This document explains the options possible for solving a DE by the method of undetermined coefficients. The references: G.B.Gustafson, Manuscript on undetermined coefficients. The basic trial solution method is presented. Kummer's method is discussed in detail. www.math.utah.edu/~gustafso/ Edwards and Penney, Differential Equations and Linear Algebra Here, undetermined coefficients are discussed in 5.5. Especially page 341, the table of trial solutions Basic Trial solution Method: This method uses an initial or predicted trial solution based upon taking repeated derivatives of the forcing term (RHS). A fixup rule is applied to obtain a corrected or final trial solution. Then the trial solution is substituted into the differential equation to find the undetermined coefficients. The answers are inserted into the trial solution to produce the particular solution to the DE. See the manuscript above for details and examples. Table Method: This method, used in E&P, determines a trial solution from a table (page 341). The trial solution is substituted into the DE, then equations are found for the undetermined coefficients. After solving for the undetermined coefficients, the answers are inserted into the trial solution to produce the particular solution to the DE. Kummers Method: This is a method of change of variable. It is used to change exponential and sine-cosine problems into polynomial problems, and as such, it is useful for justifying the table method in E&P and the basic trial solution method in the manuscript. You are advised to learn the basic trial solution method and the table method first, and then if time allows, learn also Kummer's method. TABLE METHOD. The method has four major steps. STEP 1. Break the right side of the DE into atoms (def below). Break the given equation into several atomic DEs (same DE, atom on the right side). The particular solution yp is the sum of the answers for the atomic equations. STEP 2. Determine the PREDICTED trial solution and the CORRECTED trial solution for each atomic DE. See below for details. STEP 3. Insert the corrected trial solution y into the atomic DE and get equations for the undetermined coefficients (see below for def). STEP 4. Solve for the coefficients. Insert the answers into the trial solution to obtain the formula for the particular solution of an atomic DE. Do it for all atoms, then add the answers to obtain the particular solution yp of the original DE. Definition (ATOM). A function g(x) is called an ATOM provided it has one of the forms (where m is nonnegative, k is any real number) (1) g(x)=polynomial (2) g(x)=(polynomial)e^(kx) (3) g(x)=(polynomial)e^(kx) cos(mx) (4) g(x)=(polynomial)e^(kx) sin(mx) PREDICTED trial solution. Assume the right side of the DE is an atom. Differentiate the atom and collect all independent functions that appear, so that a linear combination of these independent functions produces each of the differentiated functions. Let the PREDICTED trial solution be a linear combination of these independent functions with coefficients d0, d1, d2, ... called UNDETERMINED COEFFICIENTS. The predicted trial solution will be CORRECTED before it is used. CORRECTED trial solution. Apply the FIXUP RULE: Let yh denote the general solutuion of the homogeneous differential equation. Multiply the predicted trial solution by x, until it no longer contains a term duplicate to a term in yh. The table in E&P page 331 writes down certain atoms and the corrected trial solution (x^s multiplies the predicted trial soluiton). Here's an example that illustrates the ideas. ===================================================== Solve y'''+16y'=2+x^3+x e^x + cos(4x) + x^2 sin(4x) ===================================================== The homogeneous solution is yh=c1+c2 cos 4x + c3 sin 4x. The atoms are the right sides of the atomic DEs y1''' + 16 y1'=2+x^3 Type (1) atom y2''' + 16 y2'=x e^x Type (2) atom y3''' + 16 y3'=cos(4x) Type (3) atom y4''' + 16 y4'=x^2 sin(4x) Type (4) atom Then yp = y1+y2+y3+y4 is the desired particular solution. PREDICTED trial solution for y1''' + 16 y1'=2+x^3 Use y1=(d0+d1 x + d2 x^2 + d3 x^3) because the derivatives of RHS=2+x^3 are all linear combinations of the independent functions 1, x, x^2, x^3. CORRECTED trial solution. Use y1=x(d0+d1 x + d2 x^2 + d3 x^3) because the homogeneous solution duplicates (term c1) a term in y1 (term d0). The predicted trial solution was multiplied by x to obtain the corrected trial solution. PREDICTED trial solution for y2''' + 16 y2'=x e^x Use y2=(d0+d1 x)e^x because derivatives of RHS=xe^x are linear combinations of the independent functions e^x and xe^x. CORRECTED trial solution. Use y2=predicted trial solution=(d0+d1 x)e^x. No fixup is needed, because the homogeneous solution yh does not duplicate terms. PREDICTED trial solution for y3''' + 16 y3'=cos(4x) Use y3=d1 cos(4x) + d2 sin(4x) because derivatives of RHY=cos 4x are linear combinations of the independent functions cos 4x and sin 4x. CORRECTED trial solution. Use y3=d1 x cos(4x) + d2 x sin(4x) because the homogeneous solution yh duplicates terms in y3. Multiply by x to eliminate the duplicates. PREDICTED trial solution for y4''' + 16 y4'=x^2 sin(4x) Use y4=(d1+d2 x + d3 x^2) cos(4x) + (d4+d5 x + d6 x^2) sin(4x) because the derivatives of RHS=x^2 sin 4x are linear combinations of the independent functions cos 4x, x cos 4x, x^2 cos 4x, sin 4x, x sin 4x, x^2 sin 4x. CORRECTED trial solution. Use y4=x(d1+d2 x + d3 x^2) cos(4x) + x(d4+d5 x + d6 x^2) sin(4x) because yh duplicates terms of y4. The predicted trial solution was multiplied by x to eliminate duplicates. UNDETERMINED COEFFICIENTS y1''' + 16 y1'=2+x^3 Type (1) atom Use y1=x(d0+d1 x + d2 x^2 + d3 x^3) y2''' + 16 y2'=x e^x Type (2) atom Use y2=(d0+d1 x)e^x y3''' + 16 y3'=cos(4x) Type (3) atom Use y3=d1 x cos(4x) + d2 x sin(4x) y4''' + 16 y4'=x^2 sin(4x) Type (4) atom Use y4=(d1 + d2 x + d3 x^2) x cos(4x) + (d4 + d5 x + d6 x^2) x sin(4x) In each atomic DE, the undetermined coefficients d0,d1,d2, ... are solved for separately. The answers are then inserted into the trial solutions giving answers for y1,y2,y3,y4. Then yp=y1+y2+y3+y4 It is normal to re-use symbols in this manner, so as not to invent new notation for each atomic DE to be solved. Don't you already re-use x in the integrand of each integration problem?