Method of undetermined coefficients

Consider a linear non-homogeneous ordinary differential equation of the form

${\displaystyle \sum _{i=0}^{n}c_{i}y^{(i)}+y^{(n+1)}=g(x)}$

where ${\displaystyle y^{(i)}}$ denotes the i-th derivative of ${\displaystyle y}$, and ${\displaystyle c_{i}}$ denotes a function of ${\displaystyle x}$.

The method of undetermined coefficients provides a straightforward method of obtaining the solution to this ODE when two criteria are met:[2]

1. ${\displaystyle c_{i}}$ are constants.
2. g(x) is a constant, a polynomial function, exponential function ${\displaystyle e^{\alpha x}}$, sine or cosine functions ${\displaystyle \sin {\beta x}}$ or ${\displaystyle \cos {\beta x}}$, or finite sums and products of these functions (${\displaystyle {\alpha }}$, ${\displaystyle {\beta }}$ constants).

The method consists of finding the general homogeneous solution ${\displaystyle y_{c}}$ for the complementary linear homogeneous differential equation

${\displaystyle \sum _{i=0}^{n}c_{i}y^{(i)}+y^{(n+1)}=0,}$

and a particular integral ${\displaystyle y_{p}}$ of the linear non-homogeneous ordinary differential equation based on ${\displaystyle g(x)}$. Then the general solution ${\displaystyle y}$ to the linear non-homogeneous ordinary differential equation would be

${\displaystyle y=y_{c}+y_{p}.}$[3]

If ${\displaystyle g(x)}$ consists of the sum of two functions ${\displaystyle h(x)+w(x)}$ and we say that ${\displaystyle y_{p_{1}}}$ is the solution based on ${\displaystyle h(x)}$ and ${\displaystyle y_{p_{2}}}$ the solution based on ${\displaystyle w(x)}$. Then, using a superposition principle, we can say that the particular integral ${\displaystyle y_{p}}$ is

${\displaystyle y_{p}=y_{p_{1}}+y_{p_{2}}.}$[3]

In order to find the particular integral, we need to 'guess' its form, with some coefficients left as variables to be solved for. This takes the form of the first derivative of the complementary function. Below is a table of some typical functions and the solution to guess for them.

Function of x	Form for y
${\displaystyle ke^{ax}\!}$	${\displaystyle Ce^{ax}\!}$
${\displaystyle kx^{n},\;n=0,1,2,\ldots \!}$	${\displaystyle \sum _{i=0}^{n}K_{i}x^{i}\!}$
${\displaystyle k\cos(ax){\text{ or }}k\sin(ax)\!}$	${\displaystyle K\cos(ax)+M\sin(ax)\!}$
${\displaystyle ke^{ax}\cos(bx){\text{ or }}ke^{ax}\sin(bx)\!}$	${\displaystyle e^{ax}(K\cos(bx)+M\sin(bx))\!}$
${\displaystyle \left(\sum _{i=0}^{n}k_{i}x^{i}\right)\cos(bx){\text{ or }}\ \left(\sum _{i=0}^{n}k_{i}x^{i}\right)\sin(bx)\!}$	${\displaystyle \left(\sum _{i=0}^{n}Q_{i}x^{i}\right)\cos(bx)+\left(\sum _{i=0}^{n}R_{i}x^{i}\right)\sin(bx)}$
${\displaystyle \left(\sum _{i=0}^{n}k_{i}x^{i}\right)e^{ax}\cos(bx){\text{ or }}\left(\sum _{i=0}^{n}k_{i}x^{i}\right)e^{ax}\sin(bx)\!}$	${\displaystyle e^{ax}\left(\left(\sum _{i=0}^{n}Q_{i}x^{i}\right)\cos(bx)+\left(\sum _{i=0}^{n}R_{i}x^{i}\right)\sin(bx)\right)}$

If a term in the above particular integral for y appears in the homogeneous solution, it is necessary to multiply by a sufficiently large power of x in order to make the solution independent. If the function of x is a sum of terms in the above table, the particular integral can be guessed using a sum of the corresponding terms for y.[1]

Example 1

Find a particular integral of the equation

${\displaystyle y''+y=t\cos t.}$

The right side t cos t has the form

${\displaystyle P_{n}e^{\alpha t}\cos {\beta t}}$

with n = 2, α = 0, and β = 1.

Since α + iβ = i is a simple root of the characteristic equation

${\displaystyle \lambda ^{2}+1=0}$

we should try a particular integral of the form

${\displaystyle {\begin{aligned}y_{p}&=t\left[F_{1}(t)e^{\alpha t}\cos {\beta t}+G_{1}(t)e^{\alpha t}\sin {\beta t}\right]\\&=t\left[F_{1}(t)\cos t+G_{1}(t)\sin t\right]\\&=t\left[\left(A_{0}t+A_{1}\right)\cos t+\left(B_{0}t+B_{1}\right)\sin t\right]\\&=\left(A_{0}t^{2}+A_{1}t\right)\cos t+\left(B_{0}t^{2}+B_{1}t\right)\sin t.\end{aligned}}}$

Substituting y_p into the differential equation, we have the identity

${\displaystyle {\begin{aligned}t\cos t&=y_{p}''+y_{p}\\&=\left[\left(A_{0}t^{2}+A_{1}t\right)\cos t+\left(B_{0}t^{2}+B_{1}t\right)\sin t\right]''+\left[\left(A_{0}t^{2}+A_{1}t\right)\cos t+\left(B_{0}t^{2}+B_{1}t\right)\sin t\right]\\&=\left[2A_{0}\cos t+2\left(2A_{0}t+A_{1}\right)(-\sin t)+\left(A_{0}t^{2}+A_{1}t\right)(-\cos t)+2B_{0}\sin t+2\left(2B_{0}t+B_{1}\right)\cos t+\left(B_{0}t^{2}+B_{1}t\right)(-\sin t)\right]\\&\qquad +\left[\left(A_{0}t^{2}+A_{1}t\right)\cos t+\left(B_{0}t^{2}+B_{1}t\right)\sin t\right]\\&=[4B_{0}t+(2A_{0}+2B_{1})]\cos t+[-4A_{0}t+(-2A_{1}+2B_{0})]\sin t.\end{aligned}}}$

Comparing both sides, we have

${\displaystyle {\begin{cases}1=4B_{0}\\0=2A_{0}+2B_{1}\\0=-4A_{0}\\0=-2A_{1}+2B_{0}\end{cases}}}$

which has the solution

${\displaystyle A_{0}=0,\quad A_{1}=B_{0}={\frac {1}{4}},\quad B_{1}=0.}$

We then have a particular integral

${\displaystyle y_{p}={\frac {1}{4}}t\cos t+{\frac {1}{4}}t^{2}\sin t.}$

Example 2

Consider the following linear nonhomogeneous differential equation:

${\displaystyle {\frac {dy}{dx}}=y+e^{x}.}$

This is like the first example above, except that the nonhomogeneous part (${\displaystyle e^{x}}$) is not linearly independent to the general solution of the homogeneous part (${\displaystyle c_{1}e^{x}}$); 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:

${\displaystyle y_{p}=Axe^{x}.}$

By substituting this function and its derivative into the differential equation, one can solve for A:

${\displaystyle {\frac {d}{dx}}\left(Axe^{x}\right)=Axe^{x}+e^{x}}$

${\displaystyle Axe^{x}+Ae^{x}=Axe^{x}+e^{x}}$

${\displaystyle A=1.}$

So, the general solution to this differential equation is:

${\displaystyle y=c_{1}e^{x}+xe^{x}.}$

Example 3

Find the general solution of the equation:

${\displaystyle {\frac {dy}{dt}}=t^{2}-y}$

${\displaystyle t^{2}}$ is a polynomial of degree 2, so we look for a solution using the same form,

${\displaystyle y_{p}=At^{2}+Bt+C,}$

Plugging this particular function into the original equation yields,

${\displaystyle 2At+B=t^{2}-(At^{2}+Bt+C),}$

${\displaystyle 2At+B=(1-A)t^{2}-Bt-C,}$

${\displaystyle (A-1)t^{2}+(2A+B)t+(B+C)=0.}$

which gives:

${\displaystyle A-1=0,\quad 2A+B=0,\quad B+C=0.}$

Solving for constants we get:

${\displaystyle y_{p}=t^{2}-2t+2}$

To solve for the general solution,

${\displaystyle y=y_{p}+y_{c}}$

where ${\displaystyle y_{c}}$ is the homogeneous solution ${\displaystyle y_{c}=c_{1}e^{-t}}$, therefore, the general solution is:

${\displaystyle y=t^{2}-2t+2+c_{1}e^{-t}}$

• Boyce, W. E.; DiPrima, R. C. (1986). Elementary Differential Equations and Boundary Value Problems (4th ed.). John Wiley & Sons. ISBN 0-471-83824-1.
• Riley, K. F.; Bence, S. J. (2010). Mathematical Methods for Physics and Engineering. Cambridge University Press. ISBN 978-0-521-86153-3.
• Tenenbaum, Morris; Pollard, Harry (1985). Ordinary Differential Equations. Dover. ISBN 978-0-486-64940-5.
• de Oliveira, O. R. B. (2013). "A formula substituting the undetermined coefficients and the annihilator methods". Int. J. Math. Educ. Sci. Technol. 44 (3): 462–468. Bibcode:2013IJMES..44..462R. doi:10.1080/0020739X.2012.714496. S2CID 55834468.