.. p.2/5

Similar documents
数学演習:微分方程式

() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y

( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +

December 28, 2018

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

chap1.dvi

高等学校学習指導要領

高等学校学習指導要領

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

DE-resume

取扱説明書 [N-03A]

1 1 x y = y(x) y, y,..., y (n) : n y F (x, y, y,..., y (n) ) = 0 n F (x, y, y ) = 0 1 y(x) y y = G(x, y) y, y y + p(x)y = q(x) 1 p(x) q(

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

08-Note2-web

2 1 κ c(t) = (x(t), y(t)) ( ) det(c (t), c x (t)) = det (t) x (t) y (t) y = x (t)y (t) x (t)y (t), (t) c (t) = (x (t)) 2 + (y (t)) 2. c (t) =

Z: Q: R: C: sin 6 5 ζ a, b

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y



x x x 2, A 4 2 Ax.4 A A A A λ λ 4 λ 2 A λe λ λ2 5λ + 6 0,...λ 2, λ 2 3 E 0 E 0 p p Ap λp λ 2 p 4 2 p p 2 p { 4p 2 2p p + 2 p, p 2 λ {

201711grade1ouyou.pdf

(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)

II K116 : January 14, ,. A = (a ij ) ij m n. ( ). B m n, C n l. A = max{ a ij }. ij A + B A + B, AC n A C (1) 1. m n (A k ) k=1,... m n A, A k k

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.

4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t

<4D F736F F D B B BB2D834A836F815B82D082C88C602E646F63>

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d

I 1

9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 =

= M + M + M + M M + =.,. f = < ρ, > ρ ρ. ρ f. = ρ = = ± = log 4 = = = ± f = k k ρ. k

取扱説明書[N906i]

曲面のパラメタ表示と接線ベクトル

平成17年度 マスターセンター補助事業

24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x

TCSE4~5

m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)

Part () () Γ Part ,

II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

重力方向に基づくコントローラの向き決定方法

振動と波動

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

2

ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +

Chap10.dvi

pdf


v_-3_+2_1.eps


genron-3

DVIOUT

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s


入試の軌跡

K E N Z OU

( ) ) ) ) 5) 1 J = σe 2 6) ) 9) 1955 Statistical-Mechanical Theory of Irreversible Processes )

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

LCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t)

meiji_resume_1.PDF

Note.tex 2008/09/19( )

5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

Gmech08.dvi

6. Euler x

) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)

x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin

< F31332D817992B48DC A8CCB8E9F81458CA28E942E6A7464>

1 yousuke.itoh/lecture-notes.html [0, π) f(x) = x π 2. [0, π) f(x) = x 2π 3. [0, π) f(x) = x 2π 1.2. Euler α

.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,


表1-表4_No78_念校.indd

chap9.dvi

webkaitou.dvi

D:/BOOK/MAIN/MAIN.DVI

( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )

( : December 27, 2015) CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x) f (x) y = f(x) x ϕ(r) (gradient) ϕ(r) (gradϕ(r) ) ( ) ϕ(r)

< C93878CBB926E8C9F93A289EF8E9197BF2E786264>

☆joshin_表_0524.ai

III Kepler ( )

Gmech08.dvi

2 7 V 7 {fx fx 3 } 8 P 3 {fx fx 3 } 9 V 9 {fx fx f x 2fx } V {fx fx f x 2fx + } V {{a n } {a n } a n+2 a n+ + a n n } 2 V 2 {{a n } {a n } a n+2 a n+

( )

nosenote3.dvi

III Kepler ( )

A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1

untitled



24.15章.微分方程式

ISTC 3

But nothing s unconditional, The Bravery R R >0 = (0, ) ( ) R >0 = (0, ) f, g R >0 f (0, R), R >

Fubini

all.dvi

[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt


,.,. 2, R 2, ( )., I R. c : I R 2, : (1) c C -, (2) t I, c (t) (0, 0). c(i). c (t)., c(t) = (x(t), y(t)) c (t) = (x (t), y (t)) : (1)

Transcription:

IV. p./5

.. p.2/5

.. 8 >< >: d dt y = a, y + a,2 y 2 + + a,n y n + f (t) d dt y 2 = a 2, y + a 2,2 y 2 + + a 2,n y n + f 2 (t). d dt y n = a n, y + a n,2 y 2 + + a n,n y n + f n (t) (a i,j ) p.2/5

.. 8 >< >: d dt y = a, y + a,2 y 2 + + a,n y n + f (t) d dt y 2 = a 2, y + a 2,2 y 2 + + a 2,n y n + f 2 (t). d dt y n = a n, y + a n,2 y 2 + + a n,n y n + f n (t) (a i,j ) y(t) = (y (t), y 2 (t),..., y n (t)) p.2/5

.. 8 >< >: d dt y = a, y + a,2 y 2 + + a,n y n + f (t) d dt y 2 = a 2, y + a 2,2 y 2 + + a 2,n y n + f 2 (t). d dt y n = a n, y + a n,2 y 2 + + a n,n y n + f n (t) (a i,j ) y(t) = (y (t), y 2 (t),..., y n (t)) f i (t) ( ), (i =,..., n) 8 >< >: d dt y = a, y + a,2 y 2 + + a,n y n d dt y 2 = a 2, y + a 2,2 y 2 + + a 2,n y n. d dt y n = a n, y + a n,2 y 2 + + a n,n y n () ( ) p.2/5

.. y = y y 2. B @. y n, A = C A a, a,2 a,n a 2, a 2,2 a 2,n B...... @.. C A a n, a n,2 a n,n p.3/5

.. y = y y 2. B @. y n, A = C A a, a,2 a,n a 2, a 2,2 a 2,n B...... @.. C A a n, a n,2 a n,n () = Ay (2) dt p.3/5

.. y = y y 2. B @. y n, A = C A a, a,2 a,n a 2, a 2,2 a 2,n B...... @.. C A a n, a n,2 a n,n () = Ay (2) (2) n y (t),y 2 (t),...,y n (t) dt W(y,y 2,...,y n )(t) = (y (t) y 2 (t) y n (t)) p.3/5

.. y = y y 2. B @. y n, A = C A a, a,2 a,n a 2, a 2,2 a 2,n B...... @.. C A a n, a n,2 a n,n () = Ay (2) (2) n y (t),y 2 (t),...,y n (t) dt W(y,y 2,...,y n )(t) = (y (t) y 2 (t) y n (t)) n (2) p.3/5

.. p.4/5

.. [ ] p.4/5

.. [ ] () A {λ,..., λ n } λ i (i =,..., n) u i (2) y(t) = e λ it u i p.4/5

.. [ ] () A {λ,..., λ n } λ i (i =,..., n) u i y(t) = e λ it u i (2) (2) {u,u 2,...,u n } (2) y(t) = c e λ t u + c 2 e λ 2t u 2 + + c n e λ nt u n (c, c 2,..., c n ) p.4/5

.. 8 >< >: d dt y = y + y 2 + y 3 d dt y 2 = y y 2 d dt y 3 = y y 3 i.e. dt = B @ C A y p.5/5

.. 8 >< >: d dt y = y + y 2 + y 3 d dt y 2 = y y 2 d dt y 3 = y y 3 i.e. dt = B @ C A y [ ] ( ± i) B @ C A, ±i B @ C A, B @ C A p.5/5

.. 8 >< >: d dt y = y + y 2 + y 3 d dt y 2 = y y 2 d dt y 3 = y y 3 i.e. dt = B @ C A y [ ] ( ± i) B @ C A, ±i B @ C A, B @ C A ( + i) ( i) y(t) = c e t B @ C A + c 2e it B @ C A + c 3e it B @ C A p.5/5

.. 8 >< >: d dt y = y + y 2 + y 3 d dt y 2 = y y 2 d dt y 3 = y y 3 i.e. dt = B @ C A y [ ] ( ± i) B @ C A, ±i B @ C A, B @ C A ( + i) ( i) y(t) = c e t B @ C A + c 2e it B @ C A + c 3e it B @ C A c = c, c 2 = c 2 + c 2, c 3 = i(c 2 c 3 ) p.5/5

.. 8 >< >: d dt y = y + y 2 + y 3 d dt y 2 = y y 2 d dt y 3 = y y 3 i.e. dt = B @ C A y [ ] ( ± i) B @ C A, ±i B @ C A, B @ C A ( + i) ( i) y(t) = c e t B @ C A + c 2e it B @ C A + c 3e it B @ C A c = c, c 2 = c 2 + c 2, c 3 = i(c 2 c 3 ) sin t cos t sin t cos t y(t) = c e t B @ C A + c B 2 @ cos t C A + c B 3 @ sin t C A cos t sin t p.5/5

.. [ ] () dt = @ 6 3 Ay, (2) 2 dt = @ 3 2 Ay p.6/5

.. [ ] () dt = @ 6 3 Ay, (2) 2 dt = @ 3 2 Ay [ ] () y(t) = c e 4t @ 3 A + c 2 e 3t @ A 2 p.6/5

.. [ ] () dt = @ 6 3 Ay, (2) 2 dt = @ 3 2 Ay [ ] () y(t) = c e 4t @ 3 A + c 2 e 3t @ A 2 (2) y(t) = c e ( 2+i)t @ 2 A + c 2 e ( 2 i)t @ 2 A + i i p.6/5

.. [ ] () dt = @ 6 3 Ay, (2) 2 dt = @ 3 2 Ay [ ] () y(t) = c e 4t @ 3 A + c 2 e 3t @ A 2 (2) y(t) = c e ( 2+i)t @ 2 A + c 2 e ( 2 i)t @ 2 A + i i c = c + c 2, c 2 = i(c c 2 ) p.6/5

.. [ ] () dt = @ 6 3 Ay, (2) 2 dt = @ 3 2 Ay [ ] () y(t) = c e 4t @ 3 A + c 2 e 3t @ A 2 (2) y(t) = c e ( 2+i)t @ 2 A + c 2 e ( 2 i)t @ 2 A + i i c = c + c 2, c 2 = i(c c 2 ) y(t) = c e 2t @ 2cos t A + c 2e 2t @ 2sin t A cos t sin t cos t + sin t p.6/5

..2 p.7/5

..2 A A e ta (= exp ta) = E + ta + t2 2 A2 + + tn n! An + = X n= n! (ta)n p.7/5

..2 A A e ta (= exp ta) = E + ta + t2 2 A2 + + tn n! An + = e A = E, e sa e ta = e (s+t)a, e ta = X n= ne tao n! (ta)n p.7/5

..2 A A e ta (= exp ta) = E + ta + t2 2 A2 + + tn n! An + = e A = E, e sa e ta = e (s+t)a, e ta = X n= ne tao n! (ta)n d dt eta = Ae ta = e ta A p.7/5

..2 A A e ta (= exp ta) = E + ta + t2 2 A2 + + tn n! An + = e A = E, e sa e ta = e (s+t)a, e ta = X n= ne tao n! (ta)n d dt eta = Ae ta = e ta A = Ay (3) dt c = (c, c 2,..., c n ) y(t) = e ta c p.7/5

..2 A A e ta (= exp ta) = E + ta + t2 2 A2 + + tn n! An + = e A = E, e sa e ta = e (s+t)a, e ta = X n= ne tao n! (ta)n d dt eta = Ae ta = e ta A = Ay (3) dt c = (c, c 2,..., c n ) y(t) = e ta c e ta (3) ( ) p.7/5

..2 dt = @ Ay p.8/5

..2 dt = @ Ay [ ] exp t @ A = E + t @ A + t2 2 @ A 2 + + tn n! @ A n + p.8/5

..2 dt = @ Ay [ ] exp t @ A = E + t @ A + t2 2 = E + t @ A + t2 2 @ 2 A + + tn @ n A + n! @ 2 A + + tn @ n A + n! p.8/5

..2 dt = @ Ay [ ] exp t @ A = E + t @ A + t2 2 = E + t @ A + t2 2 P t n P t n n! = Bn= n= @ P n= (n )! t n n! @ 2 A + + tn @ n A + n! @ 2 A + + tn @ n A + n! C A p.8/5

..2 dt = @ Ay [ ] exp t @ A = E + t @ A + t2 2 = E + t @ A + t2 2 P t n P t n n! = Bn= n= @ P n= (n )! t n n! @ 2 A + + tn @ n A + n! @ 2 A + + tn @ n A + n! C A = @ et te t A e t p.8/5

..2 dt = @ Ay [ ] exp t @ A = E + t @ A + t2 2 = E + t @ A + t2 2 P t n P t n n! = Bn= n= @ P n= (n )! y(t) = c @ et A + c 2 @ tet A e t t n n! @ 2 A + + tn @ n A + n! @ 2 A + + tn @ n A + n! C A = @ et te t A e t p.8/5

..2 A J P A = PJP p.9/5

..2 A J P A = PJP A n = (PJP ) n = PJP PJP PJP = PJ n P p.9/5

..2 A J P A = PJP A n = (PJP ) n = PJP PJP PJP = PJ n P A e ta = E+tPJP + t2 2 PJ2 P + + tn n! PJn P + = P X n=! n! (tj)n P = Pe tj P p.9/5

..2 A J P A = PJP A n = (PJP ) n = PJP PJP PJP = PJ n P A! X e ta = E+tPJP + t2 2 PJ2 P + + tn n! PJn P + = P n! (tj)n P = Pe tj P n= (3) y(t) = e ta c = Pe tj P c p.9/5

..2 A J P A = PJP A n = (PJP ) n = PJP PJP PJP = PJ n P A! X e ta = E+tPJP + t2 2 PJ2 P + + tn n! PJn P + = P n! (tj)n P = Pe tj P n= (3) y(t) = e ta c = Pe tj P c [ ] dt = @ Ay p.9/5

..2 A J P A = PJP A n = (PJP ) n = PJP PJP PJP = PJ n P A! X e ta = E+tPJP + t2 2 PJ2 P + + tn n! PJn P + = P n! (tj)n P = Pe tj P n= (3) y(t) = e ta c = Pe tj P c [ ] dt = @ Ay [ ] @ 2 A = @ 2 A @ A @ 2 2 2 2 2 2 2 A p.9/5

..2 A J P A = PJP A n = (PJP ) n = PJP PJP PJP = PJ n P A! X e ta = E+tPJP + t2 2 PJ2 P + + tn n! PJn P + = P n! (tj)n P = Pe tj P n= (3) y(t) = e ta c = Pe tj P c [ ] dt = @ Ay [ ] @ 2 A = @ 2 A @ A @ 2 2 2 2 y(t)= @ 2 A@ 2 A@ 2 A@ c A e 2t 2 2 2 2 c 2 2 2 2 2 A p.9/5

..2 A J P A = PJP A n = (PJP ) n = PJP PJP PJP = PJ n P A! X e ta = E+tPJP + t2 2 PJ2 P + + tn n! PJn P + = P n! (tj)n P = Pe tj P n= (3) y(t) = e ta c = Pe tj P c [ ] dt = @ Ay [ ] @ 2 A = @ 2 A @ 2 A @ 2 A 2 2 2 2 2 2 y(t)= @ 2 A@ 2 A@ 2 A@ c A = c @ 2 + 2 e2t A+c 2 @ 2 2 e2t A e 2t 2 2 2 2 c 2 2 2 e2t 2 + 2 e2t p.9/5

..3 p./5

..3 [ ] p./5

..3 [ ] = Ay + f(t) (4) dt p./5

..3 [ ] = Ay + f(t) (4) dt «e ta dt Ay = d e ta y (4) dt p./5

..3 [ ] = Ay + f(t) (4) dt «e ta dt Ay = d e ta y (4) dt d dt e ta y = e ta f(t) p./5

..3 [ ] = Ay + f(t) (4) dt «e ta dt Ay = d e ta y (4) dt d dt e ta y = e ta f(t) e ta y = Z t t e τa f(τ)dτ + c, c p./5

..3 [ ] = Ay + f(t) (4) dt «e ta dt Ay = d e ta y (4) dt d dt e ta y = e ta f(t) e ta y = Z t t e τa f(τ)dτ + c, c y = e ta c + Z t t e (t τ)a f(τ)dτ p./5

..3 dt = @ 2 Ay + 2 @ e2t 2e 2t A p./5

..3 dt = @ 2 Ay + 2 @ e2t 2e 2t A [ ] exp t@ 2 A=e 2t @ cos t 2 sin t sin t A, cos t 8 < exp : τ 9 @ 2 = A ; @ e2τ 2e 2τ cos τ + 2 sin τ A= @ A sin τ + 2cos τ p./5

..3 dt = @ 2 Ay + 2 @ e2t 2e 2t A [ ] exp t@ 2 A=e 2t @ cos t 2 sin t sin t A, cos t 8 < exp : τ 9 @ 2 = A ; @ e2τ 2e 2τ cos τ + 2 sin τ A= @ A sin τ + 2cos τ y = exp t @ 2 A @ c A+exp t @ 2 A 2 2 c 2 Z t 8 < exp : τ 9 @ 2 = A @ e2τ A dτ ; 2e 2τ p./5

..3 dt = @ 2 Ay + 2 @ e2t 2e 2t A [ ] exp t@ 2 A=e 2t @ cos t 2 sin t sin t A, cos t 8 < exp : τ 9 @ 2 = A ; @ e2τ 2e 2τ cos τ + 2 sin τ A= @ A sin τ + 2cos τ y = exp t @ 2 A @ c A+exp t @ 2 A 2 c 2 2 = e 2t @ cos t sin t A @ c A + e 2t @ cos t sin t cos t sin t c 2 8 9 < exp : τ @ 2 = A ; sin t sin t 2 cos t + 2 A @ A cos t cos t + 2 sin t Z t @ e2τ 2e 2τ A dτ p./5

..3 dt = @ 2 Ay + 2 @ e2t 2e 2t A [ ] exp t@ 2 A=e 2t @ cos t 2 sin t sin t A, cos t 8 < exp : τ 9 @ 2 = A ; @ e2τ 2e 2τ cos τ + 2 sin τ A= @ A sin τ + 2cos τ 8 9 y = exp t @ 2 A @ c A+exp t @ 2 Z t < A exp 2 c 2 2 : τ @ 2 = A ; = e 2t @ cos t sin t A @ c A + e 2t @ cos t sin t sin t 2 cos t + 2 A @ A sin t cos t c 2 sin t cos t cos t + 2 sin t = e 2t @ cos t sin t A @ c A + e 2t @ 2 A (c = c + 2, c 2 = c 2 ) sin t cos t c 2 @ e2τ 2e 2τ A dτ p./5

..3 [ ] p.2/5

..3 [ ] dt = @ 2 Ay+ 2 @ e2t 2e 2t 8 < (D 2)y A + y 2 = e 2t : y + (D 2)y 2 = 2e 2t D = d «dt p.2/5

..3 [ ] dt = @ 2 Ay+ 2 @ e2t 2e 2t 8 < (D 2)y A + y 2 = e 2t : y + (D 2)y 2 = 2e 2t D = d «dt [ ] (D 2) (D 2) 2 y + (D 2)y 2 = (D 2)e 2t = p.2/5

..3 [ ] dt = @ 2 Ay+ 2 @ e2t 2e 2t 8 < (D 2)y A + y 2 = e 2t : y + (D 2)y 2 = 2e 2t D = d «dt [ ] (D 2) (D 2) 2 y + (D 2)y 2 = (D 2)e 2t = (D 2 4D + 5)y = 2e 2t p.2/5

..3 [ ] dt = @ 2 Ay+ 2 @ e2t 2e 2t 8 < (D 2)y A + y 2 = e 2t : y + (D 2)y 2 = 2e 2t D = d «dt [ ] (D 2) (D 2) 2 y + (D 2)y 2 = (D 2)e 2t = (D 2 4D + 5)y = 2e 2t y = e 2t (c cos t + c 2 sin t) 2e 2t p.2/5

..3 [ ] dt = @ 2 Ay+ 2 @ e2t 2e 2t 8 < (D 2)y A + y 2 = e 2t : y + (D 2)y 2 = 2e 2t D = d «dt [ ] (D 2) (D 2) 2 y + (D 2)y 2 = (D 2)e 2t = (D 2 4D + 5)y = 2e 2t y = e 2t (c cos t + c 2 sin t) 2e 2t y 2 = e 2t (D 2)y = e 2t (c sin t c 2 cos t) + e 2t p.2/5

..3 [ ] dt = @ 2 Ay+ 2 @ e2t 2e 2t 8 < (D 2)y A + y 2 = e 2t : y + (D 2)y 2 = 2e 2t D = d «dt [ ] (D 2) (D 2) 2 y + (D 2)y 2 = (D 2)e 2t = (D 2 4D + 5)y = 2e 2t y = e 2t (c cos t + c 2 sin t) 2e 2t y 2 = e 2t (D 2)y = e 2t (c sin t c 2 cos t) + e 2t y = e 2t @ cos t sin t A @ c A + e 2t @ 2 A sin t cos t c 2 p.2/5

..3 [ ] 8 < (D + )y + (D + 3)y 2 = t D = d «:(D )y + (2D 2)y 2 = 2t, dt p.3/5

..3 [ ] 8 < (D + )y + (D + 3)y 2 = t D = d «:(D )y + (2D 2)y 2 = 2t, dt [ ] (D + ) (D (D ) 2 y 2 = + 3t p.3/5

..3 [ ] 8 < (D + )y + (D + 3)y 2 = t D = d «:(D )y + (2D 2)y 2 = 2t, dt [ ] (D + ) (D (D ) 2 y 2 = + 3t y 2 = c e t + c 2 te t + 3t + 7 p.3/5

..3 [ ] 8 < (D + )y + (D + 3)y 2 = t D = d «:(D )y + (2D 2)y 2 = 2t, dt [ ] (D + ) (D (D ) 2 y 2 = + 3t y 2 = c e t + c 2 te t + 3t + 7 y 2 (D )y = 2c 2 e t + 8t + 8 p.3/5

..3 [ ] 8 < (D + )y + (D + 3)y 2 = t D = d «:(D )y + (2D 2)y 2 = 2t, dt [ ] (D + ) (D (D ) 2 y 2 = + 3t y 2 = c e t + c 2 te t + 3t + 7 y 2 (D )y = 2c 2 e t + 8t + 8 y = ce t 2c 2 te t 8t 6 p.3/5

..3 [ ] 8 < (D + )y + (D + 3)y 2 = t D = d «:(D )y + (2D 2)y 2 = 2t, dt [ ] (D + ) (D (D ) 2 y 2 = + 3t y 2 = c e t + c 2 te t + 3t + 7 y 2 (D )y = 2c 2 e t + 8t + 8 y = ce t 2c 2 te t 8t 6 y c «y = 2 c 2 2c e t 2c 2 te t 8t 6 p.3/5

2. p.4/5

2. 8 < : d 2 y dt 2 y + dz dt z = e t 2 dt 2y + d2 z dt 2 z=, y() =, y () =, z() =, z () = p.4/5

2. 8 < : d 2 y dt 2 y + dz dt z = e t 2 dt 2y + d2 z dt 2 z=, y() =, y () =, z() =, z () = [ ] 8 < (s 2 )Y (s) + (s )Z(s) = s+ Y = L(y), Z = L(z) : 2(s + )Y (s) + (s 2 )Z(s) s=, p.4/5

2. 8 < : d 2 y dt 2 y + dz dt z = e t 2 dt 2y + d2 z dt 2 z=, y() =, y () =, z() =, z () = [ ] 8 < (s 2 )Y (s) + (s )Z(s) = s+ Y = L(y), Z = L(z) : 2(s + )Y (s) + (s 2 )Z(s) s=, Y, Z 8 < Y (s)= : Z(s)= 2 s+ s s 2 + s 2 s+ + s s 2 + s 2 + p.4/5

2. 8 < : d 2 y dt 2 y + dz dt z = e t 2 dt 2y + d2 z dt 2 z=, y() =, y () =, z() =, z () = [ ] 8 < (s 2 )Y (s) + (s )Z(s) = s+ Y = L(y), Z = L(z) : 2(s + )Y (s) + (s 2 )Z(s) s=, Y, Z 8 < Y (s)= : Z(s)= 2 s+ s s 2 + s 2 s+ + s s 2 + s 2 + y(t) = e t cos t, z(t) = 2 et 2 e t + cos t sin t p.4/5

2. [ ] 8 < y y + z = y() = 2, z() = : 5y + z 3z=, p.5/5

2. [ ] 8 < y y + z = y() = 2, z() = : 5y + z 3z=, 8 < y(t)= 2 e2t (4 cos2t 3 sin2t) : z(t)= 2 e2t (2 cos2t + sin2t) p.5/5