Formulas Cullen Zill AEM Chapters 3
ay 00 + by 0 + cy = 0
I. (Linear, Homogeneous, Constant Coefficients)
Characteristic Equation
Three Cases:
ar2 + br + c = 0
1. ∆ > 0 Real distinct roots r1 6= r2
⇒
y = erx
y = c1 er1 x + c2 er2 x
(general solution)
II. (Euler Equation, x > 0) ax2 y 00 + bxy 0 + cy = 0
⇒
ar2 + (b − a)r + c = 0
y = c1 erx + c2 xerx
y = c1 eαx cos(βx) + c2 eαx sin(βx)
3. ∆ < 0 Complet roots r = α±iβ ⇒ (general solution)
Characteristic Equation
try
has roots r1 , r2 . Discriminant: ∆ = b2 − 4ac.
⇒ (general solution)
2. ∆ = 0 Real double root r = r1 = r2
⇒
try
y = xr
has roots r1 , r2 .
Three Cases:
1. Real distinct roots r1 6= r2
2. Real double root r = r1 = r2
⇒ (general solution)
y = c1 xr1 + c2 xr2
⇒ (general solution)
3. Complet roots r = α±iβ ⇒ (general solution)
y = c1 xr + c2 ln(x)xr
y = c1 xα cos(β ln(x)) + c2 xα sin(β ln(x))
III. (Reduction of Order) Suppose that y1 is a solution of y 00 + p(x)y 0 + q(x)y = 0.
Z x
Z
1
A second solution can be found in the form y2 = y1
exp −
p(x) dx dx .
y12 (x)
IV. (Nonhomogeneous Linear)
y 00 + P (x)y 0 + Q(x)y = R(x)
yc is the general solution of
General solution:
y = yc + yp
yp is any particular solution of
and
y 00 + P (x)y 0 + Q(x)y = 0
y 00 + P (x)y 0 + Q(x)y = R(x)
There are two methods:
A. (Undetermined Coefficients) Guess the form of yp from R(x). This method requires that
P and Q to be constants and R is a sum of terms of the form xk , xk eαx , xk eαx cos(βx) or
xk eαx sin(βx) (see last page for details).
B. (Variation of Parameters) Look for a particular solution in the form yp = uy1 + vy2 .
This approach leads to
Z
yp (x) = −y1 (x)
y2 (x)R(x)
dx + y2 (x)
W (x)
Z
y1 (x)R(x)
dx,
W (x)
"
W (x) = det
#
y1 (x) y2 (x)
y10 (x) y20 (x)
A homogeneous linear differential equation with constant real coefficients of order n has the form
y (n) + an−1 y (n−1) + · · · + a0 y = 0.
(∗)
d
We can introduce the notation D =
and write the above equation as
dx
P (D)y ≡ Dn + an−1 D(n−1) + · · · + a0 y = 0.
By the fundamental theorem of algebra we can factor P (D) as
(D − r1 )m1 · · · (D − rk )mk (D2 − 2α1 D + α12 + β12 )p1 · · · (D2 − 2α` D + α`2 + β`2 )p` ,
where
k
X
j=1
mj + 2
`
X
pj = n.
j=1
There are two types of factors (D − r)k and (D2 − 2αD + α2 + β 2 )k :
The general solution of (D − r)k y = 0 is y = c1 + c2 x + · · · + ck x(k−1) erx
The general solution of (D2 − 2αD + α2 + β 2 )k y = 0 is
y = c1 + c2 x + · · · + ck x(k−1) eαx cos(βx) + d1 + d2 x + · · · + dk x(k−1) eαx sin(βx).
The general solution contains one such term for each term in the factorization.
We can also argue as before and seek solutions of (∗) in the form y = erx to get a characteristic
polynomial
rn + an−1 r(n−1) + · · · + a0 = 0.
In either case we find that the general solution consists of a sum of n expressions {yj }nj=1 where
each like xk , xk erx , xk eαx cos(βx) or xk eαx sin(βx). The yj are linearly independent and the general
solution is y = c1 y1 + c2 y2 + · · · + cn yn . In the case n = 3 the Wronskian of three functions y1 , y2 , y3
is


y1 y2 y3
W = W (y1 , y2 , y3 ) =  y10 y20 y30  .
y100 y200 y300
If the functions are solutions of a linear homogeneous ODE then the functions are linearly independent on an interval I if and only if the wronskian is not zero at a single x ∈ I (and therefore
for all x ∈ I).
Method of Undetermined Coefficints
ay 00 + by 0 + cy = f (x) (∗)
Our goal is to find the general solution of
The general solution is obtained as
y = yc + yp
where
1. yc is the general solution of the homogeneous (or complementary) problem, i.e.
yc = c1 y1 + c2 y2 where y1 , y2 are linearly independent solutions of
ay 00 + by 0 + cy = 0
with the characteristic polynomial
ar2 + br + c = 0 (†) .
2. yp is (any) particular solution of the non-homogeneous problem (∗).
The main problem then is to find yp .
The method of undetermined coefficients is only applicable if the right hand side is a sum of terms
of the following form
p(x),
p(x)eax ,
p(x)eαx cos(βx),
p(x)eαx sin(βx)
(1)
where we denote by p(x) = Cxm + · · · a general polynomial of degree m. For a sum of such terms,
yp will also be a sum of terms each of which is computed using the following
f (x) = p(x)er0 x ⇒ yp = xs (Am xm + · · · + A1 x + A0 )er0 x
1. s = 0 if r0 is not a root of the characteristic polynomial ar2 + br + c = 0 (†) .
2. s = 1 if r0 is a simple root of (†).
3. s = 2 if r0 is a double root of (†).
N.B. The above case includes the case r0 = 0 in which case the right side is p(x).

 p(x)eαx cos(βx)
or
f (x) =

αx
p(x)e sin(βx)
⇒ yp = xs (Am xm + · · · + A1 x + A0 )eαx cos(βx) + xs (Bm xm + · · · + B1 x + B0 )eαx sin(βx)
In this case r0 = α + iβ and we determine s from
1. s = 0 if r0 = α + iβ is not a root of (†).
2. s = 1 if r0 = α + iβ is a simple root of (†).
N.B. In this second case we only consider β > 0.