Tutorial 12: 2nd Order Differential Equation
Q1: Given the governing equation for RLC electrical circuit: Ld2q(t)dt2+Rdq(t)dt+1Cq(t)=E(t)Ld2q(t)dt2+Rdq(t)dt+1Cq(t)=E(t). An inductor of L=50L=50 henrys, a resistor of R=5R=5 ohms and a capacitor of C=8C=8 farads are connected in series with an emf of EE volts. At t=0t=0, the switch SS is closed. Find the charge and current at any time t>0t>0 if the voltage is supplied by (a) DC battery, E(t)=αE(t)=α volts or (b) AC generator, E(t)=be−3tE(t)=be−3t volts.
Replace αα with the last three digits of your matric number. For example, if your matric number is KHA110108, your αα is thus 108.
Replace bb with the last three digits of your matric number divided by 5. For example, if your matric number is KHA110108, your bb is thus 108/5.
Solution
Question (a)
Assume α=100α=100
50d2q(t)dt2+5dq(t)dt+18q(t)=100,q(0)=0,i(0)=050d2q(t)dt2+5dq(t)dt+18q(t)=100,q(0)=0,i(0)=0Step 1: Homogeneous Part 50d2q(t)dt2+5dq(t)dt+18q(t)=050d2q(t)dt2+5dq(t)dt+18q(t)=0 |
Step 2: Nonhomogeneous Part 50d2q(t)dt2+5dq(t)dt+18q(t)=100,r(t)=eαtPn(t)=10050d2q(t)dt2+5dq(t)dt+18q(t)=100,r(t)=eαtPn(t)=100 |
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Characteristic equation: 50m2+5m+18=0400m2+40m+1=0(20m+1)2=0m=−12050m2+5m+18=0400m2+40m+1=0(20m+1)2=0m=−120 Comment: Repeated real root (m1=m2=−0.005m1=m2=−0.005) Complementary solution: qc(t)=c1e−0.05t+c2te−0.005tqc(t)=c1e−0.05t+c2te−0.005t |
RHS is a polynomial function, 100 α=0, n=0α=0, n=0 Since α≠m1 or m2α≠m1 or m2, qp(t)=e(αt)Qn(t)=Aqp(t)=e(αt)Qn(t)=A Thus, the total particular solution is qp(t)=Aqp(t)=A Differentiate it gives, ddx(qp(t))=0d2dx2(qp(t))=0ddx(qp(t))=0d2dx2(qp(t))=0 Then, we get 50(0)+5(0)+18q=10018A=100A=80050(0)+5(0)+18q=10018A=100A=800 Actual particular solution, qp(x)=800qp(x)=800 |
The complete/general solution to 50d2q(t)dt2+5dq(t)dt+18q(t)=10050d2q(t)dt2+5dq(t)dt+18q(t)=100 is
Charge solution
qtotal(t)=qc+qp=c1e−0.05t+c2te−0.05t+800qtotal(t)=qc+qp=c1e−0.05t+c2te−0.05t+800Current solution
i(t)=dqtotaldt=−0.05c1e−0.05t+c2e−0.05t+(e−0.05t(−0.05)(c2t))=c2e−0.05t−0.05e−0.05t(c1+c2t)i(t)=dqtotaldt=−0.05c1e−0.05t+c2e−0.05t+(e−0.05t(−0.05)(c2t))=c2e−0.05t−0.05e−0.05t(c1+c2t)Consider the initial value problem,
q(0)=0⟹qtotal(0)=c1e−0.05(0)+c2(0)e−0.05(0)+800=0⟹c1=−800q(0)=0⟹qtotal(0)=c1e−0.05(0)+c2(0)e−0.05(0)+800=0⟹c1=−800 i(0)=0⟹i(0)=c2e−0.05(0)−0.05e−0.05(0)(−800+c2(0))=0⟹c2=−40i(0)=0⟹i(0)=c2e−0.05(0)−0.05e−0.05(0)(−800+c2(0))=0⟹c2=−40Actual Charge solution
q(t)=−800e−0.05t−40te−0.05t+800=−e−0.05t(800+40t)+800q(t)=−800e−0.05t−40te−0.05t+800=−e−0.05t(800+40t)+800Actual Current solution
i(t)=−40e−0.05t−0.05e−0.05t(−800−40t)=e−0.05t(−40−0.05(−800−40t))=e−0.05t(−40+40+2t)i(t)=e−0.05t(2t)i(t)=−40e−0.05t−0.05e−0.05t(−800−40t)=e−0.05t(−40−0.05(−800−40t))=e−0.05t(−40+40+2t)i(t)=e−0.05t(2t)Question (b)
Assume b=20b=20
50d2q(t)dt2+5dq(t)dt+18q(t)=20e−3t,q(0)=0,i(0)=050d2q(t)dt2+5dq(t)dt+18q(t)=20e−3t,q(0)=0,i(0)=0Step 1: Homogeneous Part 50d2q(t)dt2+5dq(t)dt+18q(t)=050d2q(t)dt2+5dq(t)dt+18q(t)=0 |
Step 2: Nonhomogeneous Part 50d2q(t)dt2+5dq(t)dt+18q(t)=20e−3t,r(t)=eαtPn(t)=20e−3t50d2q(t)dt2+5dq(t)dt+18q(t)=20e−3t,r(t)=eαtPn(t)=20e−3t |
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Characteristic equation: 50m2+5m+18=0400m2+40m+1=0(20m+1)2=0m=−12050m2+5m+18=0400m2+40m+1=0(20m+1)2=0m=−120 Comment: Repeated real root (m1=m2=−0.005m1=m2=−0.005) Complementary solution: qc(t)=c1e−0.05t+c2te−0.005tqc(t)=c1e−0.05t+c2te−0.005t |
RHS is a exponential function, 100 α=−3, n=0α=−3, n=0, Since α≠m1 or m2α≠m1 or m2, qp(t)=e(αt)Qn(t)=Ae−3tqp(t)=e(αt)Qn(t)=Ae−3t Thus, the total particular solution is qp(t)=Ae−3tqp(t)=Ae−3t Differentiate it gives, ddx(qp(t))=−3Ae−3td2dx2(qp(t))=9Ae−3tddx(qp(t))=−3Ae−3td2dx2(qp(t))=9Ae−3t Then, we get 50(9Ae−3t)+5(−3Ae−3t)+18(Ae−3t)=20e−3te−3t(450A−15A+18A)=e−3t(20)A=0.0459650(9Ae−3t)+5(−3Ae−3t)+18(Ae−3t)=20e−3te−3t(450A−15A+18A)=e−3t(20)A=0.04596 Actual particular solution, qp(x)=0.04596e−3tqp(x)=0.04596e−3t |
The complete/general solution to 50d2q(t)dt2+5dq(t)dt+18q(t)=20e−3t50d2q(t)dt2+5dq(t)dt+18q(t)=20e−3t is
Charge solution
qtotal(t)=qc+qp=c1e−0.05t+c2te−0.05t+0.04596e−3tqtotal(t)=qc+qp=c1e−0.05t+c2te−0.05t+0.04596e−3tCurrent solution
i(t)=dqtotaldt=−0.05c1e−0.05t+c2e−0.05t+(e−0.05t(−0.05)(c2t))+(−3(0.04596)e−3t)=c2e−0.05t−0.05e−0.05t(c1+c2t)−0.1379e−3ti(t)=dqtotaldt=−0.05c1e−0.05t+c2e−0.05t+(e−0.05t(−0.05)(c2t))+(−3(0.04596)e−3t)=c2e−0.05t−0.05e−0.05t(c1+c2t)−0.1379e−3tConsider the initial value problem,
q(0)=0⟹qtotal(0)=c1e−0.05(0)+c2(0)e−0.05(0)+0.04596e−3(0)=0⟹c1=−0.04596q(0)=0⟹qtotal(0)=c1e−0.05(0)+c2(0)e−0.05(0)+0.04596e−3(0)=0⟹c1=−0.04596 i(0)=0⟹i(0)=c2e−0.05(0)−0.05e−0.05(0)(−0.04596+c2(0))−0.1379e−3(0)=0i(0)=0⟹i(0)=c2e−0.05(0)−0.05e−0.05(0)(−0.04596+c2(0))−0.1379e−3(0)=0 ⟹c2=0.1356⟹c2=0.1356Actual Charge solution
q(t)=−0.04596e−0.05t+0.1356te−0.05t+0.04596e−3tq(t)=−0.04596e−0.05t+0.1356te−0.05t+0.04596e−3tActual Current solution
i(t)=0.1356e−0.05t−0.05e−0.05t(−0.04596+0.1356t)−0.1379e−3ti(t)=e−0.05t(0.1379−0.00678t)−0.1379e−3ti(t)=0.1356e−0.05t−0.05e−0.05t(−0.04596+0.1356t)−0.1379e−3ti(t)=e−0.05t(0.1379−0.00678t)−0.1379e−3tQ2: The vibration transmission from the effect of equipment/ machine vibration to its structure (e.g. washing machine attached to the ground or engine attached to the car structure) can be modelled as 1 DOF spring-damper-mass vibration problem. It can be categorised into two conditions as follows.
(a) Transient Condition (Free Vibration) | (b) Steady State Condition (Forced Vibration) |
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Governing equation: md2x(t)dt2+cdx(t)dt+kx(t)=0md2x(t)dt2+cdx(t)dt+kx(t)=0 (machine at rest) x(0)=x0, ˙x(0)=˙x0x(0)=x0, ˙x(0)=˙x0 |
Governing equation: md2x(t)dt2+cdx(t)dt+kx(t)=Pcosωtmd2x(t)dt2+cdx(t)dt+kx(t)=Pcosωt (machine is rotating with cyclic/harmonic force) x(0)=x0, ˙x(0)=˙x0x(0)=x0, ˙x(0)=˙x0 |
The homogeneous 2nd order ODE is known as characteristic equation because it represents the chactacteristic of many systems. It has the complementary solution (ycyc). | The solution of nonhomogeneous 2nd order ODE is the summation of complementary solution (ycyc) and particular solution (ypyp). |
Note: In vibration field, the characteristic eqn. determines the dynamic characteristic of the vibrating system such as the natural frequency which causes mechanical resonance phenomenon. By understanding the dynamic behaviour of the system though the formulation of ODE, engineer can design a safer and reliable structure/ machine. In electrical field, engineers utilize the electrical resonance in radio tuning application through the formulation of ODE’s characteristic eqn. The detail of these are out of the scope in this study. Students are encouraged to utilize the basic of the mathematical tool learned in this course for their future engineering application.
Let the governing equation for a vibrating car structure:
2d2x(t)dt2+7dx(t)dt+8x(t)=F(t); where F(t) is the forcing function and x(0)=2, ˙x(0)=02d2x(t)dt2+7dx(t)dt+8x(t)=F(t); where F(t) is the forcing function and x(0)=2, ˙x(0)=0Find the total solution for the 2nd order ODE equation if the forcing function is given as follows:
(a) No excitation, F(t)=0F(t)=0 and it is subjected to initial condition.
(b) Repeat the same problem in 2(a) with various combinations of damping, i.e. 2d2x(t)dt2+8dx(t)dt+8x(t)=F(t)2d2x(t)dt2+8dx(t)dt+8x(t)=F(t).
(c) Repeat the same problem in 2(a) with various combinations of damping, i.e. 2d2x(t)dt2+9dx(t)dt+8x(t)=F(t)2d2x(t)dt2+9dx(t)dt+8x(t)=F(t).
Solution
Question (a)
Homogeneous Part, 2d2x(t)dt2+9dx(t)dt+8x(t)=0;x(0)=2,˙x(0)=02d2x(t)dt2+9dx(t)dt+8x(t)=0;x(0)=2,˙x(0)=0
Characteristic equation:
2m2+7m+8=0m=−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√1542m2+7m+8=0m=−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√154Comment: A pair of complex conjugates roots (m1=−74+i√154; m2=−74−i√154m1=−74+i√154; m2=−74−i√154)
Complementary solution:
x(t)=c1e(−74+i)t+c2e(−74−i)t=e(−74)t(c1e(i√154)t+c2e(−i√154)t)=e(−74+i)t(Acos√154t+Bsin√154t)x(t)=c1e(−74+i)t+c2e(−74−i)t=e(−74)t(c1e(i√154)t+c2e(−i√154)t)=e(−74+i)t(Acos√154t+Bsin√154t)where e±ix=cosx±isinx, A=c1+c2, B=i(c1−c2)e±ix=cosx±isinx, A=c1+c2, B=i(c1−c2)
Initial Value Problem
x(t)=c1e(−74+i√154)t+c2e(−74−i√154)t˙x(t)=(−74+i√154)c1e(−74+i√154)t+(−74−i√154)c1e(−74−i√154)tx(t)=c1e(−74+i√154)t+c2e(−74−i√154)t˙x(t)=(−74+i√154)c1e(−74+i√154)t+(−74−i√154)c1e(−74−i√154)t x(0)=c1e(−74+i√154)(0)+c2e(−74−i√154)(0)=2⟹c1+c2=2x(0)=c1e(−74+i√154)(0)+c2e(−74−i√154)(0)=2⟹c1+c2=2(1) ˙x(0)=(−74+i√154)c1e(−74+i√154)(0)+(−74−i√154)c1e(−74−i√154)(0)⟹(−74+i√154)(2−c2)+(−74−i√154)c2=0˙x(0)=(−74+i√154)c1e(−74+i√154)(0)+(−74−i√154)c1e(−74−i√154)(0)⟹(−74+i√154)(2−c2)+(−74−i√154)c2=0(2)Solving (1) and (2) simultaneously, c1=1−1.807i; c2=1+1.807ic1=1−1.807i; c2=1+1.807i
x(t)=(1−1.807i)e(−74+i√154)t+(1+1.807i)e(−74−i√154)tx(t)=(1−1.807i)e(−74+i√154)t+(1+1.807i)e(−74−i√154)tQuestion (b)
Homogeneous Part, 2d2x(t)dt2+9dx(t)dt+8x(t)=0;x(0)=2,˙x(0)=02d2x(t)dt2+9dx(t)dt+8x(t)=0;x(0)=2,˙x(0)=0
Characteristic equation:
2m2+8m+8=0m=−8±√82−4(2)(8)2(2)=−8±√04=−84=−22m2+8m+8=0m=−8±√82−4(2)(8)2(2)=−8±√04=−84=−2Comment: Repeated real root (m1=m2=−2m1=m2=−2)
Complementary solution: x(t)=c1e−2t+c2te−2tx(t)=c1e−2t+c2te−2t
Initial Value Problem
x(t)=c1e−2t+c2te−2t˙x(t)=−2c1e−2t+c2t(−2)e−2t+e−2tc2x(t)=c1e−2t+c2te−2t˙x(t)=−2c1e−2t+c2t(−2)e−2t+e−2tc2 x(0)=c1e(−2)(0)+c2(0)e(−2)(0)=2⟹c1=2x(0)=c1e(−2)(0)+c2(0)e(−2)(0)=2⟹c1=2 ˙x(0)=−2(2)e(−2)(0)+c2(0)(−2)e(−2)(0)+e(−2)(0)c2=0⟹c2=4˙x(0)=−2(2)e(−2)(0)+c2(0)(−2)e(−2)(0)+e(−2)(0)c2=0⟹c2=4 x(t)=2e−2t+4te−2tx(t)=2e−2t+4te−2tQuestion (c)
Homogeneous Part, 2d2x(t)dt2+9dx(t)dt+8x(t)=0;x(0)=2,˙x(0)=02d2x(t)dt2+9dx(t)dt+8x(t)=0;x(0)=2,˙x(0)=0
Characteristic equation:
2m2+9m+8=0m=−9±√92−4(2)(8)2(2)=−9±√1742m2+9m+8=0m=−9±√92−4(2)(8)2(2)=−9±√174Comment: Real and distinct root (m1=−1.2192; m2=−3.2808m1=−1.2192; m2=−3.2808)
Complementary solution: x(t)=c1e(−1.2192)t+c2e(−3.2808)tx(t)=c1e(−1.2192)t+c2e(−3.2808)t
Initial Value Problem
x(t)=c1e(−1.2192)t+c2e(−3.2808)t˙x(t)=−1.2192c1e(−1.2192)t−3.2808c2e(−3.2808)tx(t)=c1e(−1.2192)t+c2e(−3.2808)t˙x(t)=−1.2192c1e(−1.2192)t−3.2808c2e(−3.2808)t x(0)=c1e(−1.2192)(0)+c2e(−3.2808)(0)=2⟹c1+c2=2x(0)=c1e(−1.2192)(0)+c2e(−3.2808)(0)=2⟹c1+c2=2 ˙x(0)=−1.2192c1e(−1.2192)(0)−3.2808c2e(−3.2808)(0)=0⟹−1.2192c1−3.2808c2e=0˙x(0)=−1.2192c1e(−1.2192)(0)−3.2808c2e(−3.2808)(0)=0⟹−1.2192c1−3.2808c2e=0Solving (1) and (2) simultaneously, c1=3.1828; c2=−1.1828c1=3.1828; c2=−1.1828
x(t)=3.1828e(−1.2192)t−1.1828e(−3.2808)tx(t)=3.1828e(−1.2192)t−1.1828e(−3.2808)tQ3: Continue the problem 2. Let the governing equation for a vibrating car structure: 2d2x(t)dt2+7dx(t)dt+8x(t)=F(t)2d2x(t)dt2+7dx(t)dt+8x(t)=F(t); where F(t)F(t) is the forcing function and x(0)=2, ˙x(0)=0x(0)=2, ˙x(0)=0. Find the total solution for the 2nd order ODE equation if the forcing function is given as follows:
(a) Engine excitation F(t)=5cos10tF(t)=5cos10t
(b) Engine excitation F(t)=8sin8tF(t)=8sin8t
(c) Engine excitation F(t)=e−10tF(t)=e−10t
(d) Engine excitation F(t)=e−10tcos10tF(t)=e−10tcos10t [Hint/Alternative: Superposition]
(e) Engine excitation F(t)=5cos10t+e−10tF(t)=5cos10t+e−10t
(f) Road excitation F(t)=10F(t)=10
(g) Road excitation F(t)=5t2+7t+9F(t)=5t2+7t+9
(h) Road excitation F(t)=6tet+3tF(t)=6tet+3t
Hint: Student just need to show an example for the solution of homogenous part once and do not need to repeat the same step in other examples if it is needed. To further master the skill to solve2nd order ODE problem, students can repeat Q3(a-h) for various combinations of damping as shown in Q2(b) and Q2(c) respectively.
Solution
Question (a)
Step 1: Homogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=02d2x(t)dt2+7dx(t)dt+8x(t)=0 |
Step 2: Nonhomogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=5cos10t2d2x(t)dt2+7dx(t)dt+8x(t)=5cos10t, r(t)=eαtPn(t)=5cos10t=5Re[ei10t]r(t)=eαtPn(t)=5cos10t=5Re[ei10t] |
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Characteristic equation: 2m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√1542m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√154 Comment: A pair of complex conjugates roots (m1=−74+i√154, m2=−74−i√154m1=−74+i√154, m2=−74−i√154) Complementary solution: x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t)x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t) |
For exponential function, 5Re[ei10t]5Re[ei10t], α=10i, n=0α=10i, n=0 Since α≠m1 or m2α≠m1 or m2, Assume xp,real=Re[xp]xp,real=Re[xp], xp(t)=e(αt)Qn(t)=Ae(10it)xp(t)=e(αt)Qn(t)=Ae(10it) Differentiate it gives,ddx(xp(t))=10iAe(10it)d2dx2(xp(t))=−100Ae(10it)ddx(xp(t))=10iAe(10it)d2dx2(xp(t))=−100Ae(10it) Then, we get 2d2x(t)dt2+7dx(t)dt+8x(t)=5cos10t2(−100Ae(10it))+7(10iAe(10it))+8(Ae(10it))=5cos10te(10it)(−192A+70iA)=5e(10it)2d2x(t)dt2+7dx(t)dt+8x(t)=5cos10t2(−100Ae(10it))+7(10iAe(10it))+8(Ae(10it))=5cos10te(10it)(−192A+70iA)=5e(10it) Comparing the coefficient, (−192A+70iA)=5⟹A=570i−192(−192A+70iA)=5⟹A=570i−192 Actual particular solution, xp(t)=570i−192e(10it)=570i−192(cos10t+isin10t)−70i−192−70i−192=(−960cos10t+350sin10t)+i(−350cos10t−960sin10t)41764xp(t)=570i−192e(10it)=570i−192(cos10t+isin10t)−70i−192−70i−192=(−960cos10t+350sin10t)+i(−350cos10t−960sin10t)41764 xp,real(t)=Re[xp]=−0.02299cos10t+0.00838sin10txp,real(t)=Re[xp]=−0.02299cos10t+0.00838sin10t |
The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=5cos10t2d2x(t)dt2+7dx(t)dt+8x(t)=5cos10t is xtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)−0.02299cos10t+0.00838sin10txtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)−0.02299cos10t+0.00838sin10t.
Initial Value Problem
x(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)−0.02299cos10(0)+0.00838sin10(0)=2⟹c1+c2=2.02299x(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)−0.02299cos10(0)+0.00838sin10(0)=2⟹c1+c2=2.02299 ˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)+0.2299sin10t+0.0838cos10t˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)+0.2299sin10(0)+0.0838cos10(0)˙x(0)(2.02299i√154−c2i2√154)−74(2.02299)+0.838=0˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)+0.2299sin10t+0.0838cos10t˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)+0.2299sin10(0)+0.0838cos10(0)˙x(0)(2.02299i√154−c2i2√154)−74(2.02299)+0.838=0 c2=3.4564−1.9588i−1.9365i×ii=1.7849i+1.0115;c1=1.0115−1.7849ic2=3.4564−1.9588i−1.9365i×ii=1.7849i+1.0115;c1=1.0115−1.7849i x(t)=e(−74)t((1.0115−1.7849i)e(i√154)t+(1.0115−1.7849i)e(i√154)t)−0.02299cos10t+0.00838sin10tx(t)=e(−74)t((1.0115−1.7849i)e(i√154)t+(1.0115−1.7849i)e(i√154)t)−0.02299cos10t+0.00838sin10tQuestion (b)
Step 1: Homogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=02d2x(t)dt2+7dx(t)dt+8x(t)=0 |
Step 2: Nonhomogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=8sin8t2d2x(t)dt2+7dx(t)dt+8x(t)=8sin8t, r(t)=eαtPn(t)=8sin8t=8Im[ei8t]r(t)=eαtPn(t)=8sin8t=8Im[ei8t] |
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Characteristic equation: 2m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√1542m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√154 Comment: A pair of complex conjugates roots (m1=−74+i√154, m2=−74−i√154m1=−74+i√154, m2=−74−i√154) Complementary solution: x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t)x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t) |
For exponential function, 8Im[ei8t]8Im[ei8t], α=8i, n=0α=8i, n=0 Since α≠m1 or m2α≠m1 or m2, Assume xp,real=Im[xp]xp,real=Im[xp], xp(t)=e(αt)Qn(t)=Ae(8it)xp(t)=e(αt)Qn(t)=Ae(8it) Differentiate it gives, ddx(xp(t))=8iAe(8it)d2dx2(xp(t))=−64Ae(8it)ddx(xp(t))=8iAe(8it)d2dx2(xp(t))=−64Ae(8it) Then, we get 2d2x(t)dt2+7dx(t)dt+8x(t)=8sin8t2(−64Ae(8it))+7(8iAe(8it))+8(Ae(8it))=8sin8te(8it)(56i−120)A=8e(8it)2d2x(t)dt2+7dx(t)dt+8x(t)=8sin8t2(−64Ae(8it))+7(8iAe(8it))+8(Ae(8it))=8sin8te(8it)(56i−120)A=8e(8it) Comparing the coefficient, (56i−120)A=8⟹A=856i−120(56i−120)A=8⟹A=856i−120 Actual particular solution, xp(t)=856i−120e(8it)=856i−120(cos8t+isin8t)−56i−120−56i−120=(448sin8t−960cos8t)+i(−448cos8t−960sin8t)17536xp(t)=856i−120e(8it)=856i−120(cos8t+isin8t)−56i−120−56i−120=(448sin8t−960cos8t)+i(−448cos8t−960sin8t)17536 xp,real(t)=Im[xp]=−7274cos8t−15274sin8txp,real(t)=Im[xp]=−7274cos8t−15274sin8t |
The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=8sin8t2d2x(t)dt2+7dx(t)dt+8x(t)=8sin8t is xtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)−7274cos8t−15274sin8txtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)−7274cos8t−15274sin8t
Initial Value Problem
x(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)−7274cos8(0)−15274sin8(0)=2⟹c1+c2=2.0255x(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)−7274cos8(0)−15274sin8(0)=2⟹c1+c2=2.0255 ˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)−56274cos8t−90274sin8t˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)+56274cos8(0)−90274sin8(0)=0˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)−56274cos8t−90274sin8t˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)+56274cos8(0)−90274sin8(0)=0 c2=3.9826−1.9612i−1.9364i×ii=2.0567i+1.0128;c1=1.028−2.0567ic2=3.9826−1.9612i−1.9364i×ii=2.0567i+1.0128;c1=1.028−2.0567i x(t)=e(−74)t((1.0128−2.0567i)e(i√154)t+(1.0128+2.0567i)e(i√154)t)−7274cos8t−15274sin8tx(t)=e(−74)t((1.0128−2.0567i)e(i√154)t+(1.0128+2.0567i)e(i√154)t)−7274cos8t−15274sin8tQuestion (c)
Step 1: Homogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=02d2x(t)dt2+7dx(t)dt+8x(t)=0 |
Step 2: Nonhomogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=e−10t2d2x(t)dt2+7dx(t)dt+8x(t)=e−10t, $$r( t) =e^{\alpha t} P_{n}( t) =e^{-10t} |
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Characteristic equation: 2m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√1542m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√154 Comment: A pair of complex conjugates roots (m1=−74+i√154, m2=−74−i√154m1=−74+i√154, m2=−74−i√154) Complementary solution: x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t)x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t) |
For exponential function, α=−10, n=0α=−10, n=0 Since α≠m1 or m2α≠m1 or m2, xp(t)=e(αt)Qn(t)=Ae(−10t)xp(t)=e(αt)Qn(t)=Ae(−10t) Differentiate it gives, ddx(xp(t))=−10Ae(−10t)d2dx2(xp(t))=100Ae(−10t)ddx(xp(t))=−10Ae(−10t)d2dx2(xp(t))=100Ae(−10t) Then, we get 2d2x(t)dt2+7dx(t)dt+8x(t)=e(−10t)2(100Ae(−10t))+7(−10Ae(−10t))+8(Ae(−10t))=e(−10t)e(8it)(200−70+8)A=e(−10t)2d2x(t)dt2+7dx(t)dt+8x(t)=e(−10t)2(100Ae(−10t))+7(−10Ae(−10t))+8(Ae(−10t))=e(−10t)e(8it)(200−70+8)A=e(−10t) Comparing the coefficient, 138A=1⟹A=1138138A=1⟹A=1138 Actual particular solution, xp(t)=e(αt)Qn(t)=1138e(−10t)xp(t)=e(αt)Qn(t)=1138e(−10t) |
The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=e−10t2d2x(t)dt2+7dx(t)dt+8x(t)=e−10t is xtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)+1138e−10txtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)+1138e−10t
Initial Value Problem
x(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)+1138e−10(0)=2⟹c1+c2=1.9928x(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)+1138e−10(0)=2⟹c1+c2=1.9928 ˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)−10138e−10t˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)−10138e−10(0)=0˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)−10138e−10t˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)−10138e−10(0)=0 c2=3.5599−1.9295i−1.9365i×ii=1.8383i+0.9964;c1=0.9964−1.8383ic2=3.5599−1.9295i−1.9365i×ii=1.8383i+0.9964;c1=0.9964−1.8383i x(t)=e(−74)t((0.9964−1.8383i)e(i√154)t+(0.9964+1.8383i)e(i√154)t)+1138e−10tx(t)=e(−74)t((0.9964−1.8383i)e(i√154)t+(0.9964+1.8383i)e(i√154)t)+1138e−10tQuestion (d)
Step 1: Homogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=02d2x(t)dt2+7dx(t)dt+8x(t)=0 |
Step 2: Nonhomogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=5e−10tcos10t,r(t)=eαtPn(t)=5e−10tcos10t=5e−10tRe[ei10t]=Re[5e(−10+10i)t]2d2x(t)dt2+7dx(t)dt+8x(t)=5e−10tcos10t,r(t)=eαtPn(t)=5e−10tcos10t=5e−10tRe[ei10t]=Re[5e(−10+10i)t] |
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Characteristic equation: 2m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√1542m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√154 Comment: A pair of complex conjugates roots (m1=−74+i√154, m2=−74−i√154m1=−74+i√154, m2=−74−i√154) Complementary solution: x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t)x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t) |
For exponential function, Re[5e(−10+10i)t]Re[5e(−10+10i)t], α=−10+10i, n=0α=−10+10i, n=0 Since α≠m1 or m2α≠m1 or m2, xp(t)=e(αt)Qn(t)=Ae(−10t)xp(t)=e(αt)Qn(t)=Ae(−10t) Assume xp,real=Re[xp]xp,real=Re[xp], xp(t)=e(αt)Qn(t)=Ae(−10+10i)txp(t)=e(αt)Qn(t)=Ae(−10+10i)t Differentiate it gives, ddx(xp(t))=(−10+10i)Ae(−10+10i)td2dx2(xp(t))=(−200i)Ae(−10+10i)tddx(xp(t))=(−10+10i)Ae(−10+10i)td2dx2(xp(t))=(−200i)Ae(−10+10i)t Then, we get 2d2x(t)dt2+7dx(t)dt+8x(t)=5e−10tcos10t2((−200i)Ae(−10+10i)t)+7((−10+10i)Ae(−10+10i)t)+8(Ae(−10+10i)t)=5e(−10+10i)te(−10+10i)t(2(−200i)+7(−10+10i)+8)A=5e(−10+10i)t2d2x(t)dt2+7dx(t)dt+8x(t)=5e−10tcos10t2((−200i)Ae(−10+10i)t)+7((−10+10i)Ae(−10+10i)t)+8(Ae(−10+10i)t)=5e(−10+10i)te(−10+10i)t(2(−200i)+7(−10+10i)+8)A=5e(−10+10i)t Comparing the coefficient, (2(−200i)+7(−10+10i)+8)A=5⟹A=5−62−330i(2(−200i)+7(−10+10i)+8)A=5⟹A=5−62−330i Actual particular solution, xp(t)=e(αt)Qn(t)=5−62−330ie(−10+10i)t=5−62−330ie(−10)t(cos10t+isin10t)×−63+330i−63+330i=(−1650sin10t−310cos10t)+i(1650cos10t−310sin10t)112744e(−10)txp(t)=e(αt)Qn(t)=5−62−330ie(−10+10i)t=5−62−330ie(−10)t(cos10t+isin10t)×−63+330i−63+330i=(−1650sin10t−310cos10t)+i(1650cos10t−310sin10t)112744e(−10)t xp,real(t)=Re[xp]=(−0.01463sin10t−0.002750cos10t)e(−10)txp,real(t)=Re[xp]=(−0.01463sin10t−0.002750cos10t)e(−10)t |
The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=5e−10tcos10t2d2x(t)dt2+7dx(t)dt+8x(t)=5e−10tcos10t is xtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)+(−0.01463sin10t−0.002750cos10t)e(−10)txtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)+(−0.01463sin10t−0.002750cos10t)e(−10)t
Initial Value Problem
(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)+(−0.01463sin10(0)−0.002750cos10(0))e(−10)0=2⟹c1+c2=2.0028(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)+(−0.01463sin10(0)−0.002750cos10(0))e(−10)0=2⟹c1+c2=2.0028 \begin{aligned} \dot{x}_{total} & =e^{\left( -\frac{7}{4}\right) t}\left( i\frac{\sqrt{15}}{4} c_{1} e^{\left( i\frac{\sqrt{15}}{4}\right) t} -i\frac{\sqrt{15}}{4} c_{2} e^{\left( -i\frac{\sqrt{15}}{4}\right) t}\right) -\frac{7}{4} e^{\left( -\frac{7}{4}\right) t}\left( c_{1} e^{\left( i\frac{\sqrt{15}}{4}\right) t} +c_{2} e^{\left( -i\frac{\sqrt{15}}{4}\right) t}\right)\\ & +( -0.1463\cos 10t+0.0275\sin 10t) e^{( -10) t} +( 0.1463\sin 10t+0.0275\cos 10t) e^{( -10) t}\\ \dot{x}( 0) & =e^{\left( -\frac{7}{4}\right) 0}\left( i\frac{\sqrt{15}}{4} c_{1} e^{\left( i\frac{\sqrt{15}}{4}\right) 0} -i\frac{\sqrt{15}}{4} c_{2} e^{\left( -i\frac{\sqrt{15}}{4}\right) 0}\right) -\frac{7}{4} e^{\left( -\frac{7}{4}\right) 0}\left( c_{1} e^{\left( i\frac{\sqrt{15}}{4}\right) 0} +c_{2} e^{\left( -i\frac{\sqrt{15}}{4}\right) 0}\right)\\ & +( -0.1463\cos 10( 0) +0.0275\sin 10( 0)) e^{( -10) 0} +( 0.1463\sin 10( 0) +0.0275\cos 10( 0)) e^{( -10) 0} =0\begin{aligned}\dot{x}_{total} & =e^{\left( -\frac{7}{4}\right) t}\left( i\frac{\sqrt{15}}{4} c_{1} e^{\left( i\frac{\sqrt{15}}{4}\right) t} -i\frac{\sqrt{15}}{4} c_{2} e^{\left( -i\frac{\sqrt{15}}{4}\right) t}\right) -\frac{7}{4} e^{\left( -\frac{7}{4}\right) t}\left( c_{1} e^{\left( i\frac{\sqrt{15}}{4}\right) t} +c_{2} e^{\left( -i\frac{\sqrt{15}}{4}\right) t}\right)\\
& +( -0.1463\cos 10t+0.0275\sin 10t) e^{( -10) t} +( 0.1463\sin 10t+0.0275\cos 10t) e^{( -10) t}\\
\dot{x}( 0) & =e^{\left( -\frac{7}{4}\right) 0}\left( i\frac{\sqrt{15}}{4} c_{1} e^{\left( i\frac{\sqrt{15}}{4}\right) 0} -i\frac{\sqrt{15}}{4} c_{2} e^{\left( -i\frac{\sqrt{15}}{4}\right) 0}\right) -\frac{7}{4} e^{\left( -\frac{7}{4}\right) 0}\left( c_{1} e^{\left( i\frac{\sqrt{15}}{4}\right) 0} +c_{2} e^{\left( -i\frac{\sqrt{15}}{4}\right) 0}\right)\\
& +( -0.1463\cos 10( 0) +0.0275\sin 10( 0)) e^{( -10) 0} +( 0.1463\sin 10( 0) +0.0275\cos 10( 0)) e^{( -10) 0} =0 c2=3.6237−1.9392i−1.9365×ii=1.8713+1.0014i;c1=3.8741−1.0014ic2=3.6237−1.9392i−1.9365×ii=1.8713+1.0014i;c1=3.8741−1.0014i x(t)=e(−74)t((3.8741−1.0014i)e(i√154)t+(1.8713+1.0014i)e(i√154)t)+(−0.01463sin10t−0.002750cos10t)e(−10)tx(t)=e(−74)t((3.8741−1.0014i)e(i√154)t+(1.8713+1.0014i)e(i√154)t)+(−0.01463sin10t−0.002750cos10t)e(−10)t
Question (e)
Step 1: Homogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=0 |
Step 2: Nonhomogeneous Part $$2\frac{d\power{2}x(t) |
dt\power{2}}+7\frac{dx(t) | dt}+8x(t)=5\cos{10t}+e\power{-10t},<br/>r( t)=e^{\alpha t} P_{n}( t) =5\cos 10t+e^{-10t}$$ |
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Characteristic equation: 2m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√154 Comment: A pair of complex conjugates roots (m1=−74+i√154, m2=−74−i√154) Complementary solution: x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t) |
For exponential function, e−10t, α=−10, n=0 Since α≠m1 or m2, xp1(t)=e(αt)Qn(t)=Ae(−10t) For cosine function, 5cos10t=Re[5e10it], α=10i, n=0 Since α≠m1 or m2, xp2(t)=e(αt)Qn(t)=Be(10it) Thus, the total particular solution is xp(t)=Ae(−10t)+Be(10it) Assume xp,real(t)=Re[xp], xp(t)=e(αt)Qn(t)=Ae(−10t)+Be(10it) Differentiate it gives, ddx(xp(t))=−10Ae(−10t)+10iBe10itd2dx2(xp(t))=100Ae(−10t)−100Be10it Then, we get 2d2x(t)dt2+7dx(t)dt+8x(t)=5cos10t+e−10t2(100Ae(−10t)−100Be10it)+7(−10Ae(−10t)+10iBe10it)+8(Ae(−10t)+Be(10it))=5e(10i)t+e−10te(−10)t(138A)+e(10it)(70iB−192B)=5e(10i)t+e−10t Comparing the coefficient, (70i−192)B=5⟹B=570i−192138A=1⟹A=1138 Actual particular solution, xp(t)=1138e(−10t)+570i−192e(10it)=1138e(−10t)+570i−192−70i−192−70i−192(cos10t+isin10t)=1138e(−10t)+i(−0.00838cos10t−0.02299sin10t)+(0.00838sin10t−0.02299cos10t) xp,real(t)=Re[xp]=1138e(−10t)−0.02299cos10t+0.00838sin10t |
The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=5cos10t+e−10t is xtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)+1138e(−10t)−0.02299cos10t+0.00838sin10t
Initial Value Problem
(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)+1138e(−10)0−0.02299cos10(0)+0.00838sin10(0)=2⟹c1+c2=2.0157 ˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)−10138e(−10t)+0.2299sin10t+0.0838cos10t˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)−10138e(−10)0+0.2299sin10(0)+0.0838cos10(0)=0 c2=3.5161−1.9517i−1.9517i×ii=1.0078+1.8740i;c1=1.0078−1.8740i x(t)=e(−74)t((1.0078−1.8740i)e(i√154)t+(1.0078+1.8740i)e(i√154)t)+1138e(−10t)−0.02299cos10t+0.00838sin10tQuestion (f)
Step 1: Homogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=0 |
Step 2: Nonhomogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=10,r(t)=eαtPn(t)=10 |
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Characteristic equation: 2m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√154 Comment: A pair of complex conjugates roots (m1=−74+i√154, m2=−74−i√154) Complementary solution: x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t) |
For polynomial function, 10, α=0, n=0 Since α≠m1 or m2, xp=e(αt)Qn(t)=A Differentiate it gives, ddx(xp(t))=0d2dx2(xp(t))=0 Then, we get 2d2x(t)dt2+7dx(t)dt+8x(t)=102(0)+7(0)+8A=10A=1.25 Actual particular solution,xp(t)=1.25 |
The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=10 is xtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)+1.25
Initial Value Problem
x(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)+1.25=2⟹c1+c2=0.75 ˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)=0˙x(0)=(i√154c1−i√154c2)−74(c1+c2)=0 i√154(0.75−c2)−i√154c2−74(0.75)=0⟹c2=0.3750+0.6778i; c1=0.3750−0.6778i x(t)=e(−74)t((0.3750−0.6778i)e(i√154)t+(0.3750+0.6778i)e(i√154)t)+1.25Question (g)
Step 1: Homogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=0 |
Step 2: Nonhomogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=5t2+7t+9, r(t)=eαtPn(t)=5t2+7t+9 |
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Characteristic equation: 2m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√154 Comment: A pair of complex conjugates roots (m1=−74+i√154, m2=−74−i√154) Complementary solution: x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t) |
For polynomial function, 5t2+7t+9, α=0, n=2 Since α≠m1 or m2, xp=e(αt)Qn(t)=At2+Bt+C Differentiate it gives,<br/>ddx(xp(t))=2At+Bd2dx2(xp(t))=2A Then, we get 2d2x(t)dt2+7dx(t)dt+8x(t)=5t2+7t+92(2A)+7(2At+B)+8(At2+Bt+C)=5t2+7t+94A+7B+8C+14At+8Bt+8A2t=5t2+7t+9 Comparing coefficients, t2:8A=5⟹A=0.625 t:14A+8B=7⟹14(0.625)+8B=7⟹B=−0.2188 t0:4A+7B+8C=9⟹4(0.625)+8)−0.2188)+8C=9⟹C=1.0040$<br/>Actualparticularsolution,<br/>x_{p}( t) =0.625t^{2} -0.2188B+1.0040$$ |
The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=5t2+7t+9 is xtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)+0.625t2−0.2188t+1.0040
Initial Value Problem
x(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)+0.625(0)2−0.2188(0)+1.0040=2⟹c1+c2=0.996 ˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)+1.25t−0.2188˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)+1.25(0)−0.2188=0˙x(0)=(i√154c1−i√154c2)−74(c1+c2)−0.2188=0 ⟹c2=1.9618−0.9644i−1.9365i=0.4980+1.0131i; c1=0.4980−1.0131i x(t)=e(−74)t((0.4980−1.0131i)e(i√154)t+(0.4980+1.0131i)e(i√154)t)+0.625t2−0.2188t+1.0040Question (h)
Step 1: Homogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=0 |
Step 2: Nonhomogeneous Part 2d2x(t)dt2+7dx(t)dt+8x(t)=6tet+3t, r(t)=eαtPn(t)=6tet+3t |
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Characteristic equation: 2m2+7m+8=0m=−−7±√72−4(2)(8)2(2)=−7±√−154=−74±i√154 Comment: A pair of complex conjugates roots (m1=−74+i√154, m2=−74−i√154) Complementary solution: x(t)=e(−74)t(c1e(i√154)t+c2e(−i√154)t) |
ui(t)=−∫r(t)x2(t)W(y1,y2)dt=−∫(6tet+3t)e(−74−i√154)t−i2√154e−144tdt=−12√15i∫te(2.75−i√154)tdt−6√15i∫te(2.75−i√154)tdt Integration by part, 12√15i∫t⏟ue(2.75−i√154)tdt⏟dv=−12√15i[t⏟ue(2.75−i√154)t(2.75−i√154)⏟v−∫e(2.75−i√154)t(2.75−i√154)⏟vdt⏟du]=−12√15i[te(2.75−i√154)t(2.75−i√154)−e(2.75−i√154)t(2.75−i√154)2] 6√15i∫te(2.75−i√154)tdt=−6√15i[t⏟ue(1.75−i√154)t(1.75−i√154)⏟v−∫e(1.75−i√154)t(1.75−i√154)⏟vdt⏟du]=−6√15i[te(1.75−i√154)t(1.75−i√154)−e(1.75−i√154)t(1.75−i√154)2] 1(t)=−12√15i[te(2.75−i√154)t(2.75−i√154)−e(2.75−i√154)t(2.75−i√154)2]−6√15i[te(1.75−i√154)t(1.75−i√154)−e(1.75−i√154)t(1.75−i√154)2] u2(t)=∫r(t)x1(t)W(x1,x2)dt=∫(6tet+3t)e(−74+i√154)t−i2√154e−144tdt=12√15i∫te(2.75+i√154)t+6√15i∫te(1.75+i√154)tdt Integration by part, 12√15i∫t⏟ue(2.75+i√154)tdt⏟dv=12√15i[t⏟ue(2.75+i√154)t(2.75+i√154)⏟v−∫e(2.75+i√154)t(2.75+i√154)⏟vdt⏟du]=12√15i[te(2.75+i√154)t(2.75+i√154)−e(2.75+i√154)t(2.75+i√154)2] 6√15i∫te(1.75+i√154)tdt=6√15i[t⏟ue(1.75+i√154)t(1.75+i√154)⏟v−∫e(1.75+i√154)t(1.75+i√154)2⏟vdt⏟du]=6√15i[te(1.75+i√154)t(1.75+i√154)−e(1.75+i√154)t(1.75+i√154)2] 6√15i∫te(1.75+i√154)tdt=6√15i[t⏟ue(1.75+i√154)t(1.75+i√154)⏟v−∫e(1.75+i√154)t(1.75+i√154)2⏟vdt⏟du]=6√15i[te(1.75+i√154)t(1.75+i√154)−e(1.75+i√154)t(1.75+i√154)2] 2(t)=12√15i[te(2.75+i√154)t(2.75+i√154)−e(2.75+i√154)t(2.75+i√154)2+6√15i[te(1.75+i√154)t(1.75+i√154)−e(1.75+i√154)t(1.75+i√154)2] xp=u1(t)x1(t)+u2(t)x2(t) where: u1(t)x1(t)={−12√15i[te(2.75−i√154)t(2.75−i√154)−e(2.75−i√154)t(2.75−i√154)2]−6√15i[te(1.75−i√154)t(1.75−i√154)−e(1.75−i√154)t(1.75−i√154)2]}e(−74+i√154)t=−12√15i[tet(2.75−i√154)−et(2.75−i√154)2]−6√15i[t1(1.75−i√154)−1(1.75−i√154)2] $$ \begin{aligned} |
u_{1}( t) x_{1}( t) & ={\frac{12}{\sqrt{15}} i[ t\frac{e^{\left( 2.75+i\frac{\sqrt{15}}{4}\right) t}}{\left( 2.75+i\frac{\sqrt{15}}{4}\right)} -\frac{e^{\left( 2.75+i\frac{\sqrt{15}}{4}\right) t}}{\left( 2.75+i\frac{\sqrt{15}}{4}\right)^{2}}\& +\frac{6}{\sqrt{15}} i\left[ t\frac{e^{\left( 1.75+i\frac{\sqrt{15}}{4}\right) t}}{\left( 1.75+i\frac{\sqrt{15}}{4}\right)} -\frac{e^{\left( 1.75+i\frac{\sqrt{15}}{4}\right) t}}{\left( 1.75+i\frac{\sqrt{15}}{4}\right)^{2}}\right]} e^{\left( -\frac{7}{4} -i\frac{\sqrt{15}}{4}\right) t}\& =\frac{12}{\sqrt{15}} i[ t\frac{e^{t}}{\left( 2.75+i\frac{\sqrt{15}}{4}\right)} -\frac{e^{t}}{\left( 2.75+i\frac{\sqrt{15}}{4}\right)^{2}}\& +\frac{6}{\sqrt{15}} i\left[ t\frac{1}{\left( 1.75+i\frac{\sqrt{15}}{4}\right)} -\frac{1}{\left( 1.75+i\frac{\sqrt{15}}{4}\right)^{2}}\right]\end{aligned}<br/><br/> xp(t)=u1(t)x1(t)+u2(t)x2(t)&=−12√15i[tet(2.75−i√154)−et(2.75−i√154)2]&+12√15i[tet(2.75+i√154)−et(2.75+i√154)2]&−6√15i[t1(1.75−i√154)−1(1.75−i√154)2]&+6√15i[t1(1.75+i√154)−1(1.75+i√154)2]&=−12√15i[tet(2.75−i√154)8.5−et(6.625+i1.375√15)72.25]&+12√15i[tet(2.75−i√154)8.5−et(6.625−i1.375√15)72.25]&−6√15i[t(1.75+i√154)4−(2.125+i0.875√15)16]&+6√15i[t(1.75−i√154)4−(2.125−i0.875√15)16]&=[−24√15itet(i√154)8.5+24√15iet(i1.375√15)8.5]&+[−12√15it(i√154)4+12√15i(i0.875√15)16]&=[+1217tet−132289et+34et−2132]$$ |
The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=6tet+3t is xtotal=xc+xp=e(−74)t(c1e(i√154)t+c2e(−i√154)t)+1217tet−132289et+34et−2132
Initial Value Problem
x(0)=e(−74)0(c1e(i√154)0+c2e(−i√154)0)+1217(0)e0−132289e0+34e0−2132=2⟹c1+c2=2.363 ˙xtotal=e(−74)t(i√154c1e(i√154)t−i√154c2e(−i√154)t)−74e(−74)t(c1e(i√154)t+c2e(−i√154)t)+1217(et(1+t))−132289et+34et˙x(0)=e(−74)0(i√154c1e(i√154)0−i√154c2e(−i√154)0)−74e(−74)0(c1e(i√154)0+c2e(−i√154)0)+1217(e0(1+0))−132289e0+34e0=0˙x(0)=(i√154c1−i√154c2)−74(c1+c2)+0.9991=0 ⟹ c1=1.1815−1.6195i, c2=1.1815+1.6195i x(t)=e(−74)t((1.1815−1.6195)e(i√154)t+(1.1815+1.6195)e(i√154)t)+1217tet−132289et+34et−2132