The Exercise Problems in Chapter 1 are open-ended questions and are provided to encourage

Chapter 1 Solutions to Exercise Problems The Exercise Problems in Chapter 1 are open-ended questions and are provided to encourage students to reect freely, hence there are no xed answers.1 (Introduction to Integrative Engineering A Computational Approach to Biomedical Problems, 1e Guigen Zhang) (Solution Manual all Chapters) 1 / 4

Chapter 2 Solutions to Exercise Problems 1.order, dimension, linear or nonlinear, time independent or dependent, and type of coecient, etc. Also identify the dependent and independent variables.(a) d 2 x dt 2

• 2

dx dt

• 7tx= 0

Answer:

2 nd order ODE, linear, time dependent, and with non-constant coecients. It is a 0D equation becausexis dependent variable and there is no independent dimensional vari- able.tis independent time variable and it is a non-constant coecient ofx.(b) @u @t 5 @ 2 u @y 2 = (1 + 5t) sin(y)

Answer:

2 nd order PDE, linear, time dependent, and with constant coecient. It is a 1D equation.uis dependent variable,yis independent dimensional variable andtis independent time variable.(c) d 3 w dx 3 +w dw dx

• 9w= 0

Answer:

3 rd order ODE, nonlinear, time independent, and with constant coecient. It is a 1D equation.wis dependent variable,xis independent dimensional variable. It contains a product term ofwand @w @x , hence nonlinear.(d) d 2 x dt 2 2 1 t 2 x= 0

Answer:

2 nd order ODE, linear, time dependent, and with non-constant coecient. It is a 0D equation.xis dependent variable with no independent dimensional variable andtis independent time variable. 1=t 2 is the coecient ofx, hence non-constant.

2 2 / 4

3 (e) @v @t +

@v @x

2

• 8v=x+t

2

• 8xt

Answer:

1 st order PDE, nonlinear, time dependent, and with constant coecient. It is a 1D equa- tion.vis dependent variable,xis independent dimensional variable andtis independent time variable. It has the square term of @v @x , hence nonlinear.(f) @w @t

@ @x

2w @ 2 w @x 2

8w= 0

Answer:

3 rd order PDE, nonlinear, time dependent, and with constant coecient. It is a 1D equa- tion.wis dependent variable,xis independent dimensional variable andtis independent time variable. It contains the product term ofwand @w 2 @x 2, hence nonlinear.

2.PDE, order, dimension, linearity, time dependency, and type of coecient. Also identify the dependent and independent variables.(a) d 3

dt 3

• 5+ sin() = 0

Answer:

3 rd order ODE, linear, time dependent, and with constant coecient. It is a 0D equation.is dependent variable with no independent dimensional variable andtis independent time variable.(b) @ 2 u @x 2 2 @u @x

• 15

@u @y = 1 +x+y

Answer:

2 nd order PDE, linear, time independent, and with constant coecient. It is a 2D equation.uis dependent variable,xandyare the independent dimensional variables.(c) (x; y; z) @ 2 u @t 2 +(x; y; z) @u @t T(x; y; z)

@ 2 u @x 2 + @ 2 u @y 2 + @ 2 u @z 2

=g(x; y; z)

Answer:

2 nd order PDE, linear, time dependent, and with non-constant coecient. It is a 3D equation.uis dependent variable,x; yandzare the independent dimensional variables, andtis independent time variable.; andTare non-constant coecients.(d) d 2 v dt 2 2 dv dt +t 2 v= 0 3 / 4

4CHAPTER 2. SOLUTIONS TO EXERCISE PROBLEMS

Answer:

2 nd order ODE, linear, time dependent, and with non-constant coecient. It is a 0D equation.vis dependent variable with no independent dimensional variable andtis independent time variable.t 2 is the coecient ofv, hence non-constant.(e) (x)c(x)

@y @t

3

@ @x

(x) @y @x

= 0

Answer:

2 nd order PDE, nonlinear, time dependent, and with non-constant coecient. It is a 1D equation.yis dependent variable andxis independent dimensional variable andtis independent time variable.(f) d 2 x dt 2

• 2

d 4 x dt 4

• 2015x

2 = sin(t)

Answer:

4 th order ODE, nonlinear, time dependent, and with constant coecient. It is a 0D equation.xis dependent variable with no independent dimensional variable andtis independent time variable.

3.

equation in Exercise 1:

(a) x(t) =te t

Answer:

Substitutingx(t) =te t into 1(a), we have (t+ 2)e t

• 2(1 +t)e

t

• 7t

2 e t = 7t 2 e t

• 3te

t

• 4e

t 6= 0 thus, it is not a solution.(b) u(y; t) =tsin(y)

Answer:

Substitutingu(y; t) =tsin(y) into 1(b), we have sin(y) + 5tsin(y) = (1 + 5t) sin(y) thus, it is a solution.(c) w(x) =x 3

• 5x

Answer:

Substitutingw(x) =x 3

• 5xinto 1(c), we have
• + (x

3

• 5x)(3x

2

• 5) + 9(x

3

• 5x)6= 0

thus, it is not a solution.

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