Variation of Parameters

Main idea

Take $y_c = c_1y_1 + c_2y_2$ and let the constants become functions: $y_p = u_1y_1 + u_2y_2$ . Forcing the equation to hold pins down $u_1'$ and $u_2'$ , so two integrals give a particular solution for any forcing $g(t)$ , not just the nice ones.

y_p = -y_1\int\frac{y_2\,{\htmlData{sym=1}{g}}}{\htmlData{sym=0}{{W}}}\,dt + y_2\int\frac{y_1\,{\htmlData{sym=1}{g}}}{\htmlData{sym=0}{{W}}}\,dt

What the symbols mean

$W$
Wronskian
y₁y₂′ − y₁′y₂.
$g$
Forcing
With the equation written as y″ + p y′ + q y = g (leading coefficient 1).

Explanation

Variation of parameters is simply a method for finding a system's response to any outside push at all, even ones the guessing method can't handle. The name means exactly what it does: the constants (parameters) $c_1$ and $c_2$ in $y_c$ are allowed to vary with time. So you build the response to the push out of the motions the system already makes on its own, just in amounts that change from moment to moment. The formula looks scary, but it's only a few parts: $y_1$ and $y_2$ (the two homogeneous solutions you already found), $g$ (the push), and $W$ (the Wronskian, which you met last lesson). Each integral works out how much of $y_1$ or $y_2$ you need at each moment. Two integrals and you're done; bear with the setup, it's shorter than it looks.

The method that always works. Undetermined coefficients needs $g$ to be a polynomial, an exponential, a sine or cosine, or a product of these. Variation of parameters doesn't care what $g$ is: once you know $y_c$ , it produces a $y_p$ for any continuous forcing, $\sec t$ , $\ln t$ , $\frac1t$ , anything. In that sense it's the universal method for linear nonhomogeneous equations. The price is two integrals, which can get ugly, so when $g$ has one of the nice forms, undetermined coefficients is usually faster.

Most of the time you'll simply be asked to solve a nonhomogeneous second order linear equation, either with a method that's named or with whichever you choose, depending on how generous your professor is feeling. Knowing both means you're covered either way.

Where it comes from. It's the same move as reduction of order. The homogeneous solution is $c_1y_1 + c_2y_2$ with constant $c_1, c_2$ , and constants can only ever give $0$ on the right side. So let them vary: try $y_p = u_1(t)y_1 + u_2(t)y_2$ . Two unknown functions is more freedom than one equation needs, so we're allowed to ask for something extra. We ask for $u_1'y_1 + u_2'y_2 = 0$ , which keeps $y_p'$ free of $u'$ terms and stops $u''$ from ever showing up. Substituting into $y'' + py' + qy = g$ , the terms without derivatives on $u$ cancel (because $y_1$ and $y_2$ are solutions), and what's left is

u_1'y_1' + u_2'y_2' = g.

Those two conditions are a 2×2 linear system for $u_1'$ and $u_2'$ . Notice that its determinant is the Wronskian $W$ , the same one that let us match initial conditions. It's never zero, so the system always has a solution, and solving it gives the formula above.

A common mistake. In an equation like $t^2y'' + ty' - y = t^3$ , it's tempting to read off $g = t^3$ because that's what's sitting on the right. It isn't: the formula was derived with leading coefficient 1, so divide through first. Here that gives $g = t$ , and skipping the division would leave your answer off by a factor of $t^2$ .

Steps:

Divide by the leading coefficient so the equation reads $y'' + py' + qy = g$ .
Find $y_1, y_2$ and their Wronskian $W$ .
Compute $u_1 = -\int\frac{y_2g}{W}\,dt$ and $u_2 = \int\frac{y_1g}{W}\,dt$ (constants not needed).
$y_p = u_1y_1 + u_2y_2$ , and the general solution is $y = y_c + y_p$ .

See it

A spring driven by u″ + 4u = 4 sec 2t, a push that grows without bound as t approaches π/4 ≈ 0.785 (drawn at a quarter of its size). No guess from the undetermined coefficients table fits sec 2t, but variation of parameters still finds the response, u(t) = cos 2t ln(cos 2t) + 2t sin 2t. Notice that the mass is pushed out past 1 m by t = 0.7 as the force blows up.

Worked examples

Example 1

Find a particular solution of $y'' + y = \sec t$ .

1
The roots of $r^2 + 1 = 0$ are $\pm i$ , so $y_1 = \cos t$ and $y_2 = \sin t$ , with $W = \cos^2 t + \sin^2 t = 1$ . The secant isn't on the undetermined coefficients table, so this is a job for variation of parameters.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

A $1$ kg mass on a spring with stiffness $k = 4$ N/m and no damping is driven by a force $F(t) = 4\sec 2t$ newtons for $0 \le t < \frac{\pi}{4}$ . It starts at rest at its equilibrium position. Find its position $u(t)$ .

Problem 2+5 XP

What is the Wronskian of $\cos t$ and $\sin t$ ?

Problem 3+5 XP

Find a particular solution of $y'' - y = e^t$ (any one works).

Problem 4+5 XP

Use variation of parameters with $y_1 = 1$ , $y_2 = t$ to find a particular solution of $y'' = 6t$ .

Problem 5+5 XP

When is variation of parameters needed instead of undetermined coefficients?

Problem 6+5 XP

Use variation of parameters to find the most general solution of $y'' + 9y = \sec 3x$ on $|x| < \frac\pi6$ . Use $c_1$ and $c_2$ for the constants.

Problem 7+5 XP

Find a particular solution of $y'' + 4y = -8\sec 2t$ .

Problem 8+5 XP

Find a particular solution of $y'' - 6y' + 9y = \dfrac{2e^{3t}}{t^2 + 1}$ .

Problem 9+5 XP

Find a particular solution of $y'' + 10y' + 25y = \dfrac{e^{-5x}}{x^3}$ for $x > 0$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Solve $y'' - 2y' + y = \dfrac{e^t}{t}$ for $t > 0$ , with $y(1) = 0$ and $y'(1) = 2e$ .

Where it's used

Variation of parameters leads to Green's functions, which give the response of a system to any input by integrating its response to a single kick.

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