why the result is not the particular solution $t$ that undetermined coefficients gives.

Show the answer

With $y_1 = \cos t$ , $y_2 = \sin t$ and $W = 1$ , the kernel is $K(t, s) = \cos s\sin t - \cos t\sin s = \sin(t - s)$ , which gives the formula. For $g(s) = s$ , substitute $u = t - s$ and use $\int_0^t \sin u\,du = 1 - \cos t$ and $\int_0^t u\sin u\,du = \sin t - t\cos t$ . Then

\begin{aligned} y_p(t) &= \int_0^t (t - u)\sin u\,du \\[0.4em] &= t - \sin t. \end{aligned}

Both $t - \sin t$ and $t$ solve $y'' + y = t$ , and they differ by $-\sin t$ , a homogeneous solution. The definite-integral version is the one that starts at rest, with $y_p(0) = 0$ and $y_p'(0) = 1 - \cos 0 = 0$ , while the function $t$ has slope $1$ at $0$ .

The functions $y_1 = t$ and $y_2 = t^2$ solve $t^2y'' - 2ty' + 2y = 0$ on $(0, \infty)$ . Find a particular solution of $t^2y'' - 2ty' + 2y = t^3$ there, and show what goes wrong if the forcing $t^3$ is used in the formulas without dividing by $t^2$ .

Show the answer

Dividing by $t^2$ gives the normalized forcing $g = t$ . The Wronskian is $W = t \cdot 2t - 1 \cdot t^2 = t^2$ , so $u_1' = -t^2 \cdot t/t^2 = -t$ and $u_2' = t \cdot t/t^2 = 1$ . Taking $u_1 = -\tfrac12t^2$ and $u_2 = t$ gives

y_p = -\tfrac12t^3 + t^3 = \tfrac12t^3.

As a check, $t^2 \cdot 3t - 2t \cdot \tfrac32t^2 + 2 \cdot \tfrac12t^3 = 3t^3 - 3t^3 + t^3 = t^3$ . With the unnormalized forcing $t^3$ the formulas give $u_1' = -t^3$ and $u_2' = t^2$ , so $u_1 = -\tfrac14t^4$ , $u_2 = \tfrac13t^3$ and the candidate is $\tfrac{1}{12}t^5$ . Substituting it into the left side gives $\tfrac{20}{12}t^5 - \tfrac{10}{12}t^5 + \tfrac{2}{12}t^5 = t^5$ , not $t^3$ .

Variation of Parameters

Key terms

Fundamental pair

Variation of parameters

Equations for the parameter derivatives

Wronskian denominator

Particular solution by definite integrals

Check yourself

How this lesson connects

Builds on

Used later in