MTH2032 Cheatsheet .pdf

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MTH2032

First Order
Separable: y 0 D f .x/g.y/
Z
Z
1
dy D f .x/ dx
g.y/
Homogeneous: y 0 D q

Linear: y 0 C py D q
De

R

p dx

Substitute y D v .

 
y
x

Exact: M C Ny 0 D 0

If q.cx; cy/ D q.x; y/:
Substitute y D vx.

If
(R
R

@M
@y

D

@N
@x

then solve

M dx D X.x/ C Cy .y/
N dy D Y .y/ C Cx .x/

Second Order Linear
Non-Homogeneous: y 00 C by 0 C cy D r.x/

Homogeneous: y 00 C by 0 C cy D 0
Characteristic equation: 2 C b C cpD 0
2
Characteristic roots: 1 ; 2 D 2b ˙ b 2 4c

Undetermined coefficients
 If r.x/ D x n ,

1. If 1 ¤ 2 and 1 ; 2 2 R then
yh D c1 e

1 x

C c2 e

yp .x/ D An x n C    C A1 x C A0

2 x

 If r.x/ D ˛e kx ,

2. If 1 D 2 then
yh D c1 e

yp .x/ D Ae kx
1 x

C c2 xe

1 x

 If r.x/ D ˛ cos !x C ˇ sin !x,
3. If  D ˛ ˙ ˇi



yh D e ˛x c1 cos ˇx C c2 sin ˇx

yp .x/ D A cos !x C B sin !x
 If r.x/ D ˛e kx cos !x C ˇe kx sin !x,
yp .x/ D Ae kx cos !x C Be kx sin !x

Variation of parameters (requires yh )
1. Solution: yp D v1 y1 C v2 y2 .

If a summand of yp satisfies L.y/ D 0, then multiply
it by x s where s is the multiplicity of the
corresponding root.

2. Where v10 and v20 are given by the system:
(
v10 y1 C v20 y2 D 0
v10 y10 C v20 y20 D r

1

|

4

MTH2032

Misc
Recasting
 nth order ODEs: 
y D F x; y; y 0 ;    ; y .n 1/

Fundamental existence-uniqueness theorem I:
scalar first order ODEs

.n/

If f and @f
are continuous in R, then there is a
@y
unique solution curve through any point in R.

Let y D y0 ; y 0 D y1 ;    ; y .n
then define

Fundamental existence-uniqueness theorem II:
systems of first order ODEs

Y D y0 ; y1 ;    ; yn

Fundamental existence-uniqueness theorem III:
scalar second order linear ODEs

Wronskian Theorem
ˇ
ˇ
ˇ
ˇ
ˇy1 y2 ˇ
W .y1 ; y2 / D ˇ 0
ˇ D y1 y20
ˇy1 y20 ˇ

1

D yn




Numerical Analysis: ynC1 D yn C hˆ yn ; h
Truncation error of order 1:
ˆ.y; 0/ D f .y/

y10 y2

Truncation error of eror 2:
1

.y; 0/ D f 0 .y/f .y/
@h
2

Suppose y1 ; y2 are solutions of L.y/ D 0.
1. y1 ; y2 are LI on I iff W .y1 ; y2 / ¤ 0 8 x 2 I

Reduction of Order

2. W .x0 / D 0 H) W .y1 ; y2 /.x/ D 0 8 x 2 I

y 00 C py 0 C qy D 0, given y1 ,

Basis Theorem

1. Substitute y2 D uy1 .

1. Two LI solutions y1 ; y2 exist for L.y/ D 0.

2. Substitute v D u0 .

2. If w1 ; w2 is a pair of LI solutions, then any
solution is a linear combination of w1 and w2 .

F.x; y 0 ; y 00 / D 0
1. Substitute z.x/ D y 0 .
F.y; y 0 ; y 00 / D 0
1. Substitute z.y/ D y 0 .
2. ODE becomes F .y; z; zz 0 / D 0

2

1,

Which corresponds to the system:
8
ˆ
y00 D y1
ˆ
ˆ
ˆ
ˆ
< y 0 D y2
1
Y0 D
::
ˆ
ˆ
:
ˆ
ˆ
ˆ
: y 0 D F .x; Y /
n 1

@f
;    ; @y@f are continuous in R, then there
If f , @y
0
n 1
is a unique solution hyper-curve through any point in
R. (Y 0 .x/ D f .x; Y / and Y .x0 / D Y0 )

If p, q and r are continuous on I which contains the
IC, then there is a unique solution y D y.x/ in the
interval.

1/

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4

MTH2032

Fourier Series, Periodic Functions f .x C p/ D f .x/ 8 x


!

1
a0 X
2 nx
2 nx
f .x/ D
C
C bn sin
an cos
2
T
T
nD1
Z
2
a0 D
f .x/ dx
T 


Z
2
2 nx
an D
f .x/ cos
dx
T 
T


Z
2 nx
2
f .x/ sin
bn D
dx
T 
T
Odd functions f .x/ D

Even functions f .x/ D f . x/
Z T
2
4
f .x/ dx
a0 D
T 0


Z T
2
4
2 nx
an D
dx
f .x/ cos
T 0
T

f . x/

a0 D 0
an D 0
4
bn D
T

Z
0

T
2



2 nx
f .x/ sin
T


dx

bn D 0

Partial Differential Equations
Linear
A

@2 u
@2 u
@u
@u
@2 u
C
B
C
C
CD
CE
C F u D f .x; y/
2
2
@x
@x @y
@y
@x
@y

Semilinear
A

@2 u
@2 u
@2 u
C
B
C
C
D f .x; y; u; ux ; uy /
@x 2
@x @y
@y 2

Quasilinear
A.x; y; u; ux ; uy /

@2 u
@2 u
@2 u
C
B.x;
y;
u;
u
;
u
/
D f .x; y; u; ux ; uy /
C
C.x;
y;
u;
u
;
u
/
x
y
x
y
@x 2
@x @y
@y 2

3

|

4

MTH2032

Fisrt Order Quasilinear: Characteristics Method
a.x; y; u/ux C b.x; y; u/uy D c.x; y; u/
Requires that
dx
dy
du
D a.x; y; u/;
D b.x; y; u/;
D c.x; y; u/
d
d
d
Or

dx
dy
du
D
D
a.x; y; u/
b.x; y; u/
c.x; y; u/

Solution:
r.s; t / D .x.s;  /; y.s;  /; u.s;  //

Wave Equation
@2 u.x; t /
@2 u.x; t /
D c2
2
@t
@x 2
Solution
u.x; t / D .x C ct / C

.x

ct /

Second Order Semilinear: Canonical Solutions
Auxillary Equation:

a

Solve for

dy
dx

2


b

dy
dx


Cc D0

dy
. Then ˛y C ˇx D constant. Set v D ˛y C ˇx and compute ux ; uxx ; uy ; uyy ; uxy . Then the
dx
PDE can be reduced to canonical form by substituting v.

4

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4


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