Finding a Particular Solution Using the Method of Undetermined Coefficients

The Method of Undetermined Coefficients

This method provides a systematic way to determine a particular solution to a second-order nonhomogeneous differential equation of the form $$ ay'' + by' + cy = f(x) $$ by leveraging the structure of the nonhomogeneous term \( f(x) \).

When \( f(x) \) has a recognizable form, the particular solution \( y_p \) can often be assumed to have a similar structure. This makes it much easier to guess the appropriate form of the solution and determine its coefficients.

As a result, this method is a powerful and efficient tool for solving second-order differential equations with specific right-hand sides.

Why do we need a particular solution? Once we’ve found a particular solution \( y_p \), we can construct the general solution to the nonhomogeneous equation by adding it to the general solution \( y_o \) of the associated homogeneous equation: $$ y = y_o + y_p $$

Case 1: The Nonhomogeneous Term is a Polynomial

If \( f(x) \) is a polynomial \( P_n(x) \) of degree \( n \):

$$ ay'' + by' + cy = P_n(x) $$

then the particular solution \( y_p \) is assumed to be:

f(x)	yp	Conditions
Pn(x)	A0 + A1x + A2x2 + ... + Anxn	if \( c \ne 0 \)
Pn(x)	x(A0 + A1x + A2x2 + ... + Anxn)	if \( c = 0 \), \( b \ne 0 \)
Pn(x)	x2(A0 + A1x + A2x2 + ... + Anxn)	if \( c = 0 \), \( b = 0 \)

In practice, if the \( y \) term is missing (\( c = 0 \)), the particular solution must be of degree \( n+1 \). If both \( y \) and \( y' \) are missing (\( c = 0 \) and \( b = 0 \)), we increase the degree to \( n+2 \).

Example

Let’s determine a particular solution \( y_p \) for the following second-order nonhomogeneous equation:

$$ y'' + y = x^2 $$

We apply the method of undetermined coefficients.

Given: \( f(x) = x^2 \), with \( a = 1 \), \( b = 0 \), and \( c = 1 \).

Since \( f(x) \) is a degree-2 polynomial, and \( c \ne 0 \), we assume a solution of the form:

$$ y_p = A_0 + A_1 x + A_2 x^2 $$

Next, we compute its derivatives:

$$ y_p' = A_1 + 2A_2 x \qquad y_p'' = 2A_2 $$

We now substitute \( y_p \) and \( y_p'' \) into the differential equation:

$$ y'' + y = x^2 $$

$$ 2A_2 + (A_0 + A_1 x + A_2 x^2) = x^2 $$

Grouping like terms:

$$ x^2(A_2) + x(A_1) + (A_0 + 2A_2) = x^2 $$

We match coefficients on both sides:

$$ \begin{cases} A_2 = 1 \\ \\ A_1 = 0 \\ \\ A_0 + 2A_2 = 0 \end{cases} $$

Explanation. Matching powers of \( x \), we find: - From \( x^2 \): \( A_2 = 1 \) - From \( x \): \( A_1 = 0 \) - Constant term: \( A_0 + 2A_2 = 0 \) ⟹ \( A_0 = -2 \)

Simplifying:

$$ \begin{cases} A_2 = 1 \\ \\ A_1 = 0 \\ \\ A_0 = -2 \end{cases} $$

So the particular solution is:

$$ y_p = -2 + x^2 $$

Note. To find the general solution to the equation \( y'' + y = x^2 \), we combine the particular solution \( y_p = x^2 - 2 \) with the general solution of the homogeneous equation \( y_o = c_1 \cos(x) + c_2 \sin(x) \):

$$ y = y_o + y_p = c_1 \cos(x) + c_2 \sin(x) + x^2 - 2 $$

Case 2: The Nonhomogeneous Term is Exponential

If the nonhomogeneous term is an exponential function:

$$ ay'' + by' + cy = ke^{\lambda x} $$

then the particular solution \( y_p \) depends on the relationship between \( \lambda \) and the roots of the homogeneous characteristic equation:

f(x)	yp	Conditions
keλx	A · eλx	if \( \lambda \) is not a root of the characteristic equation \( ay^2 + by + c = 0 \)
keλx	A · x · eλx	if \( \lambda \) is a simple root
keλx	A · xm · eλx	if \( \lambda \) is a root of multiplicity \( m \)

For a full worked example, see the solution to \( y'' - 2y' - 3y = e^{4x} \).

When the Nonhomogeneous Term Is a Polynomial Times an Exponential

When the nonhomogeneous term \( f(x) \) is the product of a polynomial of degree \( n \) and an exponential function:

$$ ay'' + by' + cy = P_n(x) \cdot e^{\lambda x} $$

the particular solution \( y_p \) can be written in the following form:

f(x)	yp	Conditions
Pn(x)·eλx	eλx·(A0 + A1x + A2x2 + ... + Anxn)	if \( \lambda \) is not a root of the characteristic equation \( ay^2 + by + c = 0 \)
Pn(x)·eλx	xm·eλx·(A0 + A1x + A2x2 + ... + Anxn)	if \( \lambda \) is a root of multiplicity \( m \) of the characteristic equation

For a worked example, see the solution of \( y'' - 2y' + y = 6x e^x \).

When the Nonhomogeneous Term Is a Sine or Cosine Function

If \( f(x) \) is a linear combination of sine and cosine:

$$ ay'' + by' + cy = k_1 \sin(\lambda x) + k_2 \cos(\lambda x) $$

where one of the coefficients \( k_1 \) or \( k_2 \) may be zero:

$$ ay'' + by' + cy = k_1 \sin(\lambda x) \quad \text{if } k_2 = 0 $$

$$ ay'' + by' + cy = k_2 \cos(\lambda x) \quad \text{if } k_1 = 0 $$

then the particular solution takes the form:

f(x)	yp	Conditions
k1·sin(λx) + k2·cos(λx)	A·sin(λx) + B·cos(λx)	\( \lambda \) is fixed; A and B are undetermined constants
k1·sin(λx) + k2·cos(λx)	x·[A·sin(λx) + B·cos(λx)]	if \( b = 0 \) and \( i\lambda \) is a root of the characteristic equation

For a concrete example, refer to the solution of \( y'' + 4y' + 13y = \sin(3x) \).

Note. Even if the forcing term contains only sine or only cosine, the particular solution typically includes both sine and cosine terms.

When the Nonhomogeneous Term Is a Polynomial Times Sine or Cosine

If \( f(x) \) is the product of a polynomial and a sinusoidal function, possibly modulated by an exponential factor:

$$ ay'' + by' + cy = P_n(x) e^{\alpha x} \cos(\beta x) $$

$$ ay'' + by' + cy = P_n(x) e^{\alpha x} \sin(\beta x) $$

then the corresponding particular solution is:

f(x)	yp	Conditions
Pn(x)·eαx·cos(βx)	(A0 + A1x + ... + Anxn)·eαx·cos(βx) + (B0 + B1x + ... + Bnxn)·eαx·sin(βx)	if \( \alpha + i\beta \) is not a root of the characteristic equation
Pn(x)·eαx·sin(βx)
Pn(x)·eαx·cos(βx)	x·(A0 + A1x + ... + Anxn)·eαx·cos(βx) + x·(B0 + B1x + ... + Bnxn)·eαx·sin(βx)	if \( \alpha + i\beta \) is a root of the characteristic equation
Pn(x)·eαx·sin(βx)

For a detailed example, see the solution of \( y'' - y = 2x \sin(x) \).

Superposition Principle

If the nonhomogeneous term is the sum of two distinct functions:

$$ ay'' + by' + cy = f_1(x) + f_2(x) $$

then the particular solution is simply the sum of the particular solutions corresponding to each term:

$$ y_p = y_{p1} + y_{p2} $$

where \( y_{p1} \) and \( y_{p2} \) are particular solutions of:

$$ ay'' + by' + cy = f_1(x) $$

$$ ay'' + by' + cy = f_2(x) $$

For an illustrative example, see the solution of \( y'' + 3y = x + 2\cos(x) \).

And so on.