Variation of Parameters

Why undetermined coefficients fails here

The right side $g(x) = \sec x$ has derivatives that generate ever more complicated expressions: $\frac{d}{dx} \sec x = \sec x \tan x, \quad \frac{d^{2}}{dx^{2}} \sec x = \sec x (\sec^{2} x + \tan^{2} x), \ldots$

Unlike polynomials, exponentials, and trig functions, the successive derivatives of $\sec x$ don't close into a finite-dimensional span. No finite guess works. Undetermined coefficients is out; we need a more general method.

What variation of parameters does

Starting from the homogeneous solution $y_h = C_1 y_1 + C_2 y_2$, the idea is to replace the constants $C_1, C_2$ with unknown functions $u_1(x), u_2(x)$: $y_p(x) = u_1(x) \, y_1(x) + u_2(x) \, y_2(x).$

We then pick the $u_i$ to satisfy the full non-homogeneous ODE, plus one convenient extra constraint that keeps the algebra tidy.

For the ODE in standard form $y'' + p(x) y' + q(x) y = g(x)$, the $u_i$ must satisfy: $u_1' y_1 + u_2' y_2 = 0$ $u_1' y_1' + u_2' y_2' = g(x)$

This is a 2x2 linear system in $u_1', u_2'$. Its determinant is the Wronskian $W(y_1, y_2) = y_1 y_2' - y_2 y_1'$ (see #195). Cramer's rule gives $u_1' = -\frac{y_2 \, g}{W}, \qquad u_2' = \frac{y_1 \, g}{W}.$

Then integrate to get $u_1, u_2$ — and $y_p = u_1 y_1 + u_2 y_2$.

Step-by-step solution

Step 1 — Standard form

$y'' + y = \sec x.$ Already standard; $p(x) = 0$, $q(x) = 1$, $g(x) = \sec x$.

Step 2 — Homogeneous solution

$r^{2} + 1 = 0 \implies r = \pm i$, so $y_1 = \cos x$, $y_2 = \sin x$.

Step 3 — Wronskian

$W = y_1 y_2' - y_2 y_1' = \cos x \cdot \cos x - \sin x \cdot (-\sin x) = \cos^{2} x + \sin^{2} x = 1.$

(Clean: the Wronskian of $\cos$ and $\sin$ is $1$. Good omen.)

Step 4 — Compute $u_1'$ and $u_2'$

$u_1' = -\frac{y_2 \, g}{W} = -\frac{\sin x \cdot \sec x}{1} = -\frac{\sin x}{\cos x} = -\tan x$ $u_2' = \frac{y_1 \, g}{W} = \frac{\cos x \cdot \sec x}{1} = 1$

Step 5 — Integrate

$u_1 = \int -\tan x \, dx = \ln |\cos x| + C_3$ $u_2 = \int 1 \, dx = x + C_4$

The constants of integration $C_3, C_4$ only contribute multiples of $y_1, y_2$ respectively — they get absorbed into $y_h$. Drop them: take $u_1 = \ln|\cos x|$ and $u_2 = x$.

Step 6 — Assemble $y_p$ and general solution

$y_p = u_1 y_1 + u_2 y_2 = \cos x \cdot \ln|\cos x| + x \sin x.$

General solution: $\boxed{\, y(x) = C_1 \cos x + C_2 \sin x + \cos x \, \ln|\cos x| + x \sin x \,}$

Verification

Differentiate $y_p = \cos x \ln|\cos x| + x \sin x$: $y_p' = -\sin x \ln|\cos x| + \cos x \cdot \frac{-\sin x}{\cos x} + \sin x + x \cos x$ $= -\sin x \ln|\cos x| - \sin x + \sin x + x \cos x = -\sin x \ln|\cos x| + x \cos x$

$y_p'' = -\cos x \ln|\cos x| - \sin x \cdot \frac{-\sin x}{\cos x} + \cos x - x \sin x$ $= -\cos x \ln|\cos x| + \frac{\sin^{2} x}{\cos x} + \cos x - x \sin x$ $= -\cos x \ln|\cos x| + \frac{\sin^{2} x + \cos^{2} x}{\cos x} - x \sin x$ $= -\cos x \ln|\cos x| + \sec x - x \sin x$

Add $y_p$: $y_p'' + y_p = (-\cos x \ln|\cos x| + \sec x - x \sin x) + (\cos x \ln|\cos x| + x \sin x) = \sec x \; \checkmark$

Why the "extra constraint" $u_1' y_1 + u_2' y_2 = 0$?

When you compute $y_p'$ using the product rule: $y_p' = u_1' y_1 + u_1 y_1' + u_2' y_2 + u_2 y_2' = (u_1' y_1 + u_2' y_2) + (u_1 y_1' + u_2 y_2').$

Setting the first parenthetical to zero forces $y_p'$ into the cleaner form $u_1 y_1' + u_2 y_2'$, which makes $y_p''$ manageable. Without this constraint, $y_p''$ would involve $u_1''$ and $u_2''$, creating a mess. It's not a sacrifice — the constraint reduces two unknowns $(u_1, u_2)$ to satisfying just two conditions, which is exactly right.

Advantages over undetermined coefficients

• Works for any $g(x)$ (as long as you can integrate $y_i \, g / W$). No table of "acceptable" right-hand sides.
• No resonance issue — the method naturally handles it because $u_i$ are functions, not constants.
• Unified — one recipe regardless of what $g$ looks like.

Downside: you have to integrate, which can be hard or intractable in closed form. For textbook $g$'s like polynomials and exponentials, undetermined coefficients is usually faster.

General-order statement

For an $n$-th order linear ODE $y^{(n)} + \ldots = g(x)$ with homogeneous basis $\{y_1, \ldots, y_n\}$, the method generalises: solve the $n \times n$ system $u_i' y_i^{(k)}, \; k = 0, \ldots, n-1$ (the Wronskian matrix is the coefficient matrix, and its determinant is $W(y_1, \ldots, y_n)$). Cramer's rule gives each $u_i'$ as a ratio involving modified Wronskians.

Common mistakes

• Forgetting to divide the RHS into standard form. If the leading coefficient is not $1$, divide first so that $g$ means what the formula expects.
• Dropping the absolute value in $\ln|\cos x|$. The domain where $\cos x > 0$ (e.g. $|x| < \pi/2$) lets you skip it, but in general keep it.
• Confusing $y_p$'s integration constants with those in $y_h$. The integration constants from $u_1, u_2$ fold into $y_h$'s constants — don't double-count.
• Computing $u_1' y_1$ instead of $u_1 y_1$ in the final $y_p$. The integrand is $u_1'$; the thing you plug into $y_p$ is $u_1$ (the integrated version).

Try it in the visualization

Switch $g(x)$ between a $\sin$ (handled by undetermined coefficients) and a $\sec$ (handled only by variation of parameters). Watch the particular solution emerge from the Wronskian integral; overlay the homogeneous + particular decomposition so you can see each piece.