In mathematics, the method of undetermined coefficients is an approach to finding a particular solution to certain inhomogeneous ordinary differential equations. In the method, a "guess" is made as to the appropriate form of the solution, and then the values of the coefficients of determined by solving a system of linear equations. The method of undetermined coefficients is closely related to the annihilator method, and can be viewed as a simple case of the method of variation of parameters. A similar method is sometimes used to find solutions to recurrence relations.
Suppose we wish to find a solution to the following linear inhomogeneous differential equation:
Because the inhomogeneous part is e3x, we guess (correctly) that the equation has a solution of the form
for some constant A. Substituting this guess into the original equation yields:
Therefore, one solution to the differential equation above is given by
The general solution is a sum of this particular solution with a general solution to the associated homogeneous equation (see the article on linear differential equations).
The first step in the method of undetermined coefficients is to guess the form of the particular solution. This guess is usually based on the inhomogeneous part of the equation:
Sometimes the guess listed above does not work, in which case it is necessary to multiply by a power of x. For example, one might guess that the equation
has a solution of the form
However, this is not correct, as can be seen by substituting this guess into the equation:
The correct guess is
which yields the solution
The annihilator method explains this phenomenon, and can be used to determine the correct guess in a wide variety of situations.
Consider the following linear inhomogeneous differential equation:
This is like the first example above, except that the inhomogeneous part () is not linearly independent to the general solution of the homogeneous part (); as a result, we have to multiply our guess by a sufficiently large power of x to make it linearly independent.
Here our guess becomes:
By substituting this function and its derivative into the differential equation, one can solve for A:
So, the general solution to this differential equation is thus:
![{\displaystyle {\begin{array}{cc}{\text{Inhomogeneous part}}&{\text{Guess}}\\[6pt]e^{rx}&Ae^{rx}\\\cos kx{\text{ or }}\sin kx&A\cos kx+B\sin kx\\[3pt]e^{rx}\cos kx{\text{ or }}e^{rx}\sin kx&Ae^{rx}\cos kx+Be^{rx}\sin kx\\[3pt]x^{n}&A_{n}x^{n}+A_{n-1}x^{n-1}+\cdots +A_{1}x+A_{0}\\[3pt]x^{n}e^{rx}&A_{n}x^{n}e^{rx}+A_{n-1}x^{n-1}e^{rx}+\cdots +A_{0}e^{rx}\\[3pt]x^{n}\cos kx&A_{n}x^{n}\cos kx+\cdots +A_{0}\cos kx+B_{n}x^{n}\sin kx+\cdots +B_{0}\sin kx\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a21d1200a1e908098563aa3f1735b9b98cadd06)
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![{\displaystyle =t[F_{1}(t)\cos {t}+G_{1}(t)\sin {t}]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/610673e0b7501e381202f639fbf8b70aeae39f08)
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![{\displaystyle =[2A_{0}\cos {t}+2(2A_{0}t+A_{1})(-\sin {t})+(A_{0}t^{2}+A_{1}t)(-\cos {t})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1038b7a3f95a64562e1335ec7165bd30e2bd2107)
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![{\displaystyle =[4B_{0}t+(2A_{0}+2B_{1})]\cos {t}+[-4A_{0}t+(-2A_{1}+2B_{0})]\sin {t}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40d8b7fb1d75393a528d5a6f722abad5b7435e7d)
