Math 308, Sections 301, 302,
Summer 2008
Lecture 10.
06/20/2008
Section 4.8 Method of Undetermined Coefficients
In this section, we give a simple procedure for finding a particular
solution to the equation
ay ′′ + by ′ + cy = g (x),
(1)
when the nonhomogeneous term g (x) is of a special form
g (x) = eαx (Pm1 (x) cos βx + Qm2 (x) sin βx),
where
Pm1 (x) = p0 x m1 + p1 x m1 −1 + p2 x m1 −2 + . . . + pm1 −1 x + pm1
is a polynomial of degree m1 and
Qm2 (x) = q0 x m2 + q1 x m2 −1 + q2 x m2 −2 + . . . + qm2 −1 x + qm2
is a polynomial of degree m2 , α, β ∈ R.
To apply the method of undetermined coefficients, we first have to
solve the auxiliary equation for the corresponding homogeneous
equation
ar 2 + br + c = 0
Let α = β = 0, then
g (x) = p0 x m + p1 x m−1 + p2 x m−2 + . . . + pm−1 x + pm1 .
We seek a particular solution of the form
yp (x) = Ax m + Bx m−1 + Cx m−2 + . . . + Dx + F ,
if r = 0 is not a root to the auxiliary equation. Here A, B, C , D,
and F are unknown numbers.
If r = 0 is one of two roots of the auxiliary equation, then the
particular solution is
yp (x) = x(Ax m + Bx m−1 + Cx m−2 + . . . + Dx + F )
To find unknowns A, B, C ,..., D, and F , we have to substitute
yp (x), yp′ (x), and yp′′ (x) into equation (1). Set the corresponding
coefficients from both sides of this equation to each other to form
a system of linear equations with unknowns A, B, C ,..., D, and F .
Solve the system of linear equation for A, B, C ,..., D, and F .
Let α 6= 0, β = 0, then
g (x) = eαx (p0 x m + p1 x m−1 + p2 x m−2 + . . . + pm−1 x + pm ).
We seek a particular solution of the form
yp (x) = eαx (Ax m + Bx m−1 + Cx m−2 + . . . + Dx + F ),
if r = α is not a root to the auxiliary equation. Here A, B, C , D,
and F are unknown numbers.
If r = α is one of two roots of the auxiliary equation, then the
particular solution is
yp (x) = eαx x(Ax m + Bx m−1 + Cx m−2 + . . . + Dx + F )
If r = α is a repeated root to the auxiliary equation, then the
particular solution is
yp (x) = eαx x 2 (Ax m + Bx m−1 + Cx m−2 + . . . + Dx + F )
Example 1. Find the general solution to the following equations
(a) y ′′ − 2y ′ = 2e−2x
(b) y ′′ − 4y ′ + 4y = 16x 2 e2x
Let α = 0, β 6= 0, then
g (x) = (p0 x m1 + p1 x m1 −1 + p2 x m1 −2 + . . . + pm1 −1 x + pm1 ) cos βx+
+(q0 x m2 + q1 x m2 −1 + q2 x m2 −2 + . . . + qm2 −1 x + qm2 ) sin βx.
We seek a particular solution of the form
yp (x) = (A0 x m + A1 x m−1 + A2 x m−2 + . . . + Am−1 x + Am ) cos βx+
+(B0 x m + B1 x m−1 + B2 x m−2 + . . . + Bm−1 x + Bm ) sin βx,
if i β is not a root to the auxiliary equation. Here
m = max{m1 , m2 }, A0 ,...,Am , B0 ,..., Bm are unknowns.
If i β is one of two roots of the auxiliary equation, then the
particular solution is
yp (x) = x(A0 x m + A1 x m−1 + A2 x m−2 + . . . + Am−1 x + Am ) cos βx+
+x(B0 x m + B1 x m−1 + B2 x m−2 + . . . + Bm−1 x + Bm ) sin βx =
Example 3. Find the solution to the initial value problem
y ′′ − 3y ′ + 2y = 4x sin x,
y (0) = 3, y ′ (0) = 2.
Let α 6= 0, β 6= 0, then
g (x) = eαx [(p0 x m1 +p1 x m1 −1 +p2 x m1 −2 +. . .+pm1 −1 x+pm1 ) cos βx+
+(q0 x m2 + q1 x m2 −1 + q2 x m2 −2 + . . . + qm2 −1 x + qm2 ) sin βx].
We seek a particular solution of the form
yp (x) = eαx [(A0 x m +A1 x m−1 +A2 x m−2 +. . .+Am−1 x+Am ) cos βx+
+(B0 x m + B1 x m−1 + B2 x m−2 + . . . + Bm−1 x + Bm ) sin βx],
if α + i β is not a root to the auxiliary equation. Here
m = max{m1 , m2 }, A0 ,...,Am , B0 ,..., Bm are unknowns.
If α + i β is one of two roots of the auxiliary equation, then the
particular solution is
yp (x) = eαx x[(A0 x m +A1 x m−1 +A2 x m−2 +. . .+Am−1 x+Am ) cos βx+
+x(B0 x m + B1 x m−1 + B2 x m−2 + . . . + Bm−1 x + Bm ) sin βx] =
Example 4. Find a general solution to the equation
y ′′ − 9y = e3x cos x.
Particular solutions to ay ′′ + by ′ + cy = g (x)
Type g (x)
I
II
III
IV
V
VI
VII
xk
x k−1
p0 + p1
+ . . . + pk
deαx
eαx (p0 x k + ... + pk )
d cos βx + f sin βx
Pm1 (x) cos βx+
+Qm2 (x) sin βx
eαx (d cos βx + f sin βx)
eαx [Pm1 (x) cos βx+
+Qm2 (x) sin βx]
yp (x)
x s (Ax k + Bx k−1 + ... + F )
x s Aeαx
x s eαx (Ax k + Bx k−1 + . . . + F )
x s (A cos βx + B sin βx)
x s {(A0 x m + ... + Am ) cos βx+
+(B0 x m + ... + Bm ) sin βx}
x s eαx (A cos βx + B sin βx)
x s eαx [(A0 x m + ... + Am ) cos βx+
+(B0 x m + ... + Bm ) sin βx]
In this table s = 0, when α + i β is not a root to the auxiliary
equation, s = 1, when α + i β is one of two roots to the auxiliary
equation, and s = 2, when β = 0 and α is a repeated root to the
auxiliary equation; m = max{m1 , m2 },
Pm1 (x) = p0 x m1 + p1 x m1 −1 + p2 x m1 −2 + . . . + pm1 −1 x + pm1 ,
Qm2 (x) = q0 x m2 + q1 x m2 −1 + q2 x m2 −2 + . . . + qm2 −1 x + qm2
Example 5. Using Table, find the form for a particular solution
yp (x) to
y ′′ + y ′ − 2y = g (x),
where g(x) equals
(a) g (x) = (2x 2 + 3)e−2x ,
(b) g (x) = x sin 2x,
(c) g (x) = ex + cos 3x.