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478 DIFFERENTIAL EQUATIONS 4. 6 d2 y dx 2 + 5dy − 6y = 0; when x = 0, y = 5 and dx dy [ dx =−1. y = 3e 2 3 x + 2e − 2 3 x] 5. 4 d2 y dt 2 − 5dy + y = 0; when t = 0, y = 1 and dt dy [ dt =−2. y = 4e 4 1 t − 3e t] 6. (9D 2 + 30D + 25)y = 0, where D ≡ d dx ; 7. when x = 0, y = 0 and dy dx = 2. [ y = 2xe − 5 3 x] d 2 x dt 2 − 6dx + 9x = 0; when t = 0, x = 2 and dt dx dt = 0. [x = 2(1 − 3t)e3t ] d 2 y 8. dx 2 + 6dy + 13y = 0; when x = 0, y = 4 and dx dy dx = 0. [ y = 2e−3x (2 cos 2x + 3 sin 2x)] 9. (4D 2 + 20D + 125)θ = 0, where D ≡ d dt ; when t = 0, θ = 3 and dθ dt = 2.5. [θ = e −2.5t (3 cos 5t + 2 sin 5t)] 50.4 Further worked problems on practical differential equations of the form a d2 y dx 2 + bdy dx + cy = 0 Problem 4. The equation of motion of a body oscillating on the end of a spring is d 2 x + 100x = 0, dt2 where x is the displacement in metres of the body from its equilibrium position after time t seconds. Determine x in terms of t given that at time t = 0, x = 2m and dx dt = 0. An equation of the form d2 x dt 2 + m2 x = 0 is a differential equation representing simple harmonic motion (S.H.M.). Using the procedure of Section 50.2: (a) d2 x + 100x = 0 in D-operator form is dt2 (D 2 + 100)x = 0. (b) The auxiliary equation is m 2 + 100 = 0, i.e. m 2 =−100 and m = √ (−100), i.e. m =±j10. (c) Since the roots are complex, the general solution is x = e 0 (A cos 10t + B sin 10t), i.e. x = (A cos 10t + B sin 10t) metres (d) When t = 0, x = 2, thus 2 = A dx =−10A sin 10t + 10B cos 10t dt When t = 0, dx dt = 0 thus 0 =−10A sin 0 + 10B cos 0, i.e. B = 0 Hence the particular solution is x = 2 cos 10t metres Problem 5. Given the differential equation d 2 V dt 2 = ω2 V, where ω is a constant, show that its solution may be expressed as: V = 7 cosh ωt + 3 sinh ωt given the boundary conditions that when t = 0, V = 7 and dV dt = 3ω. Using the procedure of Section 50.2: (a) d2 V dt 2 = ω2 V, i.e. d2 V dt 2 − ω2 V = 0 in D-operator form is (D 2 − ω 2 )v = 0, where D ≡ d dx (b) The auxiliary equation is m 2 − ω 2 = 0, from which, m 2 = ω 2 and m =±ω. (c) Since the roots are real and different, the general solution is V = Ae ωt + Be −ωt (d) When t = 0, V = 7 hence 7 = A + B (1) dV dt = Aωeωt − Bωe −ωt
When thus t = 0, dV dt = 3ω, 3ω = Aω − Bω, i.e. 3 = A − B (2) From equations (1) and (2), A = 5 and B = 2 Hence the particular solution is V = 5e ωt + 2e −ωt Since sinh ωt = 1 2 (eωt − e −ωt ) and cosh ωt = 2 1 (eωt + e −ωt ) then sinh ωt + cosh ωt = e ωt and cosh ωt − sinh ωt = e −ωt from Chapter 5. Hence the particular solution may also be written as V = 5(sinh ωt + cosh ωt) + 2(cosh ωt − sinh ωt) i.e. V = (5 + 2) cosh ωt + (5 − 2) sinh ωt i.e. V = 7 cosh ωt + 3 sinh ωt SECOND ORDER DIFFERENTIAL EQUATIONS (HOMOGENEOUS) 479 When R = 200, L = 0.20 and C = 20 × 10 −6 , then √ √√√ [ ( − 200 ) ] 200 2 0.20 ± 4 − 0.20 (0.20)(20 × 10 −6 ) m = 2 = −1000 ± √ 0 =−500 2 (c) Since the two roots are real and equal (i.e. −500 twice, since for a second order differential equation there must be two solutions), the general solution is i = (At + B)e −500t . (d) When t = 0, i = 0, hence B = 0 di dt = (At + B)(−500e−500t ) + (e −500t )(A), by the product rule When t = 0, di = 100, thus 100 =−500B + A dt i.e. A = 100, since B = 0 Hence the particular solution is Problem 6. The equation d 2 i dt 2 + R di L dt + 1 LC i = 0 represents a current i flowing in an electrical circuit containing resistance R, inductance L and capacitance C connected in series. If R = 200 ohms, L = 0.20 henry and C = 20 × 10 −6 farads, solve the equation for i given the boundary conditions that when t = 0, i = 0 and di dt = 100. Using the procedure of Section 50.2: (a) d2 i dt 2 + R L di dt + 1 LC ( D 2 + R L D + 1 LC i = 0 in D-operator form is ) i = 0 where D ≡ d dt (b) The auxiliary equation is m 2 + R L m + 1 LC = 0 √ √√√ [ (R − R ) 2 ( ) ] 1 L ± − 4(1) L LC Hence m = 2 i = 100te −500t Problem 7. The oscillations of a heavily damped pendulum satisfy the differential equation d2 x dt 2 + 6dx + 8x = 0, where x cm is dt the displacement of the bob at time t seconds. The initial displacement ( is equal to +4 cm and the initial velocity i.e. dx ) is 8 cm/s. Solve the dt equation for x. Using the procedure of Section 50.2: (a) d2 x dt 2 + 6dx + 8x = 0 in D-operator form is dt (D 2 + 6D + 8)x = 0, where D ≡ d dt (b) The auxiliary equation is m 2 + 6m + 8 = 0. Factorising gives: (m + 2)(m + 4) = 0, from which, m =−2orm =−4. (c) Since the roots are real and different, the general solution is x = Ae −2t + Be −4t . (d) Initial displacement means that time t = 0. At this instant, x = 4. I
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Page 15 and 16: Hence from equation (4): ye x = xe
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Page 51 and 52: POWER SERIES METHODS OF SOLVING ORD
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