Z
I. (RHS only contains x) y 0 = f (x)
xn+1
+ C, n 6= −1
n+1
R
R
cf (x) dx = c f (x) dx
R
R
R
(f (x) ± g(x)) dx = f (x) dx ± g(x) dx
R
cos(x) dx = sin(x) + C
R
sin(x) dx = − cos(x) + C
R
sec2 (x) dx = tan(x) + C
R
csc2 (x) dx = − cot(x) + C
R
sec(x) tan(x) dx = sec(x) + C
R
csc(x) cot(x) dx = − csc(x) + C
Z
ex dx = ex + C
Z
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Z
11.
Z
12.
Z
13.
Z
14.
Z
15.
xn dx =
f (x) dx
⇒
Z
f (y) dy −
y 0 + P (x)y = Q(x)
µ=e
Z x
µ(t)Q(t) dt + C
y = 1/µ(x)
III. (First Order Linear)
Gen. Sol.
IV. (RHS Linear in x and y)
V. (RHS only contains y/x)
R
P dx
is separable.
y 0 = f (y/x)
v = y/x ⇒ xv 0 + v = f (v)
VI. (Bernoulli)
g(x) dx = C
y 0 = f (ax + by + c)
v = ax + by + c ⇒ v 0 = a + bf (v)
is separable.
y 0 + P (x)y = Q(x)y n , n 6= 0, 1
v = y 1−n ⇒
v 0 + (1 − n)P (x)v = (1 − n)Q(x) is First Order Linear.
ax dx = ln(a)ax + C
VII. M (x, y) dx + N (x, y) dy = 0
1
dx = sin−1 (x) + C
1 − x2
f (x) =
1
√
dx = sec−1 (x) + C
|x| x2 − 1
fy − N
ex
M
e
N
⇒ µ=e
IX. (Integrating factor)
g(y) =
fy − N
ex
M
f
M
is exact if
R
f (x) dx
f(x, y) dx + N
e (x, y) dy = 0
M
⇒ µ = e−
R
g(y) dy
I. (Linear, Homogeneous, Constant Coefficients)
⇒
Characteristic Equation
try
y = erx
ar2 + br + c = 0
1. Real distinct roots r1 6= r2
⇒
⇒ (general solution)
II. (Euler Equation, x > 0) ax2 y 00 + bxy 0 + cy = 0
Characteristic Equation
ar2 + (b − a)r + c = 0
1. Real distinct roots r1 6= r2
⇒
2. Real double root r = r1 = r2
3. Complet roots r = α ± iβ
y = c1 er1 x + c2 er2 x
⇒ (general solution)
2. Real double root r = r1 = r2
3. Complet roots r = α ± iβ
has roots r1 , r2 .
(general solution)
⇒
y = c1 erx + c2 xerx
y = c1 eαx cos(βx) + c2 eαx sin(βx)
try
y = xr
has roots r1 , r2 .
y = c1 xr1 + c2 xr2
(general solution)
⇒ (general solution)
⇒ (general solution)
∂N
∂M
=
.
∂y
∂x
f(x, y) dx + N
e (x, y) dy = 0
M
VIII. (Integrating factor)
1
dx = tan−1 (x) + C
1 + x2
ay 00 + by 0 + cy = 0
y=
Z
II. (Separable) f (y)y 0 = g(x)
dx
= ln(x) + C
x
√
⇒
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. y2 = y1
IV. (Nonhomogeneous Linear)
y 00 + P (x)y 0 + Q(x)y = R(x)
yh is the general solution of
General solution:
Z
1
exp
−
y12 (x)
Z
x
p(x) ds dx .
y = yh + 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)
Two methods to find yp :
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 p(x), p(x)eαx , p(x)eαx cos(βx) or p(x)eαx sin(βx) where p(x) = Cxm + · · · is a
polynomial of degree m.
ay 00 + by 0 + cy = 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)
00
0
or
ay + by + cy =

p(x)eαx sin(βx)
⇒ yp = xs (Am xm + · · · + A1 x + A0 )eαx cos(βx)
+xs (Bm xm + · · · + B1 x + B0 )eαx sin(βx)
1. s = 0 if r0 = α + iβ is not a root of (†).
2. s = 1 if r0 = α + iβ is a simple root of (†).
B. (Variation of Parameters)
Z
Z
y2 (x)R(x)
y1 (x)R(x)
yp (x) = −y1 (x)
dx + y2 (x)
dx,
W (x)
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.
We can introduce the notation D =
(∗)
d
and write the above equation as
dx
P (D)y ≡ Dn + an−1 D(n−1) + · · · + a0 y = 0.
P (D) = (D − r1 )m1 · · · (D − rk )mk (D2 − 2α1 D + α12 + β12 )p1 · · · (D2 − 2α` D + α`2 + β`2 )p` ,
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).
∞
Z
f (t) for t ≥ 0
e−st f (t) dt
fb =
0
1
eat
t
n
1
s
I.
n!
ta
sin bt
cos bt
where Degree(P2 ) < Degree(Q).
Γ(a + 1)
(a > 0)
sa+1
s2
b
+ b2
s2
s
+ b2
II.
P (s)
A1
A2
An
=
+
+ ··· +
Q(s)
(s − r1 ) (s − r2 )
(s − rn )
sL(f ) − f (0)
f 00 (t)
s2 L(f )−sf (0)−f 0 (0)
tn f (t)
(−1)n
u(t − a)f (t − a)
u(t − a)g(t)
δ(t − a)
(f ∗ g)(t) =
f (t − τ )g(τ ) dτ
III.
dn F
(s)
dsn
Repeated Linear Factors
If Q(s) contains a factor of the form (s − r)m then you
must have the following terms
A2
Am
A1
+
+ ··· +
2
(s − r) (s − r)
(s − r)m
L(f )(s − a)
e−as
s
IV.
A Nonrepeated Quadratic Factor
If Q(s) contains a factor of the form (s2 − 2αs + α2 + β 2 ) =
(s − α)2 + β 2 then you must have the following term
A1 s + B 1
2
(s − 2αs + α2 + β 2 )
e−as L(f )
e−as L(g(t + a))
e−as
t
Z
Nonrepeated factors
If Q(s) = (s − r1 )(s − r2 ) · · · (s − rn ) and ri 6= rj for i 6= j
f 0 (t)
eat f (t)



 0 t≤a
u(t − a) =


 1 t>a
P (s)
P2 (s)
= P1 (s) +
Q(s)
Q(s)
(n = 0, 1, . . .)
sn+1
Degree P (s) > Degree Q(s) In this case first carry
out long division to obtain
1
s−a
L(f ∗ g) = L(f )L(g)
V.
Repeated Quadratic Factors
0
Z
t
f (τ ) dτ
0
If f (t + T ) = f (t) for all t
then L(f ) =
1
L(f )
s
i.e., f is T -periodic
R
T
0
e−sτ f (τ ) dτ
1 − e−T s
If Q(s) contains a factor of the form (s2 − 2αs + α2 + β 2 )m
then you must have the following terms
Am s + B m
A1 s + B1
+ ··· + 2
(s2 − 2αs + α2 + β 2 )
(s − 2αs + α2 + β 2 )m