2.2 ( y = y(x ( (x 0, y 0 y (x 0 (y 0 = y(x 0 y = y(x ( y (x 0 = F (x 0, y(x 0 = F (x 0, y 0 (x 0, y 0 ( (x 0, y 0 F (x 0, y 0 xy (x, y (, F (x, y ( (

Similar documents

(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

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

08-Note2-web

n Y 1 (x),..., Y n (x) 1 W (Y 1 (x),..., Y n (x)) 0 W (Y 1 (x),..., Y n (x)) = Y 1 (x)... Y n (x) Y 1(x)... Y n(x) (x)... Y n (n 1) (x) Y (n 1)

1.2 y + P (x)y + Q(x)y = 0 (1) y 1 (x), y 2 (x) y 1 (x), y 2 (x) (1) y(x) c 1, c 2 y(x) = c 1 y 1 (x) + c 2 y 2 (x) 3 y 1 (x) y 1 (x) e R P (x)dx y 2

() 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)

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

ax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4


,. 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,,.

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

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

chap1.dvi

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 (

pdf

x ( ) x dx = ax


DE-resume

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

y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =

1W II K =25 A (1) office(a439) (2) A4 etc. 12:00-13:30 Cafe David 1 2 TA appointment Cafe D

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 { (

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

2011de.dvi

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

mugensho.dvi

n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x

v er.1/ c /(21)

29

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

1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =

A

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

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

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

振動と波動

() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (


ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d

..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,



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)

i

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 {

DVIOUT

.1 A cos 2π 3 sin 2π 3 sin 2π 3 cos 2π 3 T ra 2 deta T ra 2 deta T ra 2 deta a + d 2 ad bc a 2 + d 2 + ad + bc A 3 a b a 2 + bc ba + d c d ca + d bc +

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

webkaitou.dvi

I 1

1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 +

熊本県数学問題正解

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

B ver B

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+

5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h 0 g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)

simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a =


i

d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>



6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P

2. 2 I,II,III) 2 x expx) = lim + x 3) ) expx) e x 3) x. ) {a } a a 2 a 3...) a b b {a } α : lim a = α b) ) [] 2 ) f x) = + x ) 4) x > 0 {f x)} x > 0,

Z: Q: R: C:

x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)

dy + P (x)y = Q(x) (1) dx dy dx = P (x)y + Q(x) P (x), Q(x) dy y dx Q(x) 0 homogeneous dy dx = P (x)y 1 y dy = P (x) dx log y = P (x) dx + C y = C exp

. sinh x sinh x) = e x e x = ex e x = sinh x 3) y = cosh x, y = sinh x y = e x, y = e x 6 sinhx) coshx) 4 y-axis x-axis : y = cosh x, y = s

A S- hara/lectures/lectures-j.html r A = A 5 : 5 = max{ A, } A A A A B A, B A A A %

M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -

6.1 (P (P (P (P (P (P (, P (, P.

6.1 (P (P (P (P (P (P (, P (, P.101

Gmech08.dvi

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.

p = mv p x > h/4π λ = h p m v Ψ 2 Ψ


TOP URL 1


5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)

.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0,

meiji_resume_1.PDF

(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1)

No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (




I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f

(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t

m d2 x = kx αẋ α > 0 (3.5 dt2 ( de dt = d dt ( 1 2 mẋ kx2 = mẍẋ + kxẋ = (mẍ + kxẋ = αẋẋ = αẋ 2 < 0 (3.6 Joule Joule 1843 Joule ( A B (> A ( 3-2

v_-3_+2_1.eps

Transcription:

(. x y y x f y = f(x y x y = y(x y x y dx = d dx y(x = y (x = f (x y = y(x x ( (differential equation ( + y 2 dx + xy = 0 dx = xy + y 2 2 2 x y 2 F (x, y = xy + y 2 y = y(x x x xy(x = F (x, y(x + y(x 2 F (x, y 2 x, y = F (x, y ( dx y = y(x ( I (solution y (x = F (x, y(x x I ( I I C y = Ce x R = (, dx = y (2

2.2 ( y = y(x ( (x 0, y 0 y (x 0 (y 0 = y(x 0 y = y(x ( y (x 0 = F (x 0, y(x 0 = F (x 0, y 0 (x 0, y 0 ( (x 0, y 0 F (x 0, y 0 xy (x, y (, F (x, y ( ( xy (2 y = Ce x C ( y = Ce x (x 0, y 0 (y 0 = Ce x 0 Ce x 0 = y 0 (, y 0 0.75 0.5 0.25 0.5 - -0.5 0.5.5 2-0.25-0.5 - -0.5 0.5.5 2-0.5-0.75 - -.3 x, y x y f(x, y C f(x, y = C C C C ( C f(x, y = e x y e x y = C y = Ce x C f(x, y = C : C C C 2 f(x, y = C f(x, y = C 2 f(x, y = C = C 2 (x, y C C 2 x y ( f(x, y = C y x y

3 y = y(x ( y(x C C y(x y = y(x e x y = C y = y(x x e x x y + e dx = e x y(x + e x y (x = 0 ( y(x y y = y(x C (2 C (2 C ( y = Ce x (2 y = Ce x C f(x, y = C x y = y(x ( xy = C (C 0 (2 x + y2 2 = C (3 x 2 2 + y2 = C (C > 0 0.75 0.75 0.75 0.5 0.5 0.5 0.25 0.25 0.25 - -0.75-0.5-0.25 0.25 0.5 0.75-0.25 - -0.75-0.5-0.25 0.25 0.5 0.75-0.25 - -0.75-0.5-0.25 0.25 0.5 0.75-0.25-0.5-0.5-0.5-0.75-0.75-0.75 - - - 2 2. ( F (x, y x y F (x, y = g(xh(y h(y 0 dx = g(xh(y (3 h(y dx = g(x x ( g(x dx = h(y dx dx = y (x h(y(x dx = h(y

4 g(x G(x, C = C 2 C h(y H(y C, C 2 H(y + C = G(x + C 2 H(y G(x = C (3 y = y(x (C h(y = 0 y y 0 y(x = y 0 (3 x + y2 2 = C (2 dx = y y = y(x (x, y y dx = y y = dx c log y = x + c y = ±e x+c = ±e c e x = Ce x (C = ±e c C 0 y = 0 C = 0 0.75 0.5 0.25 - -0.5 0.5.5 2-0.25-0.5-0.75 - x + y2 2 = C y = Ce x

5 2 t x(t x (t x(t k x (t = kx(t dx dt = kx dx x = k dt x(t = Ce kt x(0 = C t x( = Ce k = 2C e k = 2 t Ce tk = C(e k t = C2 t 3 ( α ( β ( γ ( k t ( x(t dx dt = kx k k x(t = Ce kt C = x(0 e kt = t T 2 e kt = log 2 T = k T 2 k x(t = C(e kt t T = C2 t T 3 8 β γ 3 t 3 x(t = C2 t 8 37 30 β γ 37 t 37 x(t = C2 t 30 4 ( t x(t a x (t x(t a (x(t a > 0 x (t < 0 k dx dt = k(x a x (t = k(x(t a dx x a = k dt log x a = kt + c x = x(t = a ± e c e kt = a + Ce kt (C = ±e c.

6 2 20 (t = 0 00 5 80 5 3 ( dx = y2 (2 dx = y (3 dx = xy (4 dx = e y 5 x km y(x ( k > 0 k x y(t dt = y(x ( x ky(x = y (x y(x = Ce kx 000 km 900 C = 000, e k = 0.9 y(x = 000 0.9 x 6 x(t a x (t x(t a x(t a x(t k > 0 dx dt = kx(a x ( dx x(a x = k dt = kt + c dx x(a x = ( a x + dx = a x a log x a x ±e ac = C x a x = Ceakt x = aceakt + Ce akt = a + C e akt t a C C 0 x = a a =, k =

7 0.8 0.6 0.4 0.2 2 3 4 5 7 m t v(t mv (t mg (g v(t kv(t (k mv (t = mg kv(t m mg kv dv = dt = t + c m k log mg kv = t + c C = ±e ck/m mg kv = Ce kt/m v = v(t = k (mg Ce kt/m v(0 = 0 C = mg v(t = mg k ( e kt/m v(t t mg k 4 ( dx = xy2 (2 dx = y2 y 2 (3 dx = y2 + (4 dx = ex+y 2.2 dx = F (x, y

8 y = y(x x = x 0 y(x y 0 dx = F (x, y, y(x 0 = y 0 y = y(x x t t = t 0 ( 8 t = 0 t x = x(t dx dt = kx, x(0 = C x(t = Ce kt x(0 = C = x(t = e kt x(t t 9 dx = y2, y(0 = y = y(0 = C = x + C y(x = x x = (, (x > y = y(x x < x > 5 ( (3 (5 (7 dx = xy, y(0 = (2 dx = y2, dx = y2, y(0 = 0 (4 dx = y2 +, y(0 = (6 dx = y2 y 2, y(0 = 2 (8 y(0 = dx = xy2, y(0 = dx = e y, y(0 = 0 dx = ex+y, y(0 =

9 2.3 ( a(x, b(x y = y(x dx = a(xy + b(x (4 y b(x = 0 ( dx = a(xy a(x A(x c log y = a(x dx = A(x + c C = ±e c y = Ce A(x C C(x y = C(xe A(x (4 (C (x + A (xc(xe A(x = a(xc(xe A(x + b(x C (x = b(xe A(x ( y(x = C(xe A(x = b(xe A(x dx e A(x (5 (5 ( y (x = b(xe A(x dx A (xe A(x + b(xe A(x e A(x = a(xy + b(x (5 (4 (4 (5 y = y(x (4 z(x = e A(x y(x z (x = A (xe A(x y(x + e A(x y (x = a(xe A(x y(x + e A(x {a(xy(x + b(x} = b(xe A(x z(x = b(xe A(x dx ( y(x = z(xe A(x = b(xe A(x dx e A(x (4 (5 0 = 2xy + ex2 dx dx = 2xy

0 y = Ce x2 C C(x y = C(xe x2 dx 2xy = C (xe x2 = e x2 C (x = C(x = x + C (C y(x = (x + Ce x2 y(0 = 0 C = 0 y = xe x2 t E(t I(t R L LI (t + RI(t = E(t di dt = R L I + E(t (6 L ( I(t = E(te R L t dt e R L t L I(t (i E(t = E ( ( ( I(t = Ee R L t dt e R E L t = L R e R L t + C e R L t = E R + R Ce L t (C I(0 = 0 ( 0 C = E/R I(t = E R ( e R L t (ii E(t = E cos ωt (E, ω C(t = E e R L t cos ωt dt = E ( L L Re e ( R L +iωt dt ( = E L Re e ( R L +iωt R + iω + C = E e ( R L t R cos ωt + ω sin ωt L + C L R 2 L + ω L 2 2 I(t = E L R cos ωt + ω sin ωt L + Ce R R 2 + ω L 2 L t R cos ωt + Lω sin ωt = E R 2 + L 2 ω 2 2 + Ce R L t

(C I(0 = 0 C = ER/(R 2 + L 2 ω 2 I(t = E R(cos ωt e R L t + Lω sin ωt R 2 + L 2 ω 2 E = L = R = ω = E(t ( I(t ( ( (i, (ii 0.8 0.75 0.5 0.6 0.25 0.4-0.25 2.5 5 7.5 0 2.5 5 7.5 20 0.2-0.5-0.75 2 4 6 8 0 - Euler i = x e ix = exp(ix = (ix n n=0 n! (7 x (7 i 2k = ( k cos x sin x Taylor e ix = (ix n n=0 n! = k=0 (ix 2k (2k! + (ix 2k+ (2k +! k=0 = cos x + i sin x e ix = cos x + i sin x. Euler α = a + bi (a, b R a = Re α b = Im α cos x = Re e ix, sin x = Im e ix x, y e x+iy = exp(x + iy = e x e iy = e x (cos y + i sin y e x +iy e x 2+iy 2 = e (x +x 2 +i(y +y 2 α (α 0 d dx eαx = αe αx, e αx dx = eαx α + C (C

2 ( α = a + bi (a, b R e αx = e ax (cos bx + i sin bx ( e e ax αx cos bx dx = Re α + C, ( e e ax αx sin bx dx = Im α + C (C 2 6 ( dx = y x (2 dx = y + x2 (3 7 ( (3 dx = xy + e x2 /2 dx = y x, y(0 = 0 (2 dx = y + x2, y(0 = dx = xy + /2 e x2, y(0 = 0 8 E(t = Ee at (E, a I(0 = 0 I(t E = L = R =, a = 2 t 0 I(t I(t 3 2 a(x, b(x, f(x y = y(x dx + a(x + b(xy = f(x 2 dx 2 a(x, b(x a, b dx + a + by = f(x (8 2 dx 2 3. f(x = 0 dx + a + by = 0 (9 2 dx

3 y = y(x R = (, x (9 y(x R : (9 y (x = ay (x by(x (0 y(x 2 (0 y (x y(x 3 (0 2 y(x 4 y(x (9 R C (R x, x 2 + a e ax, sin ax, cos ax C (R y(x z(x C (R a b ay(x + bz(x C (R C (R R C (R C (R D = d dx Dy(x = d dx y(x = y (x y(x, z(x C (R a, b R (y(x C (R D(ay(x + bz(x = ady(x + bdz(x D C (R C (R R D 2, D 3,... D 2 y(x 2 y (x D 0 = I (, D, D 2 C (R C (R P (D = D 2 + ad + bi (a, b R C (R C (R 2 (differential operator P (Dy(x = y (x + ay (x + by(x (9 P (D : C (R C (R (kernel Ker P (D = {y(x C (R P (Dy(x = 0} C (R Ker P (D R C (R, C (C α = a + bi (a, b R e αx = e ax (cos by + i sin by

4 C (R, C y(x z(x C (R, C α, β αy(x + βz(x C (R, C C (R, C C C (R, C C (R, C D = d dx Dy(x = d dx y(x = y (x (y(x C (R, C y(x, z(x C (R, C α, β C D(αy(x + βz(x = αdy(x + βdz(x D C a,... a m C Q(D = D m + a D m + + a m D + a m I (a,..., a m C C (R, C C (R, C C (m (differential operator Ker C Q(D = {y(x C (R, C Q(Dy(x = 0} Ker C Q(D C 2 P (D = D 2 + ad + b D λ P (λ = λ 2 + aλ + b P (D (9 (characteristic polynomial (2 P (λ = 0 α, β λ P (λ P (λ = λ 2 + aλ + b = (λ α(λ β D αi D βi y(x C (R, C (D αi(d βiy(x = (D αi(y (x βy(x = D(y (x βy(x α(y (x βy(x = y (x (α + βy (x + αβy(x = y (x + ay (x + by(x = P (Dy(x P (λ 2 P (D D αi D βi α β

5 P (λ = (λ α(λ β α β Ker C P (D = Ker C (D αi Ker C (D βi ( P (Dy(x = 0 y(x C (R, C y(x = y (x + y 2 (x, (D αiy (x = 0, (D βiy 2 (x = 0 y (x, y 2 (x C (R, C y (x, y 2 (x C (R, C (D αiy (x = 0 (D βiy 2 (x = 0 y(x = y (x + y 2 (x P (Dy(x = 0 : y(x Ker C P (D = (λ α (λ β β α β α I = (D αi (D βi ( β α β α y (x = (D βiy(x = β α β α (y (x βy(x, y 2 (x = (D αiy(x = β α β α (y (x αy(x (D αiy (x = (D αi(d βiy(x = P (Dy(x = 0, (D βiy 2 (x = (D βi(d αiy(x = P (Dy(x = 0 y (x Ker C (D αi, y 2 (x Ker C (D βi ( y(x = Iy(x = (D αiy(x β α β α (D βiy(x = y (x + y 2 (x y (x, y 2 (x y (x y 2 (x z (x Ker C (D αi z 2 (x Ker C (D βi y(x = z (x + z 2 (x u(x := y (x z (x = z 2 (x y 2 (x Ker C (D αi Ker C (D βi (:= ( u(x = Iu(x = (D αiu(x (D βiu(x = 0 β α β α z (x = y (x, z 2 (x = y 2 (x y (x y 2 (x

6 2 α = a + bi (a, b R C D αi : C (R, C C (R, C Ker C (D αi C (R, C C e αx = e ax (cos bx + i sin bx : y(x C (R, C z(x = e αx y(x y(x = e αx z(x (D αiy(x = y (x αy(x = αe αx z(x + e αx z (x αe αx z(x = e αx z (x (D αiy(x = 0 z (x = 0 z (x = 0 z(x y(x Ker C (D αi C y(x = Ce αx 2 P (λ = (λ α(λ β α β C P (D : C (R, C C (R, C Ker C P (D C 2 e αx e βx P (Dy(x = 0 y(x C (R, C y(x = C e αx + C 2 e βx C, C 2 y(x P (Dy(x = 0 α = β 3 α Ker C (D αi 2 = {y(x C (R, C (D αi 2 y(x = 0} C 2 e αx xe αx : y(x C (R, C z(x = e αx y(x 2 (D αi(e αx z(x = e αx z (x (D αi 2 y(x = (D αi(d αi(e αx z(x = (D αi(e αx z (x = e αx z (x (D αi 2 y(x = 0 z (x = 0 z (x = 0 z (x = C z(x = C x + C 2 C 2 z(x = C x + C 2 z (x = 0 Ker C (D αi 2 = {(C x + C 2 e αx C, C 2 C} y(x = (C x + C 2 e αx y(0 = C 2 y( = (C + C 2 e α C C 2 e αx xe αx Ker C (D αi 2 (9

7 a, b P (D = D 2 +ad+bi λ 2 +aλ+b = 0 3 a 2 4b > 0 α, β P (Dy(x = 0 y(x C, C 2 y(x = C e αx + C 2 e βx y(x C, C 2 C, C 2 e αx e βx y(x y(x y(0 = C + C 2 y( = C e α + C 2 e β C, C 2 C C 2 2 P (λ = 0 α, β P (Dy(x = 0 y(x Ker P (D R 2 e αx e βx P (Dy(x = 0 y(x C (R C, C 2 y(x = C e αx + C 2 e βx 2 dx 2 dx 2y = 0 λ 2 λ 2 = (λ 2(λ + = 0 C, C 2 y(x = C e 2x + C 2 e x a 2 4b = 0 α α = a 2 3 P (Dy(x = 0 y(x C, C 2 y(x = C e αx + C 2 xe αx y(x C, C 2 C, C 2 e αx xe αx y(x y(x y(0 = C y( = (C + C 2 e α C, C 2 C C 2 3 P (λ = 0 0 α P (Dy(x = 0 y(x R 2 e αx xe αx P (Dy(x = 0 y(x C (R C, C 2 y(x = C e αx + C 2 xe αx

8 3 dx + 2 2 dx + y = 0 λ 2 + 2λ + = (λ + 2 = 0 C, C 2 y(x = (C x + C 2 e x a 2 4b < 0 α, β λ 2 + aλ + b = (λ α(λ β α β β = α p, q α = p + qi e αx = e px+iqx = e px e iqx = e px (cos qx + i sin qx P (Dy(x = 0 y(x C, C 2 y(x = C e αx + C 2 e βx a, a 2, b, b 2 C = a + ib, C 2 = a 2 + ib 2 α β p, q α = p + iq, β = p iq (q > 0 y(x = C e αx + C 2 e βx = (a + ib e px (cos qx + i sin qx + (a 2 + ib 2 e px (cos qx i sin qx = e px {(a + a 2 cos qx + ( b + b 2 sin qx} + ie px {(b + b 2 cos qx + (a a 2 sin qx} y(x a = a 2, b + b 2 = 0 y(x = e px (2a cos qx 2b sin qx C = 2a, C 2 = 2b y(x = e px (C cos qx + C 2 sin qx (2

9 4 P (λ = 0 p, q (q > 0 α = p + iq, β = p iq P (Dy(x = 0 y(x R 2 e px cos qx e px sin qx P (Dy(x = 0 y(x C (R C, C 2 y(x = e px (C cos qx + C 2 sin qx 4 dx 2 + 4 dx + 5y = 0 λ 2 + 4λ + 5 = (λ + 2 i(λ + 2 + i = 0 ( 2 ± i C, C 2 y(x = e 2x (C cos x + C 2 sin x 5 m 0 k x t x = x(t q = (k m d 2 x dt 2 + q2 x = 0 λ 2 + q 2 = (λ iq(λ + iq = 0 C, C 2 x(t = C cos qt + C 2 sin qt q 2π q 9 ( (4 dx 2 dx 6y = 0 (2 4d2 y dx 2 4 dx 3y = 0 (3 dx 2 + 2 dx = 0 dx 2 + 6 dx + 9y = 0 (5 dx 2 + 4y = 0 (6 2d2 y dx 2 + 2 dx + y = 0 3.2 (9 x y(x x y(x y (x

20 6 y = y(x dx 2 dx 2y = 0 y(0 =, y (0 = 0 C, C 2 y(x = C e 2x + C 2 e x y(0 = C + C 2 =, y (0 = 2C C 2 = 0 C = 3, C 2 = 2 3 y(x = 3 e2x + 2 3 e x 7 5 x(0 0 x (0 0 d 2 x dt 2 + q2 x = 0, x(0 =, x (0 = 0 x = x(t t x(t = cos qt x 0, y 0, y dx 2 + a dx + by = 0, y(x 0 = y 0, y (x 0 = y (3 y = y(x ( w(x = y(x + x 0 w(x w(0 = y(x 0 = y 0, w (0 = y (x 0 = y x 0 = 0 a 2 4b > 0 λ 2 + aλ + b = 0 2 α, β y(x = C e αx + C 2 e βx y(0 = C + C 2 = y 0, y (0 = αc + βc 2 = y ( ( ( C = α β C 2 y 0 y

2 β α = 0 C C 2 a 2 4b < 0 λ 2 + aλ + b = 0 2 p ± iq y(x = e px (C cos qx + C 2 sin qx y(0 = C = y 0, y (0 = pc + qc 2 = y C q 0 C 2 a 2 4b = 0 α y(x = (C x + C 2 e αx y(0 = C 2 C 2 y (x = α(c x + C 2 e αx + C e αx y (0 = C + αc 2 C 8 t x = x(t x(t d 2 x dt 2 + adx dt + bx = 0, x(0 =, x (0 = 0 a b ( x(t x(t ( a 2 4b < 0 q = 2 4b a2 x(t = e a 2 t ( cos qt + a 2q sin qt. ( (2 a 2 4b = 0 x(t = e a 2 t + a. 2 t (3 a 2 4b > 0 λ 2 + aλ + b = 0 2 α, β x(t = β α (βeαt αe βt. b = 4, a =, 4, 5 y(x 0.8 0.8 0.8 0.6 0.6 0.6 0.4 0.4 0.4 0.2 0.2 0.2-0.2 2 3 4 5 6-0.2 2 3 4 5 6-0.2 2 3 4 5 6-0.4-0.4-0.4

22 0 ( dx 2 dx 6y = 0, y(0 = 0, y (0 = (2 4 d2 y dx 4 2 dx 3y = 0, y(0 =, y (0 = 2 (3 dx + 2 2 dx = 0, y(0 =, y (0 = (4 dx + 6 2 dx + 9y = 0, y(0 =, y (0 = 2 (5 dx + 4y = 0, 2 y(0 = 0, y (0 = (6 2 d2 y dx 2 + 2 dx + y = 0, y(0 =, y (0 = 3.3 f(x dx + a + by = f(x (4 2 dx y = y(x (4 y = y 0 (x y(x (4 y = y (x y (x = y(x y 0 (x dx + a + by = 0 (5 2 dx y = y (x (5 y(x = y 0 (x + y (x (4 = + (5 (4 (Heaviside D D C (R, C I 2 P (D = a 0 D m +a D m + +a m D +a m f, u C (R, C u(x = P (D f(x P (Du(x = f(x

23 u(x P (D v(x = f(x u(x v(x Ker P (D P (D u(x P (D f(x mod Ker P (D Ker P (D Ker P (D = (4 9 u(x = D f(x u (x = f(x u(x = f(x dx u(x 20 α u(x = D α f(x u (x αu(x = f(x u(x = e αx e αx f(x dx 4 α, β, γ C f(x C (R, C = Ker (D α(d β ( (D α(d β f(x = ( D α D β f(x (2 (3 D α (eγx f(x = e γx D + (γ α f(x (D α(d β f(x = α β : ( u(x ( D α f(x D β f(x (α β (D αu(x = f(x (D β((d αu(x = f(x. D β ((D α(d βu(x = ((D β(d αu(x = (D β((d αu(x = f(x. (2 v(x = f(x D + (γ α (D α(e γx v(x = (e γx v(x αe γx v(x = e γx (v (x + (γ αv(x = e γx (D + (γ αv(x = e γx f(x.

24 D α (eγx f(x = e γx v(x = e γx D + (γ α f(x. (3 u(x ((D α(d βu(x = { ( (D β (D α α β D α f(x ( } (D α (D β D β f(x (3 = {(D βf(x (D αf(x} = f(x α β 5 α 0 f(x ( D α f(x = α + α D + 2 α 3 D2 + f(x = α f(x α 2 f (x α 3 f (x f(x k D k+ f(x = f (k+ (x = 0 : t k ( t( + t + + t k = t k+ t D α ( α ( D + α D + + α k Dk = Dk+ αk+ f(x k D k+ f(x = 0 ( α ( D + α D + + α ( k Dk f(x = Dk+ f(x = f(x αk+ ( (D α α α D 2 Dk f(x αk+ = ( α ( D + α D + + α k Dk f(x = f(x ( D α f(x = α + α D + + 2 Dk f(x αk+ D k+ f(x = D k+2 f(x = = 0

25 4 ( 20 (4 ( D α D β f(x = (e βx e βx f(x dx D α ( = e αx e (β αx e βx f(x dx dx 2 (4 f(x f(x = p(xe cx (p(x c (4 4 (2,(3 5 2 dx 2 dx 2y = x λ 2 λ 2 = (λ 2(λ + = 0 P (D = D 2 D 2 = (D 2(D + 4 (3 5 P (D x = 3 ( D 2 x D + x = ( 3 2 + 4 D + 8 D2 + = ( 3 2 x + 4 3 (x = 2 x + 4 C, C 2 x 3 ( D + D2 x 22 y(x = C e 2x + C 2 e x 2 x + 4 dx 2 dx 2y = x2 e x P (D = D 2 D 2 = (D 2(D +

26 4 (3 (2 5 P (D (x2 e x = ( 3 D 2 (x2 e x D + (x2 e x = 3 ex ( D x2 D + 2 x2 = ( 3 ex + D + D 2 + x 2 + ( 3 ex 2 + 4 D 8 D2 + = 3 ex (x 2 + 2x + 2 + ( 3 ex 2 x2 + 2 x 4 = (C, C 2 ( 2 x2 2 x 3 e x 4 y(x = C e 2x + C 2 e x ( 2 x2 + 2 x + 3 e x 4 x 2 23 dx 2 + y = (x + 2 ex dx (D ((x + 2 ex = ( D D ((x + ex = (e x D D (x + = ( e x ( D 2 x2 + x = e x ( ( 6 D 2 x2 + x = e x x3 + 2 x2 (C, C 2 y(x = ( 6 x3 + 2 x2 + C x + C 2 e x 24 dx 2 2 dx + y = xe x 4 (,(2 5 (D 2 (e x = ( D D (e x x = ( e x D D 2 x = (e ( x 2 D 4 D x = ( (e x D 2 x 4 ( = e x D 2 2 x = e ( x 2 4 4 ( D 2 x 4 = (x + e x 4

27 (C, C 2 y(x = (C x + C 2 e x + (x + e x 4 f(x = ce γx (c P (λ 0 6 P (D = D 2 + ad + b γ P (De γx = P (γe γx P (γ 0 P (D eγx = P (γ eγx. : De γx = γe γx P (De γx = γ 2 e γx + aγe γx + be γx = P (γe γx ( P (γ 0 P (D P (γ eγx = e γx P (D eγx = P (γ eγx. 25 dx 2 2 dx + y = e x P (D = D 2 2D +, P ( = 4 0 6 y(x = P (D e x = P ( e x = 4 e x P (λ = (λ 2 C, C 2 y(x = (C x + C 2 e x + 4 e x f(x = p (xe a x + + p m (xe a mx (p (x,..., p m (x P (Dy j (x = p j (xe a jx y j (x j P (D P (D(y (x + + y m (x = f(x

28 26 dx + 2 dx = e x + 2e 2x P (D = D 2 + D = D(D + P (Dy (x = e x y (x P ( = 0 6 y (x = P (D e x = D e x D + e x = e x D e x D = e x ( + D + D 2 + e x x = (x + e x P (Dy 2 (x = 2e 2x y 2 (x P ( 2 = 2 6 y 2 (x = P ( 2 2e 2x = e 2x y(x = y (x + y 2 (x = (x + e x + e 2x y(x = C e x + C 2 (x + e x + e 2x = ( x + C e x + e 2x + C 2 C, C 2 C = C ( (3 dx 2 + dx 6y = x (2 dx 2 2 dx = xe2x dx 2 2 dx + y = x + (4 dx 2 + dx 6y = e 2x + x (4 f(x f(x = p(xe ax cos bx f(x = p(xe ax sin bx (p(x a, b f(x = p(xe (a+ibx (4 z(x P (D(Re z(x = Re (P (Dz(x = Re (p(xe (a+ibx = p(xe ax cos bx, P (D(Im z(x = Im (P (Dz(x = Im (p(xe (a+ibx = p(xe ax sin bx 27 dx 2 dt 2 + 2dx dt + 2x = cos ωt, x(0 = x (0 = 0. 8 a = 2, b = 2 ω > 0 cos ωt cos ωt = Re(e iωt d 2 z dt + 2dz 2 dt + 2z = eiωt

29 z = z(t x(t = Re z(t D = d dt P (D = D 2 + 2D + 2 P (iω = 2 ω 2 + 2iω 0 6 z(t = P (iω eiωt = 2 ω 2 + 2iω eiωt = 2 ω2 2iω e iωt 4 + ω 4 x(t = Re z(t = 2 ω2 2ω cos ωt + sin ωt 4 + ω4 4 + ω4 ± i C, C 2 x(t = e t (C cos t + C 2 sin t + 2 ω2 2ω cos ωt + sin ωt 4 + ω4 4 + ω4 x(0 = x (0 = 0 x (t = e t ( C cos t C 2 sin t C sin t + C 2 cos t ω(2 ω2 sin ωt + 2ω2 cos ωt 4 + ω 4 4 + ω4 C = ω2 2 4 + ω 4, C 2 = 2 + ω2 4 + ω 4 x(t = 4 + ω 4 e t {(ω 2 2 cos t (2 + ω 2 sin t} + 2 ω2 2ω cos ωt + sin ωt 4 + ω4 4 + ω4 e t t 0 a(ω (2 ω 2 a(ω = 4 + ω 4 2 + ( 2 2ω = 4 + ω 4 4 + ω 4. a(ω ω lim ω a(ω = 0 ω = π, ω = 2π x(t x x t t 28 dx + 2 + 2y = x sin x 2 dx d 2 z dx + 2 dz + 2z = 2 xeix dx

30 z = z(x P (D = D 2 + 2D + 2 P (λ = (λ + i(λ + + i z(x = P (D (xeix = ( 2i D + i (xe ix D + + i ( = eix 2i D + x D + + 2i = eix 2i eix ( D + x 2i = eix eix (x 2i 2i y(x = Im z(x = ( + 2i ( + 2i x ( + 2i 2 y(x = e x (C cos x + C 2 sin x + 2 ( + 2i D + ( 2 2i = 5 x + ( 2 5 x + 4 ( cos x + 25 5 x 2 sin x 25 x 2 + 4i 25 e ix ( 2 5 x + 4 ( cos x + 25 5 x 2 sin x 25 ( (3 dx 2 + 4 dx + 5y = cos x (2 dx 2 + y = sin x dx + y = x cos x (4 2 dx + y = 2 e x cos x 3 ( (2 (3 (4 dx + 2 dx 6y = x, y(0 = 0, y (0 = 0 dx 2 2 dx = xe2x, y(0 = 0, y (0 = dx + y = sin x, 2 y(0 = 0, y (0 = 0 dx + y = 2 e x cos x, y(0 =, y (0 = 0

2024 © docsplayer.net プライバシー | 利用規約 | フィードバック