Recipe

Apply variation of parameters to a higher-order equation

Use when: you have a nonhomogeneous linear equation of order 3 or higher, already in standard form, with a known fundamental set y_1,\ldots,y_n for the homogeneous equation

Goal: a particular solution y_p=u_1y_1+\cdots+u_ny_n, found by solving one n\times n linear system and integrating

Taught in Variation of parameters at higher order

1. 1

   Seek y_p=u_1y_1+\cdots+u_ny_n, where each u_i is now a function to be found instead of a constant.

   Example

   For y'''=6, the fundamental set solving y'''=0 is y_1=1, y_2=x, y_3=x^2 (a repeated root r=0 of multiplicity 3, Section 2's recipe), so y_p=u_1(x)(1)+u_2(x)(x)+u_3(x)(x^2)

   Why: this is the same substitution Chapter 9 used with two functions — turning the constants of the homogeneous solution into unknown functions is what leaves room for a forcing term.

2. 2

   For k=0,\ldots,n-2, require \sum_i u_i'y_i^{(k)}=0 — the derivative-k versions of each fundamental-set function, weighted by the still-unknown u_i', sum to zero.

   Example

   Here n=3, so k=0,1. For k=0: u_1'(1)+u_2'(x)+u_3'(x^2)=0. For k=1, using y_1'=0,y_2'=1,y_3'=2x: u_2'+2xu_3'=0.

   Why: without these conditions, substituting y_p back into the equation drags in every derivative of every u_i, not just u_i' — the conditions are chosen specifically to cancel that mess.

3. 3

   The last equation matches the (n-1)th derivative to g(x): \sum_i u_i'y_i^{(n-1)}=g(x). Stack it with the n-1 convenience conditions into one linear system in u_1',\ldots,u_n'.

   Example

   Here n-1=2, so with y_1''=0,y_2''=0,y_3''=2: 2u_3'=6. All three equations together, in matrix form: \begin{bmatrix}1&x&x^2\\0&1&2x\\0&0&2\end{bmatrix}\begin{bmatrix}u_1'\\u_2'\\u_3'\end{bmatrix}=\begin{bmatrix}0\\0\\6\end{bmatrix}

   Why: this coefficient matrix is exactly the Wronskian matrix — its determinant is the Wronskian itself, nonzero for a genuine fundamental set, which is what guarantees the system has a unique solution.

4. 4

   The matrix is lower triangular here, so solve from the bottom row up; in general, Cramer's rule solves for each u_i' as a ratio of determinants.

   Example

   From row 3: 2u_3'=6\Rightarrow u_3'=3. From row 2: u_2'+2x(3)=0\Rightarrow u_2'=-6x. From row 1: u_1'+(-6x)(x)+3(x^2)=0\Rightarrow u_1'=6x^2-3x^2=3x^2.

   Why: this turns the abstract linear system into explicit formulas for each u_i' in terms of x, ready to integrate.

5. 5

   Integrate every u_i' from Step 4 to get u_i, then combine as y_p=\sum_iu_iy_i.

   Example

   u_3'=3\Rightarrow u_3=3x. u_2'=-6x\Rightarrow u_2=-3x^2. u_1'=3x^2\Rightarrow u_1=x^3. Then y_p=u_1(1)+u_2(x)+u_3(x^2)=x^3-3x^3+3x^3=x^3

   Why: this is the actual particular solution the whole system was built to produce — checking, y_p'''=6, exactly the forcing term this example started from.