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Mathematics III TS 4353 Class B

Mathematics III TS 4353 Class B. Herlina Setiyaningsih Civil Engineering Department Petra Christian Universit y. Jurusan Teknik Sipil Matematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen Petra Bab 1. Operator Diferensial. D(e 3x ) = 3e 3x D 2 (e 3x ) = 9e 3x

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Mathematics III TS 4353 Class B

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  1. Mathematics IIITS 4353Class B Herlina Setiyaningsih Civil Engineering Department Petra Christian University

  2. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Operator Diferensial D(e3x) = 3e3x D2(e3x) = 9e3x D(sin 2x) = 2 cos 2x D(x2+1) = 2x D2(x2+1) = 2 D3(x2+1) = 0

  3. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Operator D Dy = y’ D2y = y” y” + py’ +qy = f(x) D2y + pDy + qy = f(x) (D2+pD+q)y = f(x) F(D)y = f(x)

  4. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Sifat 1 Operator D • F(D)eax = (D2+pD+q)eax = (a2 + pa + q) eax = F(a) eax

  5. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 1 • y” – 5y’ + 6y = ex • (D2 – 5D + 6)y = ex • yc = c1e2x + c2e3x • PUPD: y = yc + yp • y = yc + yp = c1e2x + c2e3x + ½ ex

  6. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Sifat 2 Operator D • F(D)(eax V) = (D2+pD+q)(eaxV) • D(eax V) = aeax V + eax DV = eax (D+a)V • D2(eax V) = D(aeax V + eax DV) = a2eax V + aeax DV + aeax DV + eax D2V = eax (D2 + 2aD + a2)V = eax (D+a)2V

  7. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 • F(D)(eax V) = (D2+pD+q)(eaxV) = eax (D+a)2V + p eax (D+a)V + q eax V = eax [(D+a)2 + (D+a)p + q ] V = eax F(D+a) V

  8. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 2 • y” – 2y’ + y = xex • (D2 - 2D + 1)y = xex • PR : (D2 - 2D + 1)y = 0  subs: y=ekx • PK : (k2 – 2k + 1) = 0 • k1 = k2 = m =1 • yc= ex(c1 + c2x) • PUPD: y = ex(c1 + c2x + 1/6 x3)

  9. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Sifat 3 Operator D • D(cos ax) = -a sin ax • D2(cos ax) = -a2 cos ax • F (D2)cos ax= F (-a2)cos ax

  10. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 3 • y” – 9 y = cos 2x • (D2 - 9)y = cos 2x • PR : (D2 - 9)y = 0  subs: y=ekx • PK : (k2 – 9) = 0 • (k+3)(k-3) = 0 • k1 = 3 dan k2 = -3 • yc= c1 e3x+ c2 e-3x • PUPD: y = c1 e3x+ c2 e-3x – 1/13 cos 2x

  11. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Sifat 4 Operator D • D(sin ax) = a cos ax • D2(sin ax) = -a2 sin ax • F (D2)sin ax= F (-a2)sin ax

  12. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 4 • y” + 9 y = sin 2x • (D2 + 9)y = sin 2x • PR : (D2+9)y = 0  subs: y=ekx • PK : (k2 + 9) = 0 • (k+3)(k-3) = 0 • k1,2 = ± 3i • yc= c1cos 3x + c2 sin 3x • PUPD: y = c1cos 3x + c2 sin 3x + 1/5 sin 2x

  13. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 SummarySifat-sifat Operator D

  14. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 5 • y” -2y’ + y = ex • (D2 -2D + 1)y = ex • PR : (D2 -2D + 1)y = 0  subs: y=ekx • PK : (k2 – 2k + 1) = 0 • (k-1)(k-1) = 0 • k1,2 = m = 1 • yc= ex (c1 + c2 x) Gagal!!!

  15. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 5 (Lanjutan) • PUPD: y = ex(c1 + c2x + 1/2 x2)

  16. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 6 • y” -5y’ + 6y = sin 4x • (D2 -5D + 6)y = sin 4x • PR : (D2 -5D + 6)y = 0  subs: y=ekx • PK : (k2 – 5k + 6) = 0 • (k-2)(k-3) = 0 • k1 = 2 dan k2 = 3 • yc= c1 e2x + c2 e3x

  17. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 6 (Lanjutan) PUPD: y = yc + yp = c1 e2x + c2 e3x – 1/50 sin 4x + 1/25 cos 4x

  18. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 7 • y” -2y’ - 3y = x2 • (D2 -2D - 3)y = x2 • PR : (D2 -2D - 3)y = 0  subs: y=ekx • PK : (k2 – 2k - 3) = 0 • (k-3)(k+1) = 0 • k1 = 3 dan k2 = -1 • yc= c1 e3x + c2 e-x PUPD: y = yc + yp = c1e3x + c2e-x – 1/3x2 + 4/9x – 14/27 How???

  19. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 7 (Lanjutan) Cukup, karena Dx2 = 2x D2x2 = 2 D3x2 = 0 1

  20. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 PD Linier Tingkat n • Bentuk umum (yn + pn-1yn-1 + … + p1y’ + p0y = f(x) • P. Reduksi: F(D)y = 0, subs y = ekx PK: F(k) = 0 • Akar-akar karakteristik: k1, k2, k3,…, kn-1, kn • Jika k1 ≠ k2 ≠ k3 … ≠ kn-1 ≠ kn , maka yc = c1ek1x + c2ek2x + c3ek3x + … + cn-1ekn-1x + cneknx • Jika k1 = k2 = k3 = k4 = mdan k5 ≠ k6 ≠ k7 ≠… ≠ kn , maka yc = emx(c1+ c2x + c3x2 + c4x3)+ c5ek5x +…+ cneknx • Jika k1 = k2 = k3 = k4 = a+bidan k5 = k6 = k7 = k8 = a-bi, serta k9 ≠ k10 ≠… ≠ kn maka yc = eax[(c1+ c2x + c3x2 + c4x3)cos bx +(c5+ c6x + c7x2 + c8x3)sin bx] + c9ek9x +…+ cneknx

  21. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 1 • yiv – y = e2x • PD: (D4 – 1)y = e2x • PR: (D4 – 1)y = 0 subs y = ekx • PK: k4 – 1 = 0 (k2-1)(k2+1)=0 (k+1)(k-1)(k2+1)=0 k1 = -1, k2 = 1, k3,4 = ±i yc = c1e-x + c2ex + c3cosx + c4sinx

  22. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Example 1 (Lanjutan) • PUPD: y = yc + yp y = c1e-x + c2ex + c3cosx + c4sinx + 1/15e2x

  23. Jurusan Teknik SipilMatematika III (TS 4353) Fakultas Teknik Sipil dan Perencanaan Universitas Kristen PetraBab 1 Exercise • y”’-2y”-y’+2y=2x2-6x+4; y(0)=5, y’(0) = -5, y”(0) = 1

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