Method of undetermined coefficients

In mathematics, the method of undetermined coefficients, also known as the lucky guess method, is an approach to finding a particular solution to certain inhomogeneous ordinary differential equations and recurrence relations. It is closely related to the annihilator method, but instead of using a particular kind of differential operator (the annihilator) in order to find the best possible form of the particular solution, a "guess" is made as to the appropriate form, which is then tested by differentiating the resulting equation. For complex equations, the annihilator method or variation of parameters is less time consuming to perform.

Undetermined coefficients is not as general a method as variation of parameters, since it only works for differential equations that follow certain forms.^[1]

Description of the method

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

a_ny⁽ⁿ⁾ + a_{(n − 1)}y^{(n − 1)} + ... + a₁y' + a₀y = g(x).

The method consists of finding the general homogeneous solution y_c for the complementary linear homogeneous differential equation

a_ny⁽ⁿ⁾ + a_{(n − 1)}y^{(n − 1)} + ... + a₁y' + a₀y = 0,

and a particular integral y_p of the linear non-homogeneous ordinary differential equation based on g(x). Then the general solution y to the linear non-homogeneous ordinary differential equation would be

y = y_c + y_p.^[2]

If g(x) consists of the sum of two functions h(x) + w(x) and we say that y_p1 is the solution based on h(x) and y_p2 the solution based on w(x). Then, using a superposition principle, we can say that the particular integral y_p is

y_p = y_p1 + y_p2.^[2]

Typical forms of the particular integral

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 complementary function. Below is a table of some typical functions and the solution to guess for them.

Function of x	Form for y
$k e^{a x}\!$	$C e^{a x}\!$
$k x^n \mathrm{,}\; n = 0, 1, 2,\cdots\!$	$K_n x^n + K_{n-1} x^{n-1} + \cdots + K_1 x + K_0 \!$
$k \cos(a x) \;\;\mathrm{or}\;\;k \sin(a x) \!$	$K \cos(a x) + M \sin(a x) \!$
$k e^{a x} \cos(b x) \;\;\mathrm{or}\;\; ke^{a x} \sin(b x) \!$	$e^{a x} (K \cos(b x) + M \sin(b x)) \!$
$\left(\sum_{i=1}^n k_i x^i\right) e^{a x} \cos(b x) \;\;\mathrm{or}\;\; \left(\sum_{i=1}^n k_i x^i\right) e^{a x} \sin(b x)\!$	$e^{a x} \left(\left(\sum_{i=1}^n Q_i x^i\right) \cos(b x) + \left(\sum_{i=1}^n R_i x^i\right) \sin(b x)\right)$

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 two solutions linearly 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]

Examples

Example 1

Find a particular integral of the equation

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

The right side t cos t has the form

$P_n e^{\alpha t} \cos{\beta t} \!$

with n=1, α=0, and β=1.

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

$\lambda^2 + 1 = 0 \!$

we should try a particular integral of the form

$\begin{align} y_p &= t [F_1 (t) e^{\alpha t} \cos{\beta t} + G_1 (t) e^{\alpha t} \sin{\beta t}] \\ &= t [F_1 (t) \cos{t} + G_1 (t) \sin{t}]\\ &= t [(A_0 t + A_1) \cos{t} + (B_0 t + B_1) \sin{t}] \\ &= (A_0 t^2 + A_1 t) \cos{t} + (B_0 t^2 + B_1 t) \sin{t} .\\ \end{align}$

Substituting y_p into the differential equation, we have the identity

$\begin{align}t \cos{t} &= y_p'' + y_p \\ &= [(A_0 t^2 + A_1 t) \cos{t} + (B_0 t^2 + B_1 t) \sin{t}]'' \\ &\quad + [(A_0 t^2 + A_1 t) \cos{t} + (B_0 t^2 + B_1 t) \sin{t}] \\ &= [2A_0 \cos{t} + 2(2A_0 t + A_1)(- \sin{t}) + (A_0 t^2 + A_1 t)(- \cos{t})] \\ &\quad +[2B_0 \sin{t} + 2(2B_0 t + B_1) \cos{t} + (B_0 t^2 + B_1 t)(- \sin{t})] \\ &\quad +[(A_0 t^2 + A_1 t) \cos{t} + (B_0 t^2 + B_1 t) \sin{t}] \\ &= [4B_0 t + (2A_0 + 2B_1)] \cos{t} + [-4A_0 t + (-2A_1 + 2B_0)] \sin{t}. \\ \end{align}$

Comparing both sides, we have

$\begin{array}{rrrrl} &&4B_0&&=1\\ 2A_0 &&& + 2B_1 &= 0 \\ -4A_0 &&&& = 0 \\ &-2A_1 &+ 2B_0 && = 0 \\ \end{array}$

which has the solution A₀ = 0, A₁ = 1/4, B₀ = 1/4, B₁ = 0. We then have a particular integral

$y_p = \frac {1} {4} t \cos{t} + \frac {1} {4} t^2 \sin{t}.$

Example 2

Consider the following linear inhomogeneous differential equation:

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

This is like the first example above, except that the inhomogeneous part (e^x) is not linearly independent to the general solution of the homogeneous part (c₁e^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:

y_p = Axe^x.

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

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

Axe^x + Ae^x = Axe^x + e^x

A = 1.

So, the general solution to this differential equation is thus:

y = c₁e^x + xe^x.

Example 3

Find the general solution of the equation:

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

f(t), t², is a polynomial of degree 2, so we look for a solution using the same form,

y_p = At² + Bt + C, where

$\frac{d y_p}{dt} = 2 A t + B$

Plugging this particular integral with constants A, B, and C into the original equation yields,

2At + B = t² − (At² + Bt + C), where

t² − At² = 0 and − Bt = 2At and − C = B

Replacing resulting constants,

y_p = t² − 2t + 2

To solve for the general solution,

y = y_p + y_c

where y_c is the homogeneous solution y_c = c₁e ^{− t}, therefore, the general solution is:

y = t² − 2t + 2 + c₁e ^{− t}

References

1. ^ ^{a b} Ralph P. Grimaldi (2000). "Nonhomogeneous Recurrence Relations". Section 3.3.3 of Handbook of Discrete and Combinatorial Mathematics. Kenneth H. Rosen, ed. CRC Press. ISBN 0-8493-0149-1.
2. ^ ^{a b} Dennis G. Zill (2001). A first course in differential equations - The classic 5th edition. Brooks/Cole. ISBN 0-534-37388-7.