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Tutorial 12: 2nd Order Differential Equation


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Class Recording


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)=be3tE(t)=be3t volts.

Diagram for Question 1

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)=0
Step 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
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)=c1e0.05t+c2te0.005tqc(t)=c1e0.05t+c2te0.005t
RHS is a polynomial function, 100 α=0, n=0α=0, n=0

Since αm1 or m2αm1 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=c1e0.05t+c2te0.05t+800qtotal(t)=qc+qp=c1e0.05t+c2te0.05t+800

Current solution

i(t)=dqtotaldt=0.05c1e0.05t+c2e0.05t+(e0.05t(0.05)(c2t))=c2e0.05t0.05e0.05t(c1+c2t)i(t)=dqtotaldt=0.05c1e0.05t+c2e0.05t+(e0.05t(0.05)(c2t))=c2e0.05t0.05e0.05t(c1+c2t)

Consider the initial value problem,

q(0)=0qtotal(0)=c1e0.05(0)+c2(0)e0.05(0)+800=0c1=800q(0)=0qtotal(0)=c1e0.05(0)+c2(0)e0.05(0)+800=0c1=800 i(0)=0i(0)=c2e0.05(0)0.05e0.05(0)(800+c2(0))=0c2=40i(0)=0i(0)=c2e0.05(0)0.05e0.05(0)(800+c2(0))=0c2=40

Actual Charge solution

q(t)=800e0.05t40te0.05t+800=e0.05t(800+40t)+800q(t)=800e0.05t40te0.05t+800=e0.05t(800+40t)+800

Actual Current solution

i(t)=40e0.05t0.05e0.05t(80040t)=e0.05t(400.05(80040t))=e0.05t(40+40+2t)i(t)=e0.05t(2t)i(t)=40e0.05t0.05e0.05t(80040t)=e0.05t(400.05(80040t))=e0.05t(40+40+2t)i(t)=e0.05t(2t)

Question (b)

Assume b=20b=20

50d2q(t)dt2+5dq(t)dt+18q(t)=20e3t,q(0)=0,i(0)=050d2q(t)dt2+5dq(t)dt+18q(t)=20e3t,q(0)=0,i(0)=0
Step 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)=20e3t,r(t)=eαtPn(t)=20e3t50d2q(t)dt2+5dq(t)dt+18q(t)=20e3t,r(t)=eαtPn(t)=20e3t
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)=c1e0.05t+c2te0.005tqc(t)=c1e0.05t+c2te0.005t
RHS is a exponential function, 100 α=3, n=0α=3, n=0,

Since αm1 or m2αm1 or m2,
qp(t)=e(αt)Qn(t)=Ae3tqp(t)=e(αt)Qn(t)=Ae3t

Thus, the total particular solution is
qp(t)=Ae3tqp(t)=Ae3t

Differentiate it gives,
ddx(qp(t))=3Ae3td2dx2(qp(t))=9Ae3tddx(qp(t))=3Ae3td2dx2(qp(t))=9Ae3t

Then, we get
50(9Ae3t)+5(3Ae3t)+18(Ae3t)=20e3te3t(450A15A+18A)=e3t(20)A=0.0459650(9Ae3t)+5(3Ae3t)+18(Ae3t)=20e3te3t(450A15A+18A)=e3t(20)A=0.04596

Actual particular solution,
qp(x)=0.04596e3tqp(x)=0.04596e3t

The complete/general solution to 50d2q(t)dt2+5dq(t)dt+18q(t)=20e3t50d2q(t)dt2+5dq(t)dt+18q(t)=20e3t is

Charge solution

qtotal(t)=qc+qp=c1e0.05t+c2te0.05t+0.04596e3tqtotal(t)=qc+qp=c1e0.05t+c2te0.05t+0.04596e3t

Current solution

i(t)=dqtotaldt=0.05c1e0.05t+c2e0.05t+(e0.05t(0.05)(c2t))+(3(0.04596)e3t)=c2e0.05t0.05e0.05t(c1+c2t)0.1379e3ti(t)=dqtotaldt=0.05c1e0.05t+c2e0.05t+(e0.05t(0.05)(c2t))+(3(0.04596)e3t)=c2e0.05t0.05e0.05t(c1+c2t)0.1379e3t

Consider the initial value problem,

q(0)=0qtotal(0)=c1e0.05(0)+c2(0)e0.05(0)+0.04596e3(0)=0c1=0.04596q(0)=0qtotal(0)=c1e0.05(0)+c2(0)e0.05(0)+0.04596e3(0)=0c1=0.04596 i(0)=0i(0)=c2e0.05(0)0.05e0.05(0)(0.04596+c2(0))0.1379e3(0)=0i(0)=0i(0)=c2e0.05(0)0.05e0.05(0)(0.04596+c2(0))0.1379e3(0)=0 c2=0.1356c2=0.1356

Actual Charge solution

q(t)=0.04596e0.05t+0.1356te0.05t+0.04596e3tq(t)=0.04596e0.05t+0.1356te0.05t+0.04596e3t

Actual Current solution

i(t)=0.1356e0.05t0.05e0.05t(0.04596+0.1356t)0.1379e3ti(t)=e0.05t(0.13790.00678t)0.1379e3ti(t)=0.1356e0.05t0.05e0.05t(0.04596+0.1356t)0.1379e3ti(t)=e0.05t(0.13790.00678t)0.1379e3t

Q2: 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)
Diagram for Question 2 Diagram for Question 2
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)=0

Find 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±724(2)(8)2(2)=7±154=74±i1542m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i154

Comment: A pair of complex conjugates roots (m1=74+i154; m2=74i154m1=74+i154; m2=74i154)

Complementary solution:

x(t)=c1e(74+i)t+c2e(74i)t=e(74)t(c1e(i154)t+c2e(i154)t)=e(74+i)t(Acos154t+Bsin154t)x(t)=c1e(74+i)t+c2e(74i)t=e(74)t(c1e(i154)t+c2e(i154)t)=e(74+i)t(Acos154t+Bsin154t)

where e±ix=cosx±isinx, A=c1+c2, B=i(c1c2)e±ix=cosx±isinx, A=c1+c2, B=i(c1c2)

Initial Value Problem

x(t)=c1e(74+i154)t+c2e(74i154)t˙x(t)=(74+i154)c1e(74+i154)t+(74i154)c1e(74i154)tx(t)=c1e(74+i154)t+c2e(74i154)t˙x(t)=(74+i154)c1e(74+i154)t+(74i154)c1e(74i154)t x(0)=c1e(74+i154)(0)+c2e(74i154)(0)=2c1+c2=2x(0)=c1e(74+i154)(0)+c2e(74i154)(0)=2c1+c2=2(1) ˙x(0)=(74+i154)c1e(74+i154)(0)+(74i154)c1e(74i154)(0)(74+i154)(2c2)+(74i154)c2=0˙x(0)=(74+i154)c1e(74+i154)(0)+(74i154)c1e(74i154)(0)(74+i154)(2c2)+(74i154)c2=0(2)

Solving (1) and (2) simultaneously, c1=11.807i; c2=1+1.807ic1=11.807i; c2=1+1.807i

x(t)=(11.807i)e(74+i154)t+(1+1.807i)e(74i154)tx(t)=(11.807i)e(74+i154)t+(1+1.807i)e(74i154)t

Question (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±824(2)(8)2(2)=8±04=84=22m2+8m+8=0m=8±824(2)(8)2(2)=8±04=84=2

Comment: Repeated real root (m1=m2=2m1=m2=2)

Complementary solution: x(t)=c1e2t+c2te2tx(t)=c1e2t+c2te2t

Initial Value Problem

x(t)=c1e2t+c2te2t˙x(t)=2c1e2t+c2t(2)e2t+e2tc2x(t)=c1e2t+c2te2t˙x(t)=2c1e2t+c2t(2)e2t+e2tc2 x(0)=c1e(2)(0)+c2(0)e(2)(0)=2c1=2x(0)=c1e(2)(0)+c2(0)e(2)(0)=2c1=2 ˙x(0)=2(2)e(2)(0)+c2(0)(2)e(2)(0)+e(2)(0)c2=0c2=4˙x(0)=2(2)e(2)(0)+c2(0)(2)e(2)(0)+e(2)(0)c2=0c2=4 x(t)=2e2t+4te2tx(t)=2e2t+4te2t

Question (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±924(2)(8)2(2)=9±1742m2+9m+8=0m=9±924(2)(8)2(2)=9±174

Comment: 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)t3.2808c2e(3.2808)tx(t)=c1e(1.2192)t+c2e(3.2808)t˙x(t)=1.2192c1e(1.2192)t3.2808c2e(3.2808)t x(0)=c1e(1.2192)(0)+c2e(3.2808)(0)=2c1+c2=2x(0)=c1e(1.2192)(0)+c2e(3.2808)(0)=2c1+c2=2 ˙x(0)=1.2192c1e(1.2192)(0)3.2808c2e(3.2808)(0)=01.2192c13.2808c2e=0˙x(0)=1.2192c1e(1.2192)(0)3.2808c2e(3.2808)(0)=01.2192c13.2808c2e=0

Solving (1) and (2) simultaneously, c1=3.1828; c2=1.1828c1=3.1828; c2=1.1828

x(t)=3.1828e(1.2192)t1.1828e(3.2808)tx(t)=3.1828e(1.2192)t1.1828e(3.2808)t

Q3: 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)=e10tF(t)=e10t

(d) Engine excitation F(t)=e10tcos10tF(t)=e10tcos10t [Hint/Alternative: Superposition]

(e) Engine excitation F(t)=5cos10t+e10tF(t)=5cos10t+e10t

(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]
Characteristic equation:
2m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i1542m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i154

Comment: A pair of complex conjugates roots
(m1=74+i154, m2=74i154m1=74+i154, m2=74i154)

Complementary solution:
x(t)=e(74)t(c1e(i154)t+c2e(i154)t)x(t)=e(74)t(c1e(i154)t+c2e(i154)t)
For exponential function, 5Re[ei10t]5Re[ei10t], α=10i, n=0α=10i, n=0

Since αm1 or m2αm1 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)=5A=570i192(192A+70iA)=5A=570i192

Actual particular solution,
xp(t)=570i192e(10it)=570i192(cos10t+isin10t)70i19270i192=(960cos10t+350sin10t)+i(350cos10t960sin10t)41764xp(t)=570i192e(10it)=570i192(cos10t+isin10t)70i19270i192=(960cos10t+350sin10t)+i(350cos10t960sin10t)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(i154)t+c2e(i154)t)0.02299cos10t+0.00838sin10txtotal=xc+xp=e(74)t(c1e(i154)t+c2e(i154)t)0.02299cos10t+0.00838sin10t.

Initial Value Problem

x(0)=e(74)0(c1e(i154)0+c2e(i154)0)0.02299cos10(0)+0.00838sin10(0)=2c1+c2=2.02299x(0)=e(74)0(c1e(i154)0+c2e(i154)0)0.02299cos10(0)+0.00838sin10(0)=2c1+c2=2.02299 ˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)+0.2299sin10t+0.0838cos10t˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)+0.2299sin10(0)+0.0838cos10(0)˙x(0)(2.02299i154c2i2154)74(2.02299)+0.838=0˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)+0.2299sin10t+0.0838cos10t˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)+0.2299sin10(0)+0.0838cos10(0)˙x(0)(2.02299i154c2i2154)74(2.02299)+0.838=0 c2=3.45641.9588i1.9365i×ii=1.7849i+1.0115;c1=1.01151.7849ic2=3.45641.9588i1.9365i×ii=1.7849i+1.0115;c1=1.01151.7849i x(t)=e(74)t((1.01151.7849i)e(i154)t+(1.01151.7849i)e(i154)t)0.02299cos10t+0.00838sin10tx(t)=e(74)t((1.01151.7849i)e(i154)t+(1.01151.7849i)e(i154)t)0.02299cos10t+0.00838sin10t

Question (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]
Characteristic equation:
2m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i1542m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i154

Comment: A pair of complex conjugates roots
(m1=74+i154, m2=74i154m1=74+i154, m2=74i154)

Complementary solution:
x(t)=e(74)t(c1e(i154)t+c2e(i154)t)x(t)=e(74)t(c1e(i154)t+c2e(i154)t)
For exponential function, 8Im[ei8t]8Im[ei8t], α=8i, n=0α=8i, n=0

Since αm1 or m2αm1 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)(56i120)A=8e(8it)2d2x(t)dt2+7dx(t)dt+8x(t)=8sin8t2(64Ae(8it))+7(8iAe(8it))+8(Ae(8it))=8sin8te(8it)(56i120)A=8e(8it)

Comparing the coefficient,
(56i120)A=8A=856i120(56i120)A=8A=856i120

Actual particular solution,
xp(t)=856i120e(8it)=856i120(cos8t+isin8t)56i12056i120=(448sin8t960cos8t)+i(448cos8t960sin8t)17536xp(t)=856i120e(8it)=856i120(cos8t+isin8t)56i12056i120=(448sin8t960cos8t)+i(448cos8t960sin8t)17536

xp,real(t)=Im[xp]=7274cos8t15274sin8txp,real(t)=Im[xp]=7274cos8t15274sin8t

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(i154)t+c2e(i154)t)7274cos8t15274sin8txtotal=xc+xp=e(74)t(c1e(i154)t+c2e(i154)t)7274cos8t15274sin8t

Initial Value Problem

x(0)=e(74)0(c1e(i154)0+c2e(i154)0)7274cos8(0)15274sin8(0)=2c1+c2=2.0255x(0)=e(74)0(c1e(i154)0+c2e(i154)0)7274cos8(0)15274sin8(0)=2c1+c2=2.0255 ˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)56274cos8t90274sin8t˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)+56274cos8(0)90274sin8(0)=0˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)56274cos8t90274sin8t˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)+56274cos8(0)90274sin8(0)=0 c2=3.98261.9612i1.9364i×ii=2.0567i+1.0128;c1=1.0282.0567ic2=3.98261.9612i1.9364i×ii=2.0567i+1.0128;c1=1.0282.0567i x(t)=e(74)t((1.01282.0567i)e(i154)t+(1.0128+2.0567i)e(i154)t)7274cos8t15274sin8tx(t)=e(74)t((1.01282.0567i)e(i154)t+(1.0128+2.0567i)e(i154)t)7274cos8t15274sin8t

Question (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)=e10t2d2x(t)dt2+7dx(t)dt+8x(t)=e10t,
$$r( t) =e^{\alpha t} P_{n}( t) =e^{-10t}
Characteristic equation:
2m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i1542m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i154

Comment: A pair of complex conjugates roots
(m1=74+i154, m2=74i154m1=74+i154, m2=74i154)

Complementary solution:
x(t)=e(74)t(c1e(i154)t+c2e(i154)t)x(t)=e(74)t(c1e(i154)t+c2e(i154)t)
For exponential function, α=10, n=0α=10, n=0

Since αm1 or m2αm1 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)(20070+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)(20070+8)A=e(10t)

Comparing the coefficient,
138A=1A=1138138A=1A=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)=e10t2d2x(t)dt2+7dx(t)dt+8x(t)=e10t is xtotal=xc+xp=e(74)t(c1e(i154)t+c2e(i154)t)+1138e10txtotal=xc+xp=e(74)t(c1e(i154)t+c2e(i154)t)+1138e10t

Initial Value Problem

x(0)=e(74)0(c1e(i154)0+c2e(i154)0)+1138e10(0)=2c1+c2=1.9928x(0)=e(74)0(c1e(i154)0+c2e(i154)0)+1138e10(0)=2c1+c2=1.9928 ˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)10138e10t˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)10138e10(0)=0˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)10138e10t˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)10138e10(0)=0 c2=3.55991.9295i1.9365i×ii=1.8383i+0.9964;c1=0.99641.8383ic2=3.55991.9295i1.9365i×ii=1.8383i+0.9964;c1=0.99641.8383i x(t)=e(74)t((0.99641.8383i)e(i154)t+(0.9964+1.8383i)e(i154)t)+1138e10tx(t)=e(74)t((0.99641.8383i)e(i154)t+(0.9964+1.8383i)e(i154)t)+1138e10t

Question (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)=5e10tcos10t,r(t)=eαtPn(t)=5e10tcos10t=5e10tRe[ei10t]=Re[5e(10+10i)t]2d2x(t)dt2+7dx(t)dt+8x(t)=5e10tcos10t,r(t)=eαtPn(t)=5e10tcos10t=5e10tRe[ei10t]=Re[5e(10+10i)t]
Characteristic equation:
2m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i1542m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i154

Comment: A pair of complex conjugates roots
(m1=74+i154, m2=74i154m1=74+i154, m2=74i154)

Complementary solution:
x(t)=e(74)t(c1e(i154)t+c2e(i154)t)x(t)=e(74)t(c1e(i154)t+c2e(i154)t)
For exponential function, Re[5e(10+10i)t]Re[5e(10+10i)t],
α=10+10i, n=0α=10+10i, n=0

Since αm1 or m2αm1 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)=5e10tcos10t2((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)=5e10tcos10t2((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=5A=562330i(2(200i)+7(10+10i)+8)A=5A=562330i

Actual particular solution, xp(t)=e(αt)Qn(t)=562330ie(10+10i)t=562330ie(10)t(cos10t+isin10t)×63+330i63+330i=(1650sin10t310cos10t)+i(1650cos10t310sin10t)112744e(10)txp(t)=e(αt)Qn(t)=562330ie(10+10i)t=562330ie(10)t(cos10t+isin10t)×63+330i63+330i=(1650sin10t310cos10t)+i(1650cos10t310sin10t)112744e(10)t

xp,real(t)=Re[xp]=(0.01463sin10t0.002750cos10t)e(10)txp,real(t)=Re[xp]=(0.01463sin10t0.002750cos10t)e(10)t

The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=5e10tcos10t2d2x(t)dt2+7dx(t)dt+8x(t)=5e10tcos10t is xtotal=xc+xp=e(74)t(c1e(i154)t+c2e(i154)t)+(0.01463sin10t0.002750cos10t)e(10)txtotal=xc+xp=e(74)t(c1e(i154)t+c2e(i154)t)+(0.01463sin10t0.002750cos10t)e(10)t

Initial Value Problem

(0)=e(74)0(c1e(i154)0+c2e(i154)0)+(0.01463sin10(0)0.002750cos10(0))e(10)0=2c1+c2=2.0028(0)=e(74)0(c1e(i154)0+c2e(i154)0)+(0.01463sin10(0)0.002750cos10(0))e(10)0=2c1+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=3.62371.9392i1.9365×ii=1.8713+1.0014i;c1=3.87411.0014ic2=3.62371.9392i1.9365×ii=1.8713+1.0014i;c1=3.87411.0014i x(t)=e(74)t((3.87411.0014i)e(i154)t+(1.8713+1.0014i)e(i154)t)+(0.01463sin10t0.002750cos10t)e(10)tx(t)=e(74)t((3.87411.0014i)e(i154)t+(1.8713+1.0014i)e(i154)t)+(0.01463sin10t0.002750cos10t)e(10)t

Question (e)

Step 1: Homogeneous Part
2d2x(t)dt2+7dx(t)dt+8x(t)=0
Step 2: Nonhomogeneous Part
$$2\frac{d\power{2}x(t)
dt\power{2}}+7\frac{dx(t) dt}+8x(t)=5\cos{10t}+e\power{-10t},<br/>r( t)=e^{\alpha t} P_{n}( t) =5\cos 10t+e^{-10t}$$
Characteristic equation:
2m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i154

Comment: A pair of complex conjugates roots
(m1=74+i154, m2=74i154)

Complementary solution:
x(t)=e(74)t(c1e(i154)t+c2e(i154)t)
For exponential function, e10t, α=10, n=0

Since αm1 or m2,
xp1(t)=e(αt)Qn(t)=Ae(10t)

For cosine function,
5cos10t=Re[5e10it], α=10i, n=0

Since αm1 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+e10t2(100Ae(10t)100Be10it)+7(10Ae(10t)+10iBe10it)+8(Ae(10t)+Be(10it))=5e(10i)t+e10te(10)t(138A)+e(10it)(70iB192B)=5e(10i)t+e10t

Comparing the coefficient,
(70i192)B=5B=570i192138A=1A=1138

Actual particular solution,
xp(t)=1138e(10t)+570i192e(10it)=1138e(10t)+570i19270i19270i192(cos10t+isin10t)=1138e(10t)+i(0.00838cos10t0.02299sin10t)+(0.00838sin10t0.02299cos10t)

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+e10t is xtotal=xc+xp=e(74)t(c1e(i154)t+c2e(i154)t)+1138e(10t)0.02299cos10t+0.00838sin10t

Initial Value Problem

(0)=e(74)0(c1e(i154)0+c2e(i154)0)+1138e(10)00.02299cos10(0)+0.00838sin10(0)=2c1+c2=2.0157 ˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)10138e(10t)+0.2299sin10t+0.0838cos10t˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)10138e(10)0+0.2299sin10(0)+0.0838cos10(0)=0 c2=3.51611.9517i1.9517i×ii=1.0078+1.8740i;c1=1.00781.8740i x(t)=e(74)t((1.00781.8740i)e(i154)t+(1.0078+1.8740i)e(i154)t)+1138e(10t)0.02299cos10t+0.00838sin10t

Question (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
Characteristic equation:
2m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i154

Comment: A pair of complex conjugates roots
(m1=74+i154, m2=74i154)

Complementary solution:
x(t)=e(74)t(c1e(i154)t+c2e(i154)t)
For polynomial function, 10, α=0, n=0

Since αm1 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(i154)t+c2e(i154)t)+1.25

Initial Value Problem

x(0)=e(74)0(c1e(i154)0+c2e(i154)0)+1.25=2c1+c2=0.75 ˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)=0˙x(0)=(i154c1i154c2)74(c1+c2)=0 i154(0.75c2)i154c274(0.75)=0c2=0.3750+0.6778i; c1=0.37500.6778i x(t)=e(74)t((0.37500.6778i)e(i154)t+(0.3750+0.6778i)e(i154)t)+1.25

Question (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
Characteristic equation:
2m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i154

Comment: A pair of complex conjugates roots
(m1=74+i154, m2=74i154)

Complementary solution:
x(t)=e(74)t(c1e(i154)t+c2e(i154)t)
For polynomial function, 5t2+7t+9, α=0, n=2

Since αm1 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=5A=0.625
t:14A+8B=714(0.625)+8B=7B=0.2188
t0:4A+7B+8C=94(0.625)+8)0.2188)+8C=9C=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(i154)t+c2e(i154)t)+0.625t20.2188t+1.0040

Initial Value Problem

x(0)=e(74)0(c1e(i154)0+c2e(i154)0)+0.625(0)20.2188(0)+1.0040=2c1+c2=0.996 ˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)+1.25t0.2188˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)+1.25(0)0.2188=0˙x(0)=(i154c1i154c2)74(c1+c2)0.2188=0 c2=1.96180.9644i1.9365i=0.4980+1.0131i; c1=0.49801.0131i x(t)=e(74)t((0.49801.0131i)e(i154)t+(0.4980+1.0131i)e(i154)t)+0.625t20.2188t+1.0040

Question (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
Characteristic equation:
2m2+7m+8=0m=7±724(2)(8)2(2)=7±154=74±i154

Comment: A pair of complex conjugates roots
(m1=74+i154, m2=74i154)

Complementary solution:
x(t)=e(74)t(c1e(i154)t+c2e(i154)t)
ui(t)=r(t)x2(t)W(y1,y2)dt=(6tet+3t)e(74i154)ti2154e144tdt=1215ite(2.75i154)tdt615ite(2.75i154)tdt

Integration by part,
1215itue(2.75i154)tdtdv=1215i[tue(2.75i154)t(2.75i154)ve(2.75i154)t(2.75i154)vdtdu]=1215i[te(2.75i154)t(2.75i154)e(2.75i154)t(2.75i154)2]

615ite(2.75i154)tdt=615i[tue(1.75i154)t(1.75i154)ve(1.75i154)t(1.75i154)vdtdu]=615i[te(1.75i154)t(1.75i154)e(1.75i154)t(1.75i154)2]
1(t)=1215i[te(2.75i154)t(2.75i154)e(2.75i154)t(2.75i154)2]615i[te(1.75i154)t(1.75i154)e(1.75i154)t(1.75i154)2]

u2(t)=r(t)x1(t)W(x1,x2)dt=(6tet+3t)e(74+i154)ti2154e144tdt=1215ite(2.75+i154)t+615ite(1.75+i154)tdt

Integration by part,
1215itue(2.75+i154)tdtdv=1215i[tue(2.75+i154)t(2.75+i154)ve(2.75+i154)t(2.75+i154)vdtdu]=1215i[te(2.75+i154)t(2.75+i154)e(2.75+i154)t(2.75+i154)2]

615ite(1.75+i154)tdt=615i[tue(1.75+i154)t(1.75+i154)ve(1.75+i154)t(1.75+i154)2vdtdu]=615i[te(1.75+i154)t(1.75+i154)e(1.75+i154)t(1.75+i154)2]

615ite(1.75+i154)tdt=615i[tue(1.75+i154)t(1.75+i154)ve(1.75+i154)t(1.75+i154)2vdtdu]=615i[te(1.75+i154)t(1.75+i154)e(1.75+i154)t(1.75+i154)2]

2(t)=1215i[te(2.75+i154)t(2.75+i154)e(2.75+i154)t(2.75+i154)2+615i[te(1.75+i154)t(1.75+i154)e(1.75+i154)t(1.75+i154)2]

xp=u1(t)x1(t)+u2(t)x2(t)

where:
u1(t)x1(t)={1215i[te(2.75i154)t(2.75i154)e(2.75i154)t(2.75i154)2]615i[te(1.75i154)t(1.75i154)e(1.75i154)t(1.75i154)2]}e(74+i154)t=1215i[tet(2.75i154)et(2.75i154)2]615i[t1(1.75i154)1(1.75i154)2]

$$ \begin{aligned}
u_{1}( t) x_{1}( t) & ={\frac{12}{\sqrt{15}} i[ t\frac{e^{\left( 2.75+i\frac{\sqrt{15}}{4}\right) t}}{\left( 2.75+i\frac{\sqrt{15}}{4}\right)} -\frac{e^{\left( 2.75+i\frac{\sqrt{15}}{4}\right) t}}{\left( 2.75+i\frac{\sqrt{15}}{4}\right)^{2}}\& +\frac{6}{\sqrt{15}} i\left[ t\frac{e^{\left( 1.75+i\frac{\sqrt{15}}{4}\right) t}}{\left( 1.75+i\frac{\sqrt{15}}{4}\right)} -\frac{e^{\left( 1.75+i\frac{\sqrt{15}}{4}\right) t}}{\left( 1.75+i\frac{\sqrt{15}}{4}\right)^{2}}\right]} e^{\left( -\frac{7}{4} -i\frac{\sqrt{15}}{4}\right) t}\& =\frac{12}{\sqrt{15}} i[ t\frac{e^{t}}{\left( 2.75+i\frac{\sqrt{15}}{4}\right)} -\frac{e^{t}}{\left( 2.75+i\frac{\sqrt{15}}{4}\right)^{2}}\& +\frac{6}{\sqrt{15}} i\left[ t\frac{1}{\left( 1.75+i\frac{\sqrt{15}}{4}\right)} -\frac{1}{\left( 1.75+i\frac{\sqrt{15}}{4}\right)^{2}}\right]\end{aligned}<br/><br/> xp(t)=u1(t)x1(t)+u2(t)x2(t)&=1215i[tet(2.75i154)et(2.75i154)2]&+1215i[tet(2.75+i154)et(2.75+i154)2]&615i[t1(1.75i154)1(1.75i154)2]&+615i[t1(1.75+i154)1(1.75+i154)2]&=1215i[tet(2.75i154)8.5et(6.625+i1.37515)72.25]&+1215i[tet(2.75i154)8.5et(6.625i1.37515)72.25]&615i[t(1.75+i154)4(2.125+i0.87515)16]&+615i[t(1.75i154)4(2.125i0.87515)16]&=[2415itet(i154)8.5+2415iet(i1.37515)8.5]&+[1215it(i154)4+1215i(i0.87515)16]&=[+1217tet132289et+34et2132]$$  

The complete/general solution to 2d2x(t)dt2+7dx(t)dt+8x(t)=6tet+3t is xtotal=xc+xp=e(74)t(c1e(i154)t+c2e(i154)t)+1217tet132289et+34et2132

Initial Value Problem

x(0)=e(74)0(c1e(i154)0+c2e(i154)0)+1217(0)e0132289e0+34e02132=2c1+c2=2.363 ˙xtotal=e(74)t(i154c1e(i154)ti154c2e(i154)t)74e(74)t(c1e(i154)t+c2e(i154)t)+1217(et(1+t))132289et+34et˙x(0)=e(74)0(i154c1e(i154)0i154c2e(i154)0)74e(74)0(c1e(i154)0+c2e(i154)0)+1217(e0(1+0))132289e0+34e0=0˙x(0)=(i154c1i154c2)74(c1+c2)+0.9991=0  c1=1.18151.6195i, c2=1.1815+1.6195i x(t)=e(74)t((1.18151.6195)e(i154)t+(1.1815+1.6195)e(i154)t)+1217tet132289et+34et2132

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