Variation of Parameters

Variation of parameters extends directly to higher-order linear equations (see Higher-Order Linear Equations), where an $n \times n$ linear system built from the Wronskian-like determinant of $n$ fundamental solutions plays the same role, and to first-order linear systems $\mathbf{x}' = A(t)\mathbf{x} + \mathbf{g}(t)$ (see Systems of First-Order Linear ODEs), where the fundamental matrix $\Phi(t)$ replaces the pair $(y_1,y_2)$ and the formula becomes $\mathbf{x}_p(t) = \Phi(t)\int \Phi(t)^{-1}\mathbf{g}(t)\,dt$. In physics, the same "constants become slowly varying functions" idea underlies classical perturbation theory for planetary orbits (Lagrange's original motivation) and the interaction picture in quantum mechanics, where wavefunction coefficients evolve slowly under a perturbing Hamiltonian. In electrical engineering, it computes the response of an $RLC$ circuit to an arbitrary time-varying voltage source that is not expressible via simple Laplace-transform table entries. In control theory, the variation-of-parameters (Duhamel) formula gives the state of a linear time-varying system driven by an arbitrary input, forming the basis of convolution representations of system response. In numerical analysis and applied math more broadly, the formula is the continuous analogue of Duhamel's principle, used to build solution representations for inhomogeneous PDEs (heat and wave equations) from their homogeneous Green's functions.

Ask the tutor: How does the variation-of-parameters formula change for a third-order linear equation, and what replaces the $2\times2$ Wronskian system?

The starting point is the general homogeneous solution $y_h(t) = c_1 y_1(t) + c_2 y_2(t)$, valid when $g \equiv 0$. When $g \not\equiv 0$, constants $c_1, c_2$ can no longer make the equation balance — the equation has a "budget" of forcing $g(t)$ to absorb at every instant, and fixed constants have no way to track a moving target. The fix is to let the coefficients drift in time, $c_1 \to u_1(t)$ and $c_2 \to u_2(t)$, giving them exactly enough freedom to soak up the forcing at each instant while still riding along the shapes $y_1, y_2$ that solve the homogeneous problem. There is a subtlety: differentiating $y_p = u_1 y_1 + u_2 y_2$ twice introduces four new terms involving $u_1', u_2', u_1'', u_2''$, and substituting into the ODE gives one equation in two unknown functions $u_1, u_2$ — hopelessly underdetermined without another condition. The trick is to impose a second condition by choice: demand $u_1' y_1 + u_2' y_2 = 0$. This eliminates the second-derivative terms from the substitution entirely, and what remains collapses to a clean $2\times 2$ linear algebraic system for $u_1'(t), u_2'(t)$ at each fixed $t$. That system's coefficient matrix is exactly the matrix whose determinant is the Wronskian — this is the payoff for demanding $y_1, y_2$ be linearly independent solutions in the first place. Geometrically, think of $(y_1(t), y_1'(t))$ and $(y_2(t), y_2'(t))$ as two vectors in the phase plane at each time $t$. Linear independence of $y_1, y_2$ as functions is equivalent to these two vectors never being parallel, i.e. $W(t) \neq 0$. The system for $u_1', u_2'$ asks: "in the basis given by these two non-parallel vectors, how do I resolve the pure forcing vector $(0, g(t))$?" Because the vectors are never parallel, this resolution is always possible and unique.

Example 1: Non-constant forcing requiring variation of parameters, with $g(t) = \tan t$. Solve $y'' + y = \tan t$ on the interval $I = (-\pi/2, \pi/2)$ (where $\tan t$ is continuous). The homogeneous equation $y'' + y = 0$ has fundamental set $y_1(t) = \cos t$, $y_2(t) = \sin t$. Compute the Wronskian:

So $W \equiv 1 \neq 0$ on $I$, confirming a fundamental set. Undetermined coefficients cannot handle $\tan t$ (it is not a solution of any constant-coefficient homogeneous equation), so variation of parameters is the tool required. Using the formula with $t_0 = 0$:

Integrate:

Next,

so

Assemble the particular solution:

Expand and simplify:

The $\sin t \cos t$ terms cancel exactly — a useful arithmetic check. The general solution is

and since $\sin t$ already appears in the homogeneous family, one may absorb it: $y(t) = c_1 \cos t + \tilde c_2 \sin t - \cos t \ln|\sec t + \tan t|$ with $\tilde c_2 = c_2 + 1$ arbitrary. As a sanity check, differentiate $y_p = \sin t - \cos t \ln|\sec t + \tan t|$ directly:

using $\frac{\sec t(\tan t + \sec t)}{\sec t + \tan t} = \sec t$. Then

Add: $y_p'' + y_p = \big(-\sin t + \cos t\ln|\sec t + \tan t| + \tan t\big) + \big(\sin t - \cos t \ln|\sec t+\tan t|\big) = \tan t$, confirming the solution.

Example 2: A Cauchy–Euler equation with forcing $g(t) = t\ln t$ (variable coefficients, asymmetric homogeneous solutions). Solve, for $t > 0$,

First convert to standard form by dividing by $t^2$ (valid since $t > 0$):

This is a Cauchy–Euler equation; trying $y = t^r$ in $t^2y'' - 2ty' + 2y=0$ gives $r(r-1) - 2r + 2 = 0$, i.e. $r^2 - 3r + 2 = 0$, so $r = 1, 2$. Take the fundamental set

Compute the Wronskian:

nonzero for $t > 0$, confirming linear independence there. Now apply the formula with the standard-form $g(t) = t\ln t$:

Integrate by parts ($\int t \ln t\, dt = \tfrac{t^2}{2}\ln t - \tfrac{t^2}{4}$):

Next,

so, integrating by parts ($\int \ln t\, dt = t\ln t - t$),

Assemble:

Expand:

Since $-\tfrac{3}{4}t^3$ is not a homogeneous solution (the homogeneous solutions are $t, t^2$, not $t^3$), this term must remain — a good asymmetry check distinguishing this from Example 1, where a stray homogeneous term appeared and could be absorbed. Check by direct substitution. With $y_p = \tfrac{1}{2}t^3\ln t - \tfrac34 t^3$:

Now compute $t^2 y_p'' - 2t y_p' + 2 y_p$:

This matches the original right-hand side $t^3 \ln t$ exactly, confirming the solution. The general solution is