§ 微分方程式(Differential Equation:D.E.)
D.E.:凡是含有微分的方程式。
分類:
常微分方程式(僅含有一自變數的 D.E.)
:偏微分方程式(具有一個以上自變數的 D.E.)
D. E. :
Ordinary D E O D E Partial D E P D E Total D E T D E
. . ( . . .) . .( . . .) . .( . . .)
R S|
T|
:全微分方程式(化成全微分形式的 D.E.):
Examples :
O D E y x
P D E
T D E e x dx x d y xydz
d dx
u x
u y y
. . . : sin . . . :
. . . : cos sin
=
+ =
+ +
R S|
T|
∂
∂
∂
∂
2 2
2
2 0
=0
( * P.D.E.中,必存在偏微分算符:’∂’ )
與 D.E.相關的定義:
(1) 階(order):D.E.中,因變數的最高微分階數。
ex. 2y′ +3y=4x xy′′ +2xy′ + =y 0 (y′′′ + ′′ + =)2 (y )4 y 0 (2) 次(degree):將 D.E.化成有理整式後,最高階導函數的次方。
其中有理整式:因變數及其導函數的次方均為正整數。
ex. y = ′ y
(3) 線性(linear):D.E.各項中,因變數及其導函數的次方和不大於 1(≦1);
或 D.E.各項中,因變數或其導函數至多出現一次(可以沒有)。 ex.
′′ + = + ′ =
′′ + ⋅ = ∂
∂ + ∂
∂ =
y y y y
x y x y e u
y y u
x x
x
0 0
0 2
sin
sin sin
(4) 齊次(homogeneous):D.E.各項中,均含有因變數或其導函數;
y c c x
y k x k x a line y k x not a line
= = ⋅
= = ⋅
UV W
= ⋅
0
1
2 :
或沒有不含因變數的項。( a trivial solution : y = 0 ) ex. ′′ + = ′ + = ′′ + ′ + =
′ + = ′ + =
y xy e y y x x y xy y
y y y y
3 0 x 4 4
1 0
2
2
cos ( ) cos
0 D.E.的解(以O.D.E.為例)
若將函數 y=y(x)代入 D.E.中,且能滿足該 D.E.,則稱 y=y(x)為此 D.E.的解。
(由此定義可得到解的驗證法:將函數直接代入 D.E.中)
ex.
S
If , then is the solution, where is an arbitrary constant.
= and can substitute for
is the solution of
D E y y y ce c
y ce y ce y y D E
y y ce ce
y ce D E y y
x
x x
x x
x
. .: '
, . .,
( ) ( )
. .: .
+ = =
∴ ′ = − ′
′ + = − + =
∴ = ′ + =
R ||
T |
|
−
− −
− −
−
2 0
2
2 2 2 0
2 0
2
2 2
2 2
2
Q
(一般而言,n 階 D.E.其通解中,會含有 n 個任意常數)
微分方程式的解:
(1) 形式上:顯解(explicit solution) => y=f(x), u=u(x,y).
(因變數與自變數的關係可以明確分開表示)
隱解(implicit solution) => G(x,y)=0, (x, y 的關係隱含在方程式中)
ex. explicit sol.: y=c x1 +c0; implicit sol.: sin(x+y)+exy =0 (2) 種類:
通解( general sol.):n 階 D.E.的解中,含有 n 個任意的獨立常數者,稱之為通解 或原函數。
特解( particular sol.):經由指定通解中任意常數的值而獲得的解。
奇解( singular sol.):仍是 D.E.的解,但不能經由指定通解中任意常數的值而得到者。
ex. ( )y′ +2 x y′ − =y 0 : 通解:y = c x+c2 (只有一個常數 c) 特解:y = x+1 (set c=1)
奇解:y = - x2/4
(令 p = y’,並將上述 D.E.對 x 微分、化簡)
[*] 直接積分求解 D.E.
1) 求解因變數與自變數的關係,即得出 x, y 的關係。
2) 使用的積分關係:
a f x d x d f x f x C
b f x d x f x d f x d x f x
d dx
d dx
) ( ) ( ) ( )
) ( ) ( ) ( ) ( )
z z
z
z
== ⇒= + = dx( )1 d y
dx =e−x ( )4 y′ =e2xsin3x ( )2 cos4
2
2
d y
d x = x ( )5 y′′ = x
( )3 d 2 0 d x x d y
dx
L NM O
QP
= ( )6 d xd x y⋅ ′ =2[*] 變數可分離型
1) 設 u = u(x,y)為 x, y 的函數,且可寫成 u = f(x)·h(y),
則稱 u(x,y)為變數可分離函數。
1 2 3 4 5
6 1
) 2
) )
) ln(
) sin(
)
u x y u x y u e
u xy
u x
u y
x y
=
= +
=
=
=
= +
+
) ) y
2) 變數可分離型 D.E.求解原則:將 x, y 先分開,再分別積分。
( )1 xd y+yd x= ( )0 7 xd y−yd x = 0 ( ) (2 3x y2 −xy d x) +(2x y3 2+x y3 4)d y=0 ( ) (8 x+1)y′ +(y+ = 1) 0
( ) (3 x2+1)y′ + +(1 y2)= 0 ( ) (9 y 1+x dx) +x(1+y d y) = 0
( ) sin( ) ( )
4 1 2 2
′ = +
′= ⋅ +
y x
y y x e
y
x y
; (10) y′=ex y+
( )5 x y′ + = ∧y 1 y( )1 =3, find y x( ). ( )11 x y′ − = ∧y 1 y( )2 =1, find y x( ). ( )6 y′ =(y−x)2 , [Hint:set u= −y x ] (12 x y) ′ =e−x y −y, [Hint:set u=xy]
[*] 活用微分公式 1) 基本微分公式:
*1) , .
*2) ( )
*3)
*4)
( )
, .
*5) ( ) ( ) ( )
*6)
d C C is a constant
d u u du du
d f g g d f f d g
du du du
d u u u
du u u u u C the solution
f x d x d f x d x easy but important d x y
x y
m x y n x y
n
n u
n
m n
m n
m n m n
=
+ = +
⋅ = ⋅ + ⋅ + + ⋅⋅⋅⋅⋅ + =
+ + ⋅⋅⋅⋅⋅ + =
= ⇒ = + + ⋅⋅⋅⋅⋅ + =
=
= ⋅ + ⋅
z
− −
− −
0
0 0 0
1 2 1 2
1 2
1 2
1 2
1 1
1 1
1442443
c h
x y m yd x n xd y d x y
x y
m n
m n
m n
− −
− −
⇒ ⋅ + ⋅ =
1 1
1 1
c h
2) 利用微分公式解 D.E.原則:先乘開再觀察合併。
Examples :
( )1 xd y+yd x= 0 ( )2 ′ = +
y y−x y x
( ) (3 3x2+ycos )x d x+(sinx−4y dy3) = 0 ( ) (4 yex+ey)d x+(ex+xey)dy= 0 ( )5 xy′ + + = 0 y 4
( ) (6 2xtan )y dx+sec2y d y= 0 ( ) (7 y3+2x d x) +(3xy2+1)d y= 0 ( )8 y(1+x d x) +x(1+y d y) = 0 ( )
* *
9 1
2 2
x y y integrating factor d y
x
xd y ydx
x d x
y
yd x xd y y
′ − = →
F HG I
KJ
= −F
HG I
KJ
= −[*]一階線性 D.E. D E y P x y Q x
P x Q x are both the functions of x when Q x is homogeneous
Q x is nonhomogeneous . .: ( ) ( ) [ ]
( ), ( ) .
( ) [ ]
( ) [ ]
′ + =
= →
RS
≠ →T
L A
A A 0
0
公式推導:
D E y P x y Q x d y
d x P x y Q x d y P ydx Q x d x
I I x I d y I P ydx I Qd x
I d y y d I I Qdx
d I y I Qd x I y I Qdx C y
I I Qd x C I(x)
I P yd x y d I y
we hope
. .: ( ) ( ) ( ) ( )
( )
( )
,
′ + = ⇒ + =
+ =
⇒ ⋅ + ⋅ = ⋅
⇒ ⋅ + ⋅ = ⋅
⇒ ⋅ = ⋅ ⇒ ⋅ = ⋅ + ⇒ = ⋅ ⋅ +
⋅ = ⋅
E
z z
if it exists an integrating factor = ( ), and then multiplies above D. E.,
Does function really exist ?
expected both sides divided by , we have 1442443
1
dI
I =Pd x ⇒ elnI =e
z
Pdx ⇒ I =ez
P x dx( ) =expb z
P x d x( )g
=I x( )解題步驟:1) 化成標準式: y P 的係數為 1。
2) 令 I e 3)
′ + ( )x y=Q x( ) ⇒ ′y P x d x
=exp
b z
( )g
P x dx
=
z
( )∴ =y 1I ⋅
z
I Q d x⋅ +CExamples :
(0) y′-y= (1) 1 y′ + =y x (7) y′ +2y=sin3x (2) y′ +2ytanx=sinx and y( )0 = 1 (8) d y
dt + =y e−t (3) d y
dt −2y=cos t (9) y′+2 coty x=cos x (4) x y′+(1+ )x y=ex (10) x y′+ 2y= x
(11) * ′ =
⋅ −
y y x
1 2 sin (6) D. E. :
( ) is a continuous function, find ( ).
* ′ + = = ≤ ≤
RS
<y y g x g x
T
xx x
y x y x
2 1 0 1
( ), ( ) , 1
, (12) D. E. :
( ) is a continuous function, find ( ).
* ′ + = = ≤ ≤
RS
<y y g x
T
xx
y x y x
( ) , ,
1 0 1
0 1 (5) *
c
ey+ ⋅ ′ =x yh
1[*] 電路上的應用
[*] 運算子、算符(Operator)
設 O1、O2、O3均為線性運算子
(1) 運算子的作用順序(靠近的先作用)
(O1O2O3) y = (O1O2)(O3 y)= O1[O2(O3 y)]
ex.
assume D
x D y x D y x y D x y D x y x y y
d
≡ dx
⋅ = = ′
⋅ = = ′ +
1 2 )
)
b g b g
b g b g
(2) ( O1+ O2 ) y = O1 y + O2 y
ex. (x+D y) =( )x y+( )D y=x y+ ′y (3) linear operator L
L k y k L y
L y⋅ = ⋅y L y L y k is an arbitrary constant
+ = +
RS T
1 2 1 2 .(4) 一般而言,運算子不具交換性。
( O1O2 ) ≠ ( O2O1)
但 a 為常數時,則有:(D a)=(a D) & (DD)=(DD)= D2 (5) 運算子的相等
若 L1[y] = L2[y],則稱 L1 ≡ L2。
或說:對函數 y 而言,L1 ≡ L2,即 L1,L2具相同的作用。
(*雖然 L1 ≡ L2,但未必具有相同的形式)
ex.
D x
x
x x
for x x D
D e e D
D e e D
x x
n x n x n n
2 2
2 2
c h c h
b g b g
c h c h
sin cos
sin cos sin , cos :
, , ω
ω ω ω
ω
ω ω ω
λ λ
λ λ
λ λ
λ λ
RS T UV W
= −RS T UV W
≡ −
= ≡
= ≡
(6)
if D d
d x D d
d x
a D D D D
b D D D D D
c D D D D D D D
k k
k
s r r s
r s s r r s
r s t r s r
≡ ≡
+ = +
= =
+ = +
+
, ( )
( )
( )
c h
t*D.E.的運算子表示法
D. E. : + ( ) + ( ) = ( ) + ( ) + ( ) = ( ) + ( ) + ( )
+ ( ) + ( )
( , )
′′ ′
=
⇒ = =
y p x y q x y r x D y p x Dy q x y r x
D p x D q x y r x
L x D y r x L x D D p x D q x
L x D 2
2
2
144424443 ( )
( , ) ( ) , ( , ) .
[*] 線性相依與線性獨立 線性組合:
若有 n 個已知函數:u x1( ),u x2( ),L,u xn( ),
g
n 個任意常數:c c1, 2,L,cn,
則 c u x1 1( )+c u x2 2( )+L+c u xn n( ) (注意沒有等號) 稱為
b
u u1, 2,L,un 的線性組合。線性相依:
對於方程式:c1 1y +c y2 2+L+c yn n =0 ( )A
其中c c1, 2 ,L,cn為 n 個待求常數,y y1, 2, ...,yn為 n 個任意已知函數 顯然 c1=c2 =L=cn =0為其一解,
若除了上述情形外,至少存在一 ck ≠ 0不為零,使得(A)式仍成立,於是有
yk c c y c y c y c y c y
k
k k k k n n
= −
⋅ + + + − − + + + + + 1
1 1 2 2 L 1 1 1 1 L
則稱
b
y y1, 2,L,yng
彼此為線性相依。g
或者說:
b
y y1, 2,L,yn 函數中的某一個,能夠以其餘函數的線性組合來表示。(*但不必每一個函數都用到) 線性獨立:
但如果(A)式僅存在c1 =c2 =...=cn =0的解,
則稱
b
y y1, 2, ....,yng
彼此為線性獨立。或者說:函數
b
y y1, 2, ....,yng
中的任一個,都不能夠以其餘函數的線性組合表示。*判斷下列函數為線性相依或獨立
( ) ,
( ) cos , cos , sin
( ) ,
( ) , ,
1 4
2 2
3
4 4
1 2
1 2
2 3
2
1 2
2
2
u x u x
y x y x y
u x u x
y x y x y x
= =
= = =
= =
= = =
1 2 3
x
[*] Wronsky 行列式
1) 設
b
u x u x1( ), 2( ),LL,u xn( )g
為 n 個已知函數,則稱W u x u x u x
u u u u
u u u u
u u u u
n
n
n
n n n
n n
1 2
1 2 3
1 2 3
1 1
2 1
3
1 1
( ), ( ), , ( )
( ) ( ) ( ) ( )
LL L L
M M M O M
L
b g
= ′ ′ ′
− − −
′
−
x 為
b
u x u x1( ), 2( ),LL,u xn( )g
的 Wronsky 行列式。2) 若W u ,則
b
為線性相依;若W u
b
1( ),x u2,LL,ung
=0,則b g
為線性獨立。x u un
1( ), 2,LL, 0
b g
≠ u x u1( ), 2,LL,ung
u x u1( ), 2,LL,un
Examples:( ) 使用 Wronsky 行列式判斷。
( ) ,
( ) ,
,
( ) , ,
( ) cos , cos , sin
1 4
2 3
4 4
5 2
1 2
1 2
2
1 2
1 2 3
2
1 2
2 3
2
1 2
u x u x
u x u x
u e u e
y x y x y x
y x y x y
x x
= =
= =
= =
= = =
= = =
λ λ
(*)結論:若是線性相依: y k ;
或是若
y W y y y y
y y
k y y k y y
1 2 1 2
1 2
1 2
2 2
2 2
= ⋅ ∴ = 0
′ ′ = ⋅
⋅ ′ ′ = , ( , )
y y1 k
2
= :常數,則( ,y y1 2)為線性相依。
[*] 二階常係數齊次 D.E.
D E y A y B y A B are both constants
D AD B y
. . : ′′ + ′ + = , ,
+ + =
0
2 0
.
我們假設解具 y=eλx的形式,代入原 D.E.,可得 特性方程式:(characteristic or auxiliary equation)
λ2 +Aλ+ =B 0 一元二次方程式 其中 λ1 2 2 4
, = − ±A 2A − B
0
) 上述方程式的根有下述三種可能的組合:
(1) 相異實根:λ1≠λ2 ⇒ y= c e1 λ1x+c e2 λ2x (2) 相等實根:λ1=λ2 =λ ⇒ y= eλ0x(c0+c x1 )
(3) 共軛虛根:λ1= +p jq,λ2 = −p jq ⇒ y= ep x(a⋅cosqx b+ ⋅sinqx
Examples :
( )1 y′′ + ′ +5y 6y=0 , I C y. .: ( )0 =16. , y′( )0 =0. ( )2 y′′ −4y′ +4y=0 , I C y. .: ( )0 =1, y′( )0 =4. ( )3 y′′ −2y′ +10y=0 , I C y. .: ( )0 =4, y′( )0 =1. ( )4 y′′ −2y=0
( )5 y′′ +4y=0 ( )6 y′ =0 ( )7 y′′ =0 ( )8 y′′ +3y′ =0
[*] 高階常係數齊次 D.E.
D E a y a y a y a y
a D a D a D a y
L D y
L chracteristic equation or auxiliary equation
n n
n n
n n
n n
L D
. . :
( )
( ) : .
( ) ( )
( )
+ + + ′ + =
+ + + + =
=
⇒ =
− −
− −
1 1
1 0
1 1
1 0
0 0 0
0
LL
14444442LL4444443
λ
若 L( )λ = 0為特性方程式:
(1) L( )λ =
b
λ λ− 1gb
λ λ− 2g b
L λ λ− ng
∴ λ λ λ= 1, 2,L,λn 為互不相等的單根 ⇒ y=c e1 λ1x+c e2 λ2x+LL+c en λnx (2) L( )λ =
b
λ λ− 0g
k∴ λ λ= 0 為 k 次重根
⇒ y=eλ0x⋅
c
c0 +c x1 +c x2 2 +LL+ck−1xk−1h
(3) L( )λ =
b
λ− +p jqg b
⋅ λ− −p jqg
k∴ λ p jq p jq= + , − 的 k 次重根
⇒ = +
+ ⋅ +
+ ⋅ +
+ − ⋅ − + −
y e A qx B qx
x e A qx B qx
x e A qx B qx
x e A qx B qx
p x
p x
p x
k p x
k k
0 0
1 1
2
2 2
1
1 1
cos sin cos sin
cos sin
cos sin
b g
b g
b g
b g
M
Examples:
( )1
b
D+1gc
D2+5D+6h
y= 0 ( )2b
D−1gc
D2 −4D+13h
y= 0 ( )3b
D−1g b
3 D+2g
2 y= 0( )4
b
D+4gc
D2 +8D+17hd
D+ 3i
3 y= 0 ( )5c
D2+2D+2h
2y=0[*] 二階常係數線性非齊次 D.E.
D E. .:y′′ +a y′ +b y=r x( ) ⋅⋅⋅⋅⋅ A [ ] 其中 a,b 均為常數。
( )1 yh: yh′′+a yh′ +b yh =0 yh
p
:含兩個任意常數 ( )2 yp: y′′ +p a yp′ +b yp =r x( ) y :不含任意常數 則 y = yh + yp
齊次解 非齊次解
零輸入響應 零態響應
自然響應 受激(激勵)響應
主要的暫態部分 穩態部分
求解二階常係數線性非齊次 D.E. [A]式的步驟:
1) 先求齊次解:yh
2) 設法找到非齊次解:yp
3) 最後得到全解:y = yh + yp
未定係數法:(求解常係數 D.E.的非齊次解 yp) (*) 未定係數法的適用情形:
(1) 常係數 D.E.
(2) 若 r(x)經任意次微分後,僅出現有限的可能項 ex. : ea x, x2, cosωx, tan , lnx x
(*) 如何假設 yp (與 r(x)有關) 規則:
1) r(x) 經任意次微分後,所出現的有限可能項為 u1(x), u2(x), u3(x), …… , un(x) ( known )
則設: yp為(u1(x), u2(x), u3(x), …… , un(x))的線性組合,
即 yp = c1 u1(x) + c2 u2(x) + c3 u3(x) + ……+ cn un(x)
其中(c1 , c2 ,c3 , …. , cn)為等待決定的係數,故稱之為未定係數法。
2) 接著將 yp代入原 D.E.,比較係數以決定(c1 , c2 ,c3 , …. , cn)的值。
(*)
r(x) yp應假設的形式 K yp = A
eax yp = ⋅A eax
xn yp = A0+A x1 +A x2 2+ ⋅⋅⋅ + A xn n cosqx
sin qx yp = ⋅A cosqx+ ⋅B sinqx
給定的e 形式 λx 對應的λ 值
eax a
1 0
e q
e q
px
px
cos sin
x x
q p± j cosqx
sin qx ± jq xk⋅
c h
eax a 的(k+1)次重根r x u x u x u x
assume linear combination
y c u x c u x c u x
where c c c are waiting to be determinated
D
n
p n n
n
( ) k ( ) , ( ) , , ( )
( ) ( ) ( )
, , ,
→ ⋅⋅⋅⋅⋅
E
= + + ⋅⋅⋅⋅⋅ +
⋅⋅⋅
1 2
1 1 2 2
1 2 .
(*) 假設 yp 時的修正原則: (與 r(x),yh的特性根有關) 1) 若 r(x) = r1(x) + r2(x),
r1(x):yp1
r2(x):yp2 , 則 yp = yp1 + yp2
2) 若 r(x) = r1(x)× r2(x),
r1(x):yp1
r2(x):yp2 , 則 yp = yp1 × yp2
3) 若 r x( )=ea x,且λ = a亦為 y 的 k 次重根, h 即 yh =ea x
c
c0+c x1 +c x2 2+ ⋅⋅⋅ +ck−1xk−1h
, 則須設: yp =eax⋅xk例如:(合併同類項或係數)
r x e y A e
r x x y b b x
if r x r x r x y y y A e b b x
if r x r x r x
y y y A e b b x e c c x
where c A b c A b
a x
p
a x
p
p p p
a x
p p p
a x a x
1 1 1
2 2 0 1
1 2 1 2 1 0
1 2
1 2 1 0 1 0 1
0 1 0 1 1 1
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( )( ) ( )
,
= → =
= → = +
= + ∴ = + = + +
= ⋅
∴ = ⋅ = + = +
= =
1
if r x x
r x y a
r x x y b b x
y y y a b b x A A x where A a b A
p
p
p p p
( ) ( )
( )
, ;
= +
= → =
= → = +
∴ = + = + + = + = + =
1
1 1 1 0
2 2 0 1
1 2 0 0 1 0 1 0 0 0 1 b .1
if y e c c x a x b x j
r x e x
r x e y c e x
r x x j y A x B x x
y y y c e x A x B x x
h x
x
x
p
x
p
p p p
x
= + + ⋅ + ⋅ → = ±
= +
= = → = ⋅
= = ± → = ⋅ + ⋅
= + = ⋅ + ⋅ + ⋅ ⋅
2
0 1
2
1
2
1
2 2
2 2
1 2
2 2
3 3 2 2
3 2
3 3 3 3
3 3
( ) ( cos sin ) , ,
( ) sin
( ) ( )
( ) sin cos sin
cos sin λ
λ λ
b g c h b g
c h b g
⋅ 3
b g b g
Examples :
( )1 y′ + =y sinx ( )2 y′′ +3y′ +2y=e x
( )3 y′ −2y=x2 , I C. . : y( )0 =1/2
Examples :
( )1 y′′ +4y′ +3y=ex+5 ( )2 y′ +2y=e−2x
( )3 y′′ − ′ +3y 2y=3e2x ( )4 D D2
b
−1g
y= +3 e−x ( )5 y′′ +2y′ +2y=cosx ( )6 y′′ +9y=sin3x[*] 聯立方程式
( )1 ( ) , ( ).
d x
dt x y d y
dt x y
find x t y t
= +
= −
R S ||
T ||
( )2 ( ) , ( )
d x 2 dt
d y dt d x
dt d y
dt e
find x t y t
t
+ =
− =
R S ||
T ||
− .( )1 y′′ +2y′ −3y=0 , I.C.: y( )0 =0,y′( )0 =2 y= 12
c
ex−e−3xh
( )2 y′′ −6y′ +9y=0 , I.C.: y( )0 =1,y′( )1 = −5 ye3 =e3x(1 2− x)
( )3 y′′ +2y′ +5y=0 , I.C.: y( )0 =1,y′( )0 =1 y=e−x
b
cos2x+sin2xg
( )1 D. E.: D
b
+2gc
D2−4D+3h
y= 0 y=c e1 − x+c ex+c e x 22 3
3
( )2 D. E.: D
d
− 2ic
D2 +4D−1h
y=0 y=c e1 2x+c e2 (− +2 5)x+c e3 (− −2 5)x ( )3 D. E.: Db
+1 3gc
D2+5D−2h
y=0 y=c e1 −x+c e2 13x+c e3 −2x( )4 D. E.: D
b
+3gc
D2 +10D+25h
y= 0 y=c e1 − x+e− x d +d x3 5
0 1
b g
( )5 D. E.: D
b
+4gc
D2 +8D+20h
y= 0 y c ee a x b
x
x
=
+ ⋅ + ⋅
−
− 1
4
4
b
cos2 sin2xg
( )6 D. E.: D
b
−4g b
2 D+6g
3 y=0 y e c c xe d d x d x
x
x
= +
+ − + +
4
0 1
6
0 1 2
2
b g
c h
( )7 D. E.: D
c
2 −2D+2hc
D2+6D+13h
y=0 y e a x b xe a x b
x
x
= +
+ − +
1 1
3
2 2 2 2
cos sin cos sin
b g
b
xg
( )8 D. E.: D
d
+ 3i c
3 D2 −4D+7h
y=0 y e c c x c xe a x b x
x
x
= + +
+ +
− 3
0 1 2
2
2 3 3
c h
d
cos sini
( )9 D. E.: D
c
2+4D+13h
2 y=0 y e a x b xxe a x b x
x
x
= +
+ +
−
− 2
0 0
2
1 1
3 3
3 3
cos sin cos sin
b g
b g
(1) y′′ −8y=60e2x y c e c e
y Ae e
h
x x
p
x x
= +
= = −
− 1
2 2 2
2 2
2 2
15
(2) y′′ +4y′ +3y=5sin 2 x y c e c e
y x x
h
x x
p
= +
= ⋅ +
− −
−
1 2
3
1
13 8 cos2 sin2
(3) y′′ +4y′ +4y=3 y e c c x
y A
h x
p
= +
= =
−2
0 1
3 4
b g
(4) y′′ +4y=sin3x y a x b x
y A x B x x
h
p
= +
= + = −
0 0
1 5
2 2
3 3 3
cos sin
cos sin sin
(5) y′′ −3y′ +2y=ex−e−x y c e c e
y Ae x Be xe e
h
x x
p
x x x x
= +
= ⋅ + − = − − −
1 2
2
1 6
(6) D
c
2−8D+16h
y=3e4x y e c c xy Ae x x e
h x
p
x x
= +
= ⋅ =
4
0 1
4 2 3
2 2 4
b g
a x d t
t x
special cases
b x y x y
x
y x y
x r x
x
r
) ln ( )
ln( ) , ln( ) , ln( ) ) ln( ) ln ln
ln( ) ln ln ln( ) ln
= >
= = −∞ ∞ =
= +
= −
= ⋅
z
1 01 0 0
:
+ ∞
a e e e j
b e x
c e y y y
e y y y
j
x
y
y
) . , , cos si
) exp ( )
) exp (ln )
ln ln (exp( ))
ln
1 0
2 718281828 1
≈ = =
=
= ⇔ =
= ⇔ =
θ θ θ
+ n
sin( ) sin cos cos sin sin( ) sin cos cos sin cos( ) cos cos sin sin cos( ) cos cos sin sin sin cos sin( ) sin( ) cos sin sin( ) sin( ) cos cos cos( ) cos( ) sin sin cos( ) cos( )
( ) / (
A B A B A B
A B A B A B
A B A B A B
A B A B A B
A B A B A B
A B A B A B
A B A B A B
A B A B A B
assume x A B
y A B then A x y
B x
+ = ⋅ + ⋅
− = ⋅ − ⋅
+ = ⋅ − ⋅
− = ⋅ + ⋅
⋅ = + + −
⋅ = + − −
⋅ = + + −
⋅ = − + − −
= +
RS
= −T
== +1 2 1 2 1 2 1 2
2
RS
+T
+ = ⋅ +
⋅ −
− = ⋅ +
⋅ −
+ = ⋅ +
⋅ −
− = − ⋅ +
⋅ −
y
x y x y x y
x y x y x y
x y x y x y
x y x y x y
) / sin sin sin cos
sin sin cos sin cos cos cos cos cos cos sin sin
2
2 2 2
2 2 2
2 2 2
2 2 2
sin sin cos
cos cos sin cos sin
cos cos
, sin cos
2 2
2 2 1
1 2
2
1 2
2
2 2 2
2 2
θ θ θ
θ θ θ θ
θ θ θ θ
=
= − = − = −
= +
= −
1 2 2θ
d
dx f x g x d
dx f x d dxg x d
dx f g g d
dx f f d
dxg f g f g f
f g
f g f g
g g x
y y u u u x dy
dx dy du
du dx
( ) ( ) ( ) ( )
( )
( ) ( )
± = ±
⋅ = ⋅ + ⋅ ⇔ ⋅ ′= ′⋅ + ⋅ ′
L NM O QP
′ = ′ ⋅ − ⋅ ′ ⇒ ≠
= = =
2 0
chain rule : if and then
g
⋅ d F x
dx f x f x dx F x C
d
dx f t dt f x a d
dx f x dx f x d f x dx f x dx f x dx d f x dx
d f x
dx dx f x C d f x f x C C
y y x d y y d x u u x y du u
x d x u y d y u dv u v v du
a x
( ) ( ) ( ) ( )
( ) ( ) , :
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( : )
( ) ( , )
= ⇒ = +
=
= => =
=
= + => = +
= = ′ ⋅
= = ∂
∂ ⋅ + ∂
∂ ⋅
R S|
T|
⋅ = ⋅ − ⋅
z z
z z
z z
z
z z
constant.
integral constant
( integral by parts )
( ) ( )
d
dx f x t dt f x b x db
d x f x p x d p
d x x f x t dt
p x b x
p x
( , ) , ( ) , ( ) b x ( , )
( ) ( )
z
=b g
−b g
+z
∂∂d C dx d x
dx n x
n n
=
= −
0
1
d e
dx e d e
dx a e d a
dx a a
x
x chain rule
a x
a x
x
x
= → =
=ln ⋅
d x
dx x d x
dx x
d x
dx x d x
dx x
d x
dx x d x
dx x
d x
dx x d x
dx x
d x
dx x x d x
dx x x
d x
dx x x d x
dx x
chain rule
chain rule
chain rule
chain rule
chain rule
chain rule
sin cos sin
cos
cos sin cos
sin
tan sec tan
sec
cot csc cot
csc
sec sec tan sec
sec tan
csc csc cot csc
csc
= → = ⋅
= − → = − ⋅
= → = ⋅
= − → = − ⋅
= ⋅ → = ⋅ ⋅
= − ⋅ → = − ⋅ ⋅
ω ω ω
ω ω ω
ω ω ω
ω ω ω
ω ω ω ω
ω ω ω
2 2
2 2
cotω x
d x
dx x
d x
dx x
d x
dx x
d x
dx x
d x
dx x
d x
dx x
d x
dx x
d x
dx x
d x
dx x x
d x
dx x
chain rule
chain rule
chain rule
chain rule
chain rule
sin sin
( )
cos cos
( )
tan tan
( )
cot cot
( )
sec sec
(
− −
− −
− −
− −
= − → =
−
= −
− → = −
−
= + → =
+
= −
+ → = −
+
= − → =
2 2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
1
1 1
1
1 1
1
1 1
1
1 1
1 1
1
ω ω
ω
ω ω
ω
ω ω
ω
ω ω
ω ω
ω ω
ω x
d x
dx x x
d x
dx x x
chain rule
)
csc csc
( )
2
1
2
1
2
1 1
1
1 1
−
= −
− → = −
−
− −
* Taylor’s series expansion :
f x f a
n x a
f a f a
x a f a
x a f a
x a f(x) is an infinitely differentiable function
some important expansions :
e x
n
x x x
x x
n x x x x
x
n
n
n
x
n
n
n n
n
( ) ( )
! ( )
( ) ( )
! ( ) ( )
! ( ) ( )
! ( ) .
! ! ! !
sin ( )
( )! ! ! !
cos (
( )
( )
= ⋅ −
= + ′ − + ′′ − + − +
= = + + + + ⋅⋅⋅⋅⋅
= −
+ = − + − + ⋅⋅⋅⋅⋅
=
=
∞
=
∞
+
=
∞
∑
∑
∑
0
2 3
3
0
2 3
2 1
0
3 5 7
1 2 3
1 1 2 3
1
2 1 3 5 7
LL
− = − + − + ⋅⋅⋅⋅⋅
+ = + + −
+ − −
+
=
∑
∞ 12 1 2 4 61 1 1
2
1 2
3
2
0
2 4 6
2 3
)
( )! ! ! !
! !
n n
n
p
x n
x x x
x px p p
x p p p
b g b g b gb g
xLL
Binomial formula
x y C x y C x y
Leibniz s formula d
d x f g f g C f g
where C n k
n k n k
n
k n
k n
k n k
k n
k n
n k k
n
n
n
k n
k n
k n k
k n
:
' :
!
! !
( ) ( ) ( )
+ = =
⋅ = ⋅ =
=
F HG I
KJ
= −=
−
=
−
=
−
∑ ∑
∑
b g
b g b g
b g
0 0
0
x dx n x C n dx
x x C
n = n
+ + ≠
= +
z
+z
1
1 1 ( 1)
ln
e dx a e C
a dx a a C
a x a x
x x
= ⋅ +
= ⋅ +
z z
1 1 ln 1 1
1 1 1
1
1 1 1
1
1 1
2
1
2
1
2
1
2
1
2
1
2
1
− = + −
− = +
+ = + −
+ = +
− = + −
− = +
z z
z z
z z
− −
− −
− −
x
dx x C
x
dx x C
x dx x C
x dx x C
x x
dx x C
x x
dx x C
sin cos
tan cot
sec csc
e bx dx e
a b a bx b bx C
e bx dx e
a b a bx b bx C
a x
a x
a x
a x
cos cos sin
sin sin cos
= + ⋅ + ⋅ +
= + ⋅ − ⋅ +
z z
2 2
2 2
sin cos
cos sin
tan ln sec sec tan
cot ln sin csc cot
sec ln sec tan sec tan
csc ln csc cot csc cot
sin cos
cos sin
tan ln sec cot ln sin
x dx x C
x dx x C
x dx x C x dx x C
x dx x C x dx x C
x dx x x C x dx x C
x dx x x C x dx x C
x dx x C
x dx x C
x dx x C
x dx
z z
z z
z z
z z
z z
z z z z
= − +
= +
= + ⇒ = +
= + ⇒ = − +
= + + ⇒ = +
= − + + ⇒ = − +
= − ⋅ +
= ⋅ +
= +
=
2
2
2
2
1 1 1
1 1 1
ω ω ω
ω ω ω
ω ω ω
ω ω ω
ω ω ω
ω ω ω x C
x dx x x C
x dx x x C
+
= + +
= − + +
z z
sec ln sec tan
csc ln csc cot
ω ω ω ω
ω ω ω ω
1 1