Variation of parameters

Where it fits. Variation of parameters is presented in the MA10001 ODE unit as the general-purpose alternative to undetermined coefficients: reach for it when $R(x)$ is not polynomial, exponential, or sinusoidal, or when the ODE has non-constant coefficients. It leans entirely on the complementary function already found from the auxiliary-equation method, so that earlier topic is a hard prerequisite here.

The setup. Write the ODE in standard form, leading coefficient equal to one:

where: $P(x), Q(x)$ are the coefficient functions and $R(x)$ is the forcing term after dividing through by whatever multiplied $y''$ originally. Suppose the complementary function is already known,

where: $y_1, y_2$ are two linearly independent solutions of the homogeneous equation $y'' + Py' + Qy = 0$. The method looks for a particular integral of the same shape but with the constants promoted to unknown functions,

Two unknown functions need two equations. One comes from substituting $y_p$ into the ODE; the other is a free choice — imposed purely to keep the algebra from spiralling into second derivatives of $u_1, u_2$. The standard choice is $u_1'y_1 + u_2'y_2 = 0$. Together with the equation the ODE forces, this yields the pair

The picture below shows the shift in strategy this represents: undetermined coefficients hands you a fixed trial shape to match against $R(x)$; variation of parameters instead lets the "constants" of the complementary function flex into functions, solved for by the system above.

Solving the $2\times2$ system by Cramer's rule, using the Wronskian $W = y_1y_2' - y_2y_1'$ as the determinant, gives closed forms for the two derivatives:

Integrating each gives $u_1(x)$ and $u_2(x)$; the particular integral is then $y_p = u_1y_1 + u_2y_2$, and the general solution is $y = y_c + y_p$, exactly as with undetermined coefficients. Nothing here required $R(x)$ to have any particular shape — that is the entire point of the method.

Problem. Solve $y'' + y = \sec x$ using variation of parameters.

Step 1. Find the complementary function. The auxiliary equation is $m^2+1=0 \implies m = \pm i$, giving $y_1 = \cos x$, $y_2 = \sin x$, and $y_c = C_1\cos x + C_2\sin x$. Undetermined coefficients cannot be used here since $\sec x$ is not a standard trial type — this is exactly the signal to reach for variation of parameters.

Step 2. Compute the Wronskian: $W = y_1y_2' - y_2y_1' = \cos x(\cos x) - \sin x(-\sin x) = \cos^2x+\sin^2x = 1$.

Step 3. Compute $u_1'$ and $u_2'$ using the boxed formulas, with $R(x)=\sec x$: $u_1' = -y_2R/W = -\sin x\sec x = -\tan x$, and $u_2' = y_1R/W = \cos x\sec x = 1$.

Step 4. Integrate: $u_1 = \int -\tan x\,dx = \ln|\cos x|$, and $u_2 = \int 1\,dx = x$.

Step 5. Assemble the particular integral: $y_p = u_1y_1+u_2y_2 = \cos x\ln|\cos x| + x\sin x$.

Step 6. State the general solution: $\boxed{y = C_1\cos x + C_2\sin x + \cos x\ln|\cos x| + x\sin x}$.

Competitive stretch

The same machine handles repeated roots and non-trigonometric $R(x)$ together. For $y''-2y'+y = e^x/x$ ($x>0$), the auxiliary equation $(m-1)^2=0$ gives the repeated-root complementary function $y_1=e^x$, $y_2=xe^x$, with Wronskian $W = e^x(e^x+xe^x)-xe^x(e^x) = e^{2x}$. Then $u_1' = -xe^x\cdot(e^x/x)/e^{2x} = -1$ and $u_2' = e^x\cdot(e^x/x)/e^{2x} = 1/x$, so $u_1=-x$ and $u_2=\ln x$. That gives $y_p = -xe^x + xe^x\ln x = xe^x(\ln x-1)$, and the general solution is $\boxed{y=(C_1+C_2x)e^x + xe^x(\ln x-1)}$ — note the $-xe^x$ piece could be absorbed into $C_2xe^x$ but is left explicit for clarity of derivation.

• Skipping the auxiliary condition $u_1'y_1+u_2'y_2=0$. That condition is what turns one second-order ODE into a solvable pair of first-order equations for two unknowns; omit it and the system is underdetermined.
• Forgetting to reduce to standard form first. If the given equation is $a(x)y''+b(x)y'+c(x)y=R_0(x)$ with $a(x)\neq1$, divide through by $a(x)$ before reading off $R(x)=R_0(x)/a(x)$ — using $R_0$ directly gives a wrong particular integral.
• Sign slip between $u_1'$ and $u_2'$. Remember $u_1'=-y_2R/W$ carries a minus sign while $u_2'=y_1R/W$ does not; swapping the signs is the single most common arithmetic error in this method.
• Reaching for undetermined coefficients on a non-standard $R(x)$ such as $\sec x$, $\tan x$, $\ln x$, or $1/x$ instead of recognising that only variation of parameters applies to these forcing terms.

Ask the tutor: if the Wronskian setup or the sign in $u_1'=-y_2R/W$ keeps tripping you up, ask the AI tutor to walk through a fresh variation-of-parameters problem with you line by line.