Deck 12: Boundary-Value Problems in Rectangular Coordinates

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سؤال
The general solution of y+n2n2y=0,y(0)=0,y(1)=0,n=1,2,3y ^ { \prime \prime } + n ^ { 2 } n ^ { 2 } y = 0 , y ( 0 ) = 0 , y ( 1 ) = 0 , n = 1,2,3 \ldots is

A) y=0y = 0
B) y=csin(nπx)y = c \cdot \sin ( n \pi x )
C) y=ccos(nπx)y = c \cdot \cos ( n \pi x )
D) c(en+xen+x)c \left( e ^ { n + x } - e ^ { - n + x } \right)
E) c(en+x+en+x)c \left( e ^ { n + x } + e ^ { - n + x } \right)
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سؤال
The solution of the eigenvalue problem from the previous problem is

A) λ=nπ/L,X(x)=sin(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
B) λ=nπ/L,X(x)=cos(nπx/L),n=0,1,2,\lambda = n \pi / L , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
C) λ=(nπ/L),X(x)=sin(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) , X ( x ) = \sin ( n \pi x / L ) , n = 0,1,2 , \ldots
D) λ=(nπ/L)2,X(x)=sin(nπx/L),n=1,2,3,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
E) λ=(nπ/L)2,X(x)=cos(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
سؤال
The wave equation for a vibrating string is derived using the assumptions Select all that apply.

A) the string is perfectly flexible.
B) the displacements may be large.
C) the tension acts perpendicular to the string.
D) the tension is large compared with gravity.
E) the string is homogeneous.
سؤال
The differential equation 2ux2+2uy2=0\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = 0 is

A) first order, linear, homogeneous
B) first order, linear, non-homogeneous
C) second order, nonlinear
D) second order, linear, homogeneous
E) second order, linear, non-homogeneous
سؤال
The differential equation 2ux2ut=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } - \frac { \partial u } { \partial t } = u is Select all that apply.

A) nonlinear
B) linear
C) hyperbolic
D) elliptic
E) parabolic
سؤال
In the previous three problems, the solution of the original problem is

A) u(x,t)=n1cncos((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \cos ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=20Lf(x)cos((n1/2)πx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( ( n - 1 / 2 ) \pi x / L ) d x / L
B) u(x,t)=n1cnsin((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=20Lf(x)sin((n1/2)πx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( ( n - 1 / 2 ) \pi x / L ) d x / L
C) u(x,t)=n1cncos((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \cos ( ( n - 1 / 2 ) \pi x / L ) e ^ { ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=0Lf(x)cos((n1/2)πx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( ( n - 1 / 2 ) \pi x / L ) d x
D) u(x,t)=n=0cnsin((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=20Lf(x)sin((n1/2)πx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( ( n - 1 / 2 ) \pi x / L ) d x / L
E) u(x,t)=n0cnsin((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n - 0 } ^ { \infty } c _ { n } \sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=0Lf(x)sin((n1/2)πx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \sin ( ( n - 1 / 2 ) \pi x / L ) d x
سؤال
The solution of the eigenvalue problem from the previous problem is

A) λ=nπ/L,X(x)=sin(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
B) λ=nπ/L,X(x)=cos(nπx/L),n=0,1,2,\lambda = n \pi / L , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
C) λ=(n1/2)π/L,X(x)=sin((n1/2)πx/L),n=1,2,3,\lambda = ( n - 1 / 2 ) \pi / L , X ( x ) = \sin ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots
D) λ=((n1/2)π/L)2,X(x)=sin((n1/2)πx/L),n=1,2,3,\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X ( x ) = \sin ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots
E) λ=((n1/2)π/L)2,X(x)=cos((n1/2)πx/L),n=1,2,3,\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X ( x ) = \cos ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots
سؤال
In the previous three problems, the solution of the original problem is

A) u(x,t)=n=1cnsin(nπx/L)e(nπ/L)tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) t } , where cn=0Lf(x)cos(nπx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x
B) u(x,t)=n=0cncos(nπx/L)cos(nπt/L)u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) \cos ( n \pi t /L ) , where c0=0Lf(x)dx/L,cn=20Lf(x)cos(nπx/L)dx/Lc _ { 0 } = \int _ { 0 } ^ { L } f ( x ) d x / L , c _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / L
C) u(x,t)=n=1cnsin(nπx/L)sin(nπt/L)u ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sin ( n \pi t / L ) , where cn=20Lf(x)cos(nπx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / L
D) u(x,t)=n=1cnsin(nπx/L)enxt/Lu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - nxt / L } , where cn=20Lf(x)sin(nπx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / L
E) u(x,t)=n=0cncos(nπx/L)enxt/Lu ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) e ^ { - n x t / L } , where cn=0Lf(x)cos(nπx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x
سؤال
The differential equation 2ux2+2uxy+2uy2=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial x \partial y } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = u is Select all that apply.

A) linear
B) nonlinear
C) hyperbolic
D) elliptic
E) parabolic
سؤال
Consider the equation uxxut=0u _ { xx } - u _ { t } = 0 with conditions u(0,t)=0,ux=(L,t)=0,u(x,0)=f(x)u ( 0 , t ) = 0 , u _ { x } = ( L , t ) = 0 , u ( x , 0 ) = f ( x ) . When separating variables with u(x,t)=X(x)T(t)u ( x , t ) = X ( x ) T ( t ) , the resulting problems for X,TX , T are

A) X+λX=0,X(0)=0,X(L)=0,TλT=0,T(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } - \lambda T = 0 , T ( 0 ) = f ( x )
B) X+λX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } + \lambda T = 0 , T ( 0 ) = f ( x )
C) XλX=0,X(0)=0,X(L)=0,T+λT=0X ^ { \prime \prime } - \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } + \lambda T = 0
D) X+λX=0,X(0)=0,X(L)=0,TλT=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } - \lambda T = 0
E) X+λX=0,X(0)=0,X(L)=0,T+λT=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } + \lambda T = 0
سؤال
Consider the equation uxxutt=0u _ { xx } - u _ { t t } = 0 with conditions dudx(0,t)=0,dudx=(L,t)=0,u(x,0)=f(x),dudt(0)=0\frac { d u } { d x } ( 0 , t ) = 0 , \frac { d u } { d x } = ( L , t ) = 0 , u ( x , 0 ) = f ( x ) , \frac { d u } { d t } ( 0 ) = 0 . When separating variables with u(x,t)=X(x)T(t)u ( x , t ) = X ( x ) T ( t ) , the resulting problems for X,TX , T are

A) X+λX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ( 0 ) = 0
B) XλX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=0X ^ { \prime \prime } - \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
C) X+λX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
D) X+λX=0,X(0)=0,X(L)=0,TλT=0,T(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } - \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
E) X+λX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ( 0 ) = f ( x )
سؤال
In the previous two problems, the product solutions are

A) sin((n1/2)πx/L)e(n1/2)2π2t/L2\sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } \pi ^ { 2 } t / L ^ { 2 } }
B) cos((n1/2)πx/L)e(n1/2)2π2t/L2\cos ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } \pi ^ { 2 } t / L ^ { 2 } }
C) sin((n1/2)πx/L)e(n1/2)/L\sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) / L }
D) sin((n1/2)πx/L)e(n1/2)2x2t/L2\sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } }
E) cos((n1/2)πx/L)e(n1/2)2π2t/L2\cos ( ( n - 1 / 2 ) \pi x / L ) e ^ { ( n - 1 / 2 ) ^ { 2 } \pi ^ { 2 } t / L ^ { 2 } }
سؤال
The solution of the previous three problems is n=1cnXn(x)eλnkt\sum _ { n = 1 } ^ { \infty } c _ { n } X _ { n } ( x ) e ^ { - \lambda _ { n } k t } , where XnX _ { n } and λn\lambda _ { n } are given in the previous problem and

A) cn=0Lf(x)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) d x
B) cn=0Lf(x)Xn(x)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x
C) cn=0Lf(x)Xn(x)dx/0LXn(x)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x / \int _ { 0 } ^ { L } X _ { n } ( x ) d x
D) cn=0LXn2(x)dx/0Lf(x)Xn(x)dxc _ { n } = \int _ { 0 } ^ { L } X _ { n } ^ { 2 } ( x ) d x / \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x
E) cn=0Lf(x)Xn(x)dx/0LXn2(x)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x / \int _ { 0 } ^ { L } X _ { n } ^ { 2 } ( x ) d x
سؤال
After separating variables in the previous problem, the eigenvalue problem becomes

A) X+λX=0,X(0)=0,X(L)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0
B) X+λX=0,X(0)=0,X(L)=hX(L)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = - h X ( L )
C) X+λX=0,X(0)=0,X(L)=hX(0)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = - h X ^ { \prime } ( 0 )
D) X+λX=0,X(0)=0,X(L)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0
E) X+λX=0,X(0)=0,X(L)=hX(L)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = h X ( L )
سؤال
In the previous two problems, the product solutions are

A) sin(nπx/L)sin(nπ/L)\sin ( n \pi x / L ) \sin ( n \pi / L )
B) sin(nπx/L)cos(nπ/L)\sin ( n \pi x / L ) \cos ( n \pi / L )
C) cos(nπx/L)cos(nπ/L)\cos ( n \pi x / L ) \cos ( n \pi / L )
D) sin(nπx/L)enx/L\sin ( n \pi x / L ) e ^ { n x / L }
E) cos(nπx/L)ent/L\cos ( n \pi x / L ) e ^ { - n t / L }
سؤال
The solution of yn2n2y=0,y(0)=0,y(1)=0,n=1,2,3y ^ { \prime \prime } - n ^ { 2 } n ^ { 2 } y = 0 , y ( 0 ) = 0 , y ( 1 ) = 0 , n = 1,2,3 \ldots is

A) y=0y = 0
B) y=csin(nπx)y = c \cdot \sin ( n \pi x )
C) y=ccos(nπx)y = c \cdot \cos ( n \pi x )
D) c(enxtenxt)c \left( e ^ { n xt } - e ^ { - nxt } \right)
E) c(enxt+enxt)c \left( e ^ { n xt } + e ^ { - nxt } \right)
سؤال
When u(x,y)=X(x)Y(y)u ( x , y ) = X ( x ) Y ( y ) is substituted into the equation uxx+uyy=0u _ { xx } + u _ { y y } = 0 , the resulting equations for XX and YY are

A) X+λX=0,Y+λY=0X ^ { \prime } + \lambda X = 0 , Y ^ { \prime } + \lambda Y = 0
B) X+λX=0,YλY=0X ^ { \prime } + \lambda X = 0 , Y ^ { \prime } - \lambda Y = 0
C) X+λX=0,Y+λY=0X ^ { \prime \prime } + \lambda X = 0 , Y ^ { \prime \prime } + \lambda Y = 0
D) X+λX=0,YλY=0X ^ { \prime \prime } + \lambda X = 0 , Y ^ { \prime \prime } - \lambda Y = 0
E) XλX=0,Y+λY=0X ^ { \prime \prime } - \lambda X = 0 , Y ^ { \prime \prime } + \lambda Y = 0
سؤال
The solution of the eigenvalue problem in the previous problem is

A) λ=((n1/2)π/L)2,X=sin((n1/2)πx/L)\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X = \sin ( ( n - 1 / 2 ) \pi x / L )
B) λ/h=tan(λL),X=sin(λx)\sqrt { \lambda } / h = \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )
C) λ=(nπ/L)2,X=sin(nπx/L)\lambda = ( n \pi / L ) ^ { 2 } , X = \sin ( n \pi x / L )
D) λ/h=tan(λL),X=sin(λx)\sqrt { \lambda } / h = - \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )
E) λh=tan(λL),X=sin(λx)\sqrt { \lambda } h = - \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )
سؤال
The solution of y+λy=0,y(0)=0,y(π)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 if λ=0\lambda = 0 is

A) y=sinxy = \sin x
B) y=cosxy = \cos x
C) y=0y = 0
D) y=ax+by = a x + b
E) y=xπy = x - \pi
سؤال
The model describing the temperature in a rod where the temperature at the left end is zero and where there is heat transfer from the right boundary into the external medium is

A) k2ux2=2ut2,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = - h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
B) k2ux2=2ut2,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
C) k2ux2=ut,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = - h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
D) k2ux2=ut,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
E) k2ux2+2ut2=0,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = - h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
سؤال
Consider the equation ux+uyy=0u _ { x } + u _ { y y } = 0 with conditions u(0,y)=0,u=(L,y)=0,u(x,0)=f(x),u(x,H)=0u ( 0 , y ) = 0 , u = ( L , y ) = 0 , u ( x , 0 ) = f ( x ) , u ( x , H ) = 0 . When separating variables with u(x,y)=X(x)Y(y)u ( x , y ) = X ( x ) Y ( y ) , the resulting problems for X,YX , Y are

A) X+λX=0,X(0)=0,X(L)=0,Y+λY=0,Y(H)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } + \lambda Y = 0 , Y ( H ) = 0
B) X+λX=0,X(0)=0,X(L)=0,YλY=0,Y(H)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( H ) = 0
C) X+λX=0,X(0)=0,X(L)=0,YλY=0,Y(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( 0 ) = 0
D) XλX=0,X(0)=0,X(L)=0,YλY=0,Y(H)=f(0)X ^ { \prime \prime } - \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( H ) = f ( 0 )
E) X+λX=0,X(0)=0,X(L)=0,Y+λY=0,Y(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } + \lambda Y = 0 , Y ( 0 ) = f ( x )
سؤال
In the previous problem, the eigenfunction expansion if xetx e ^ { t } is

A) etn1(1)n2Lcos(nπx/L)/(nπ)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n } 2 L \cos ( n \pi x / L ) / ( n \pi )
B) etn1(1)n2Lsin(nπx/L)/(n2π2)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n } 2 L \sin ( n \pi x / L ) / \left( n ^ { 2 } \pi ^ { 2 } \right)
C) etn1(1)n2Lcos(nπx/L)/(n2π2)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n } 2 L \cos ( n \pi x / L ) / \left( n ^ { 2 } \pi ^ { 2 } \right)
D) etn1(1)n+12Lcos(nπx/L)/(nπ)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } 2 L \cos ( n \pi x / L ) / ( n \pi )
E) etn1(1)n+12Lsin(nπx/L)/(nπ)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } 2 L \sin ( n \pi x / L ) / ( n \pi )
سؤال
When u(x,y)=X(x)Y(y)u ( x , y ) = X ( x ) Y ( y ) is substituted into the equation uxuyy=0u _ { x } - u _ { y y } = 0 , the resulting equations for XX and YY are

A) X+λX=0,Y+λY=0X ^ { \prime \prime } + \lambda X = 0 , Y ^ { \prime \prime } + \lambda Y = 0
B) X+λX=0,YλY=0X ^ { \prime \prime } + \lambda X = 0 , Y ^ { \prime \prime } - \lambda Y = 0
C) XλX=0,Y+λY=0X ^ { \prime \prime } - \lambda X = 0 , Y ^ { \prime \prime } + \lambda Y = 0
D) X+λX=0,Y+λY=0X ^ { \prime } + \lambda X = 0 , Y ^ { \prime } + \lambda Y = 0
E) X+λX=0,YλY=0X ^ { \prime } + \lambda X = 0 , Y ^ { \prime } - \lambda Y = 0
سؤال
The solution of the eigenvalue problem from the previous problem is

A) λ=nπ/L,X(x)=sin(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
B) λ=nπ/L,X(x)=cos(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \cos ( n \pi x / L ) , n = 1,2,3 , \ldots
C) λ=(nπ/L)2,X(x)=sin(nπx/L),n=1,2,3,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
D) λ=(nπ/L)2,X(x)=cos(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
E) λ=(nπ/L),X(x)=sin(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) , X ( x ) = \sin ( n \pi x / L ) , n = 0,1,2 , \ldots
سؤال
The solution of the eigenvalue problem from the previous problem is

A) λ=nπ/L,X(x)=sin(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
B) λ=nπ/L,X(x)=cos(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \cos ( n \pi x / L ) , n = 1,2,3 , \ldots
C) λ=(nπ/L)2,X(x)=sin(nπx/L),n=1,2,3,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
D) λ=(nπ/L)2,X(x)=cos(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
E) λ=(nπ/L),X(x)=sin(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) , X ( x ) = \sin ( n \pi x / L ) , n = 0,1,2 , \ldots
سؤال
The solution of yn2y=0,y(0)=0,y(π)=0,n=1,2,3y ^ { \prime \prime } - n ^ { 2 } y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 , n = 1,2,3 \ldots is

A) y=0y = 0
B) y=csin(nx)y = c \cdot \sin ( n x )
C) y=ccos(nx)y = c \cdot \cos ( n x )
D) y=c(enxenx)y = c \left( e ^ { n x } - e ^ { - n x } \right)
E) y=c(enx+enx)y = c \left( e ^ { n x } + e ^ { - n x } \right)
سؤال
The differential equation 2ux22uy2=sinu\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } - \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = \sin u is Select all that apply.

A) linear
B) nonlinear
C) hyperbolic
D) elliptic
E) parabolic
سؤال
In the previous three problems, the solution of the original problem is

A) u(x,t)=n1cnsin(nπx/L)sinh(nπ(yH)/L)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L ) , where cn=0Lf(x)cos(nπx/L)dx/sinh(nπH/L)c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / \sinh ( - n \pi H / L )
B) u(x,t)=n1cnsin(nπx/L)sinh(nπ(yH)/L)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L ) , where cx=20Lf(x)sin(nπx/L)dx/(Lsinh(nπH/L))c _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )
C) u(x,t)=n1cnsin(nπx/L)cosh(nπ(yH)/L)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \cosh ( n \pi ( y - H ) / L ) , where cn=20Lf(x)cos(nπx/L)dx/(Lsinh(nπH/L))c _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )
D) u(x,t)=n1cnsin(nπx/L)cosh((nπ/L)2(yH))u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \cosh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right) , where cx=20Lf(x)sin(nπx/L)dx/(Lsinh(nπH/L))c _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )
E) u(x,t)=n=0cncos(nπx/L)sinh((nπ/L)2(yH))u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) \sinh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right) , where cn=0Lf(x)cos(nπx/L)dx/sinh(nπH/L)c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / \sinh ( - n \pi H / L )
سؤال
In the previous two problems, the product solutions are

A) sin(nπx/L)sinh(nπ(yH)/L)\sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L )
B) cos(nπx/L)sinh(nπ(yH)/L)\cos ( n \pi x / L ) \sinh ( - n \pi ( y - H ) / L )
C) cos(nπx/L)cosh((nπ/L)2(yH))\cos ( n \pi x / L ) \cosh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right)
D) sin(nπx/L)cosh((nπ/L)2(yH))\sin ( n \pi x / L ) \cosh \left( - ( n \pi / L ) ^ { 2 } ( y - H ) \right)
E) cos(nπx/L)cosh(nπ(yH)/L)\cos ( n \pi x / L ) \cosh ( - n \pi ( y - H ) / L )
سؤال
The differential equation 2ux2+2uy2=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = u is

A) first order, linear, homogeneous
B) first order, linear, non-homogeneous
C) second order, nonlinear
D) second order, linear, homogeneous
E) second order, linear, non-homogeneous
سؤال
In the problem 2ux2+xet=ut,u(0,t)=0,u(L,t)=0,u(x,0)=0\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + x e ^ { t } = \frac { \partial u } { \partial t } , u ( 0 , t ) = 0 , u ( L , t ) = 0 , u ( x , 0 ) = 0 , the eigenvalues and eigenfunctions of the underlying homogeneous problem are

A) λ=n2,X=sin(nx)\lambda = n ^ { 2 } , X = \sin ( n x )
B) λ=n2,X=cos(nx)\lambda = n ^ { 2 } , X = \cos ( n x )
C) λ=n2π2/L2,X=sin(nπx/L)\lambda = n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } , X = \sin ( n \pi x / L )
D) λ=n2π2/L2,X=cos(nπx/L)\lambda = n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } , X = \cos ( n \pi x / L )
E) λ=n2π2,X=sin(nπx)\lambda = n ^ { 2 } \pi ^ { 2 } , X = \sin ( n \pi x )
سؤال
The quantity of heat in an element of a rod of mass mm is proportional to Select all that apply.

A) mass
B) thermal conductivity
C) specific heat
D) thermal diffusivity
E) temperature
سؤال
In the previous two problems, the solution for uu takes the form

A) u(x,t)=n1un(t)sin(nπx/L)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } u _ { n } ( t ) \sin ( n \pi x / L )
B) u(x,t)=n=1un(t)cos(nπx)u ( x , t ) = \sum _ { n = 1 } ^ { \infty } u _ { n } ( t ) \cos ( n \pi x )
C) u(x,t)=n1un(t)sin(n2π2x2/L2)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } u _ { n } ( t ) \sin \left( n ^ { 2 } \pi ^ { 2 } x ^ { 2 } / L ^ { 2 } \right)
D) u(x,t)=n1un(t)cos(n2π2x2/L2)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } u _ { n } ( t ) \cos \left( n ^ { 2 } \pi ^ { 2 } x ^ { 2 } / L ^ { 2 } \right)
E) u(x,t)=n1un(t)sin(nπx)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } u _ { n } ( t ) \sin ( n \pi x )
سؤال
In the previous two problems, the product solutions are

A) sin(nπx/L)enx/L\sin ( n \pi x / L ) e ^ { n x / L }
B) sin(nπx/L)enx/L\sin ( n \pi x / L ) e ^ { - n x / L }
C) cos(nπx/L)e(nπ/D)2t\cos ( n \pi x / L ) e ^ { ( n \pi / D ) ^ { 2 } t }
D) sin(nπx/L)e(nπ/L)2t\sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t }
E) cos(nπx/L)ent/L\cos ( n \pi x / L ) e ^ { - n t / L }
سؤال
Consider the equation umut=0u _ { m } - u _ { t } = 0 with conditions u(0,t)=0,u=(L,t)=0,u(x,0)=f(x)u ( 0 , t ) = 0 , u = ( L , t ) = 0 , u ( x , 0 ) = f ( x ) . When separating variables with u(x,t)=X(x)T(t)u ( x , t ) = X ( x ) T ( t ) , the resulting problems for X,TX , T are

A) X+λX=0,X(0)=0,X(L)=0,Tt+λT=0,T(0)=f(0)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { t } + \lambda T = 0 , T ( 0 ) = f ( 0 )
B) X+λX=0,X(0)=0,X(L)=0,Tt+λT=0,T(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { t } + \lambda T = 0 , T ( 0 ) = f ( x )
C) XλX=0,X(0)=0,X(L)=0,T+λT=0X ^ { \prime \prime } - \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { \prime } + \lambda T = 0
D) X+λX=0,X(0)=0,X(L)=0,TλT=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { \prime } - \lambda T = 0
E) X+λX=0,X(0)=0,X(L)=0,T+λT=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { \prime } + \lambda T = 0
سؤال
In the previous three problems, the solution of the original problem is

A) u(x,t)=n=1cnsin(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cn=0Lf(x)cos(nπx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x
B) u(x,t)=n=1cncos(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cn=0Lf(x)sin(nπx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x
C) u(x,t)=n=1cnsin(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cn=20Lf(x)cos(nπx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / L
D) u(x,t)=n=1cnsin(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cn=20Lf(x)sin(nπx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / L
E) u(x,t)=n=0cncos(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cx=20Lf(x)cos(nπx/L)dxc _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x
سؤال
The general solution of y+n2y=0,y(0)=0,y(π)=0,n=1,2,3y ^ { \prime \prime } + n ^ { 2 } y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 , n = 1,2,3 \ldots is

A) y=0y = 0
B) y=csin(nx)y = c \cdot \sin ( n x )
C) y=ccos(nx)y = c \cdot \cos ( n x )
D) y=c(enxenx)y = c \left( e ^ { n x } - e ^ { - n x } \right)
E) y=c(enx+enx)y = c \left( e ^ { n x } + e ^ { - n x } \right)
سؤال
The solution of y+λy=0,y(0)=0,y(π)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 if λ=0\lambda = 0 is

A) y=ax+by = a x + b
B) y=xπy = x - \pi
C) y=0y = 0
D) y=sinxy = \sin x
E) y=cosxy = \cos x
سؤال
In the previous three problems, the solution for u(t)u ( t ) is

A) un(t)=(1)n2L(eten2a2t/L2)/(nπ(1+nπ/L))u _ { n } ( t ) = ( - 1 ) ^ { n } 2 L \left( e ^ { t } - e ^ { - n ^ { 2 } a ^ { 2 } t / L ^ { 2 } } \right) / ( n \pi ( 1 + n \pi / L ) )
B) ux(t)=(1)n2L(et+en2x2t/L2)/(nπ(1+n2π2/L2))u _ { x } ( t ) = ( - 1 ) ^ { n } 2 L \left( e ^ { t } + e ^ { - n ^ { 2 } x ^ { 2 } t / L ^ { 2 } } \right) / \left( n \pi \left( 1 + n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } \right) \right)
C) ux(t)=(1)n+12L(et+en2x2t/L2)/(nπ(1+n2π2/L2))u _ { x } ( t ) = ( - 1 ) ^ { n + 1 } 2 L \left( e ^ { t } + e ^ { - n ^ { 2 } x ^ { 2 } t / L ^ { 2 } } \right) / \left( n \pi \left( 1 + n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } \right) \right)
D) ux(t)=(1)n+12L(eten2x2t/L2)/(nπ(1+n2π2/L2))u _ { x } ( t ) = ( - 1 ) ^ { n + 1 } 2 L \left( e ^ { t } - e ^ { - n ^ { 2 } x ^ { 2 } t / L ^ { 2 } } \right) / \left( n \pi \left( 1 + n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } \right) \right)
E) un(t)=2L(eten2x2t/L2)/(nπ(1+n2π2/L2))u _ { n } ( t ) = 2 L \left( e ^ { t } - e ^ { - n ^ { 2 } x ^ { 2 } t / L ^ { 2 } } \right) / \left( n \pi \left( 1 + n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } \right) \right)
سؤال
The differential equation 2ux2+2uy2=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = u is Select all that apply.

A) linear
B) nonlinear
C) hyperbolic
D) elliptic
E) parabolic
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Deck 12: Boundary-Value Problems in Rectangular Coordinates
1
The general solution of y+n2n2y=0,y(0)=0,y(1)=0,n=1,2,3y ^ { \prime \prime } + n ^ { 2 } n ^ { 2 } y = 0 , y ( 0 ) = 0 , y ( 1 ) = 0 , n = 1,2,3 \ldots is

A) y=0y = 0
B) y=csin(nπx)y = c \cdot \sin ( n \pi x )
C) y=ccos(nπx)y = c \cdot \cos ( n \pi x )
D) c(en+xen+x)c \left( e ^ { n + x } - e ^ { - n + x } \right)
E) c(en+x+en+x)c \left( e ^ { n + x } + e ^ { - n + x } \right)
y=csin(nπx)y = c \cdot \sin ( n \pi x )
2
The solution of the eigenvalue problem from the previous problem is

A) λ=nπ/L,X(x)=sin(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
B) λ=nπ/L,X(x)=cos(nπx/L),n=0,1,2,\lambda = n \pi / L , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
C) λ=(nπ/L),X(x)=sin(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) , X ( x ) = \sin ( n \pi x / L ) , n = 0,1,2 , \ldots
D) λ=(nπ/L)2,X(x)=sin(nπx/L),n=1,2,3,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
E) λ=(nπ/L)2,X(x)=cos(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
λ=(nπ/L)2,X(x)=cos(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
3
The wave equation for a vibrating string is derived using the assumptions Select all that apply.

A) the string is perfectly flexible.
B) the displacements may be large.
C) the tension acts perpendicular to the string.
D) the tension is large compared with gravity.
E) the string is homogeneous.
A, D, E
4
The differential equation 2ux2+2uy2=0\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = 0 is

A) first order, linear, homogeneous
B) first order, linear, non-homogeneous
C) second order, nonlinear
D) second order, linear, homogeneous
E) second order, linear, non-homogeneous
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5
The differential equation 2ux2ut=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } - \frac { \partial u } { \partial t } = u is Select all that apply.

A) nonlinear
B) linear
C) hyperbolic
D) elliptic
E) parabolic
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6
In the previous three problems, the solution of the original problem is

A) u(x,t)=n1cncos((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \cos ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=20Lf(x)cos((n1/2)πx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( ( n - 1 / 2 ) \pi x / L ) d x / L
B) u(x,t)=n1cnsin((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=20Lf(x)sin((n1/2)πx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( ( n - 1 / 2 ) \pi x / L ) d x / L
C) u(x,t)=n1cncos((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \cos ( ( n - 1 / 2 ) \pi x / L ) e ^ { ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=0Lf(x)cos((n1/2)πx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( ( n - 1 / 2 ) \pi x / L ) d x
D) u(x,t)=n=0cnsin((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=20Lf(x)sin((n1/2)πx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( ( n - 1 / 2 ) \pi x / L ) d x / L
E) u(x,t)=n0cnsin((n1/2)πx/L)e(n1/2)2x2t/L2u ( x , t ) = \sum _ { n - 0 } ^ { \infty } c _ { n } \sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } } , where cn=0Lf(x)sin((n1/2)πx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \sin ( ( n - 1 / 2 ) \pi x / L ) d x
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7
The solution of the eigenvalue problem from the previous problem is

A) λ=nπ/L,X(x)=sin(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
B) λ=nπ/L,X(x)=cos(nπx/L),n=0,1,2,\lambda = n \pi / L , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
C) λ=(n1/2)π/L,X(x)=sin((n1/2)πx/L),n=1,2,3,\lambda = ( n - 1 / 2 ) \pi / L , X ( x ) = \sin ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots
D) λ=((n1/2)π/L)2,X(x)=sin((n1/2)πx/L),n=1,2,3,\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X ( x ) = \sin ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots
E) λ=((n1/2)π/L)2,X(x)=cos((n1/2)πx/L),n=1,2,3,\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X ( x ) = \cos ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots
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In the previous three problems, the solution of the original problem is

A) u(x,t)=n=1cnsin(nπx/L)e(nπ/L)tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) t } , where cn=0Lf(x)cos(nπx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x
B) u(x,t)=n=0cncos(nπx/L)cos(nπt/L)u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) \cos ( n \pi t /L ) , where c0=0Lf(x)dx/L,cn=20Lf(x)cos(nπx/L)dx/Lc _ { 0 } = \int _ { 0 } ^ { L } f ( x ) d x / L , c _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / L
C) u(x,t)=n=1cnsin(nπx/L)sin(nπt/L)u ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sin ( n \pi t / L ) , where cn=20Lf(x)cos(nπx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / L
D) u(x,t)=n=1cnsin(nπx/L)enxt/Lu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - nxt / L } , where cn=20Lf(x)sin(nπx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / L
E) u(x,t)=n=0cncos(nπx/L)enxt/Lu ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) e ^ { - n x t / L } , where cn=0Lf(x)cos(nπx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x
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The differential equation 2ux2+2uxy+2uy2=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial x \partial y } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = u is Select all that apply.

A) linear
B) nonlinear
C) hyperbolic
D) elliptic
E) parabolic
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Consider the equation uxxut=0u _ { xx } - u _ { t } = 0 with conditions u(0,t)=0,ux=(L,t)=0,u(x,0)=f(x)u ( 0 , t ) = 0 , u _ { x } = ( L , t ) = 0 , u ( x , 0 ) = f ( x ) . When separating variables with u(x,t)=X(x)T(t)u ( x , t ) = X ( x ) T ( t ) , the resulting problems for X,TX , T are

A) X+λX=0,X(0)=0,X(L)=0,TλT=0,T(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } - \lambda T = 0 , T ( 0 ) = f ( x )
B) X+λX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } + \lambda T = 0 , T ( 0 ) = f ( x )
C) XλX=0,X(0)=0,X(L)=0,T+λT=0X ^ { \prime \prime } - \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } + \lambda T = 0
D) X+λX=0,X(0)=0,X(L)=0,TλT=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } - \lambda T = 0
E) X+λX=0,X(0)=0,X(L)=0,T+λT=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } + \lambda T = 0
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11
Consider the equation uxxutt=0u _ { xx } - u _ { t t } = 0 with conditions dudx(0,t)=0,dudx=(L,t)=0,u(x,0)=f(x),dudt(0)=0\frac { d u } { d x } ( 0 , t ) = 0 , \frac { d u } { d x } = ( L , t ) = 0 , u ( x , 0 ) = f ( x ) , \frac { d u } { d t } ( 0 ) = 0 . When separating variables with u(x,t)=X(x)T(t)u ( x , t ) = X ( x ) T ( t ) , the resulting problems for X,TX , T are

A) X+λX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ( 0 ) = 0
B) XλX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=0X ^ { \prime \prime } - \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
C) X+λX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
D) X+λX=0,X(0)=0,X(L)=0,TλT=0,T(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } - \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
E) X+λX=0,X(0)=0,X(L)=0,T+λT=0,T(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ^ { \prime } ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ( 0 ) = f ( x )
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In the previous two problems, the product solutions are

A) sin((n1/2)πx/L)e(n1/2)2π2t/L2\sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } \pi ^ { 2 } t / L ^ { 2 } }
B) cos((n1/2)πx/L)e(n1/2)2π2t/L2\cos ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) ^ { 2 } \pi ^ { 2 } t / L ^ { 2 } }
C) sin((n1/2)πx/L)e(n1/2)/L\sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { - ( n - 1 / 2 ) / L }
D) sin((n1/2)πx/L)e(n1/2)2x2t/L2\sin ( ( n - 1 / 2 ) \pi x / L ) e ^ { ( n - 1 / 2 ) ^ { 2 } x ^ { 2 } t / L ^ { 2 } }
E) cos((n1/2)πx/L)e(n1/2)2π2t/L2\cos ( ( n - 1 / 2 ) \pi x / L ) e ^ { ( n - 1 / 2 ) ^ { 2 } \pi ^ { 2 } t / L ^ { 2 } }
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13
The solution of the previous three problems is n=1cnXn(x)eλnkt\sum _ { n = 1 } ^ { \infty } c _ { n } X _ { n } ( x ) e ^ { - \lambda _ { n } k t } , where XnX _ { n } and λn\lambda _ { n } are given in the previous problem and

A) cn=0Lf(x)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) d x
B) cn=0Lf(x)Xn(x)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x
C) cn=0Lf(x)Xn(x)dx/0LXn(x)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x / \int _ { 0 } ^ { L } X _ { n } ( x ) d x
D) cn=0LXn2(x)dx/0Lf(x)Xn(x)dxc _ { n } = \int _ { 0 } ^ { L } X _ { n } ^ { 2 } ( x ) d x / \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x
E) cn=0Lf(x)Xn(x)dx/0LXn2(x)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) X _ { n } ( x ) d x / \int _ { 0 } ^ { L } X _ { n } ^ { 2 } ( x ) d x
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After separating variables in the previous problem, the eigenvalue problem becomes

A) X+λX=0,X(0)=0,X(L)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0
B) X+λX=0,X(0)=0,X(L)=hX(L)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = - h X ( L )
C) X+λX=0,X(0)=0,X(L)=hX(0)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = - h X ^ { \prime } ( 0 )
D) X+λX=0,X(0)=0,X(L)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0
E) X+λX=0,X(0)=0,X(L)=hX(L)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = h X ( L )
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15
In the previous two problems, the product solutions are

A) sin(nπx/L)sin(nπ/L)\sin ( n \pi x / L ) \sin ( n \pi / L )
B) sin(nπx/L)cos(nπ/L)\sin ( n \pi x / L ) \cos ( n \pi / L )
C) cos(nπx/L)cos(nπ/L)\cos ( n \pi x / L ) \cos ( n \pi / L )
D) sin(nπx/L)enx/L\sin ( n \pi x / L ) e ^ { n x / L }
E) cos(nπx/L)ent/L\cos ( n \pi x / L ) e ^ { - n t / L }
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16
The solution of yn2n2y=0,y(0)=0,y(1)=0,n=1,2,3y ^ { \prime \prime } - n ^ { 2 } n ^ { 2 } y = 0 , y ( 0 ) = 0 , y ( 1 ) = 0 , n = 1,2,3 \ldots is

A) y=0y = 0
B) y=csin(nπx)y = c \cdot \sin ( n \pi x )
C) y=ccos(nπx)y = c \cdot \cos ( n \pi x )
D) c(enxtenxt)c \left( e ^ { n xt } - e ^ { - nxt } \right)
E) c(enxt+enxt)c \left( e ^ { n xt } + e ^ { - nxt } \right)
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When u(x,y)=X(x)Y(y)u ( x , y ) = X ( x ) Y ( y ) is substituted into the equation uxx+uyy=0u _ { xx } + u _ { y y } = 0 , the resulting equations for XX and YY are

A) X+λX=0,Y+λY=0X ^ { \prime } + \lambda X = 0 , Y ^ { \prime } + \lambda Y = 0
B) X+λX=0,YλY=0X ^ { \prime } + \lambda X = 0 , Y ^ { \prime } - \lambda Y = 0
C) X+λX=0,Y+λY=0X ^ { \prime \prime } + \lambda X = 0 , Y ^ { \prime \prime } + \lambda Y = 0
D) X+λX=0,YλY=0X ^ { \prime \prime } + \lambda X = 0 , Y ^ { \prime \prime } - \lambda Y = 0
E) XλX=0,Y+λY=0X ^ { \prime \prime } - \lambda X = 0 , Y ^ { \prime \prime } + \lambda Y = 0
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The solution of the eigenvalue problem in the previous problem is

A) λ=((n1/2)π/L)2,X=sin((n1/2)πx/L)\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X = \sin ( ( n - 1 / 2 ) \pi x / L )
B) λ/h=tan(λL),X=sin(λx)\sqrt { \lambda } / h = \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )
C) λ=(nπ/L)2,X=sin(nπx/L)\lambda = ( n \pi / L ) ^ { 2 } , X = \sin ( n \pi x / L )
D) λ/h=tan(λL),X=sin(λx)\sqrt { \lambda } / h = - \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )
E) λh=tan(λL),X=sin(λx)\sqrt { \lambda } h = - \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )
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19
The solution of y+λy=0,y(0)=0,y(π)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 if λ=0\lambda = 0 is

A) y=sinxy = \sin x
B) y=cosxy = \cos x
C) y=0y = 0
D) y=ax+by = a x + b
E) y=xπy = x - \pi
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20
The model describing the temperature in a rod where the temperature at the left end is zero and where there is heat transfer from the right boundary into the external medium is

A) k2ux2=2ut2,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = - h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
B) k2ux2=2ut2,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
C) k2ux2=ut,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = - h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
D) k2ux2=ut,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
E) k2ux2+2ut2=0,u(0,t)=0,ux=(L,t)=hu(L,t),h>0,u(x,0)=f(x)k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 0 , t ) = 0 , \frac { \partial u } { \partial x } = ( L , t ) = - h u ( L , t ) , h > 0 , u ( x , 0 ) = f ( x )
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21
Consider the equation ux+uyy=0u _ { x } + u _ { y y } = 0 with conditions u(0,y)=0,u=(L,y)=0,u(x,0)=f(x),u(x,H)=0u ( 0 , y ) = 0 , u = ( L , y ) = 0 , u ( x , 0 ) = f ( x ) , u ( x , H ) = 0 . When separating variables with u(x,y)=X(x)Y(y)u ( x , y ) = X ( x ) Y ( y ) , the resulting problems for X,YX , Y are

A) X+λX=0,X(0)=0,X(L)=0,Y+λY=0,Y(H)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } + \lambda Y = 0 , Y ( H ) = 0
B) X+λX=0,X(0)=0,X(L)=0,YλY=0,Y(H)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( H ) = 0
C) X+λX=0,X(0)=0,X(L)=0,YλY=0,Y(0)=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( 0 ) = 0
D) XλX=0,X(0)=0,X(L)=0,YλY=0,Y(H)=f(0)X ^ { \prime \prime } - \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } - \lambda Y = 0 , Y ( H ) = f ( 0 )
E) X+λX=0,X(0)=0,X(L)=0,Y+λY=0,Y(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , Y ^ { \prime \prime } + \lambda Y = 0 , Y ( 0 ) = f ( x )
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22
In the previous problem, the eigenfunction expansion if xetx e ^ { t } is

A) etn1(1)n2Lcos(nπx/L)/(nπ)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n } 2 L \cos ( n \pi x / L ) / ( n \pi )
B) etn1(1)n2Lsin(nπx/L)/(n2π2)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n } 2 L \sin ( n \pi x / L ) / \left( n ^ { 2 } \pi ^ { 2 } \right)
C) etn1(1)n2Lcos(nπx/L)/(n2π2)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n } 2 L \cos ( n \pi x / L ) / \left( n ^ { 2 } \pi ^ { 2 } \right)
D) etn1(1)n+12Lcos(nπx/L)/(nπ)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } 2 L \cos ( n \pi x / L ) / ( n \pi )
E) etn1(1)n+12Lsin(nπx/L)/(nπ)e ^ { t } \sum _ { n - 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } 2 L \sin ( n \pi x / L ) / ( n \pi )
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23
When u(x,y)=X(x)Y(y)u ( x , y ) = X ( x ) Y ( y ) is substituted into the equation uxuyy=0u _ { x } - u _ { y y } = 0 , the resulting equations for XX and YY are

A) X+λX=0,Y+λY=0X ^ { \prime \prime } + \lambda X = 0 , Y ^ { \prime \prime } + \lambda Y = 0
B) X+λX=0,YλY=0X ^ { \prime \prime } + \lambda X = 0 , Y ^ { \prime \prime } - \lambda Y = 0
C) XλX=0,Y+λY=0X ^ { \prime \prime } - \lambda X = 0 , Y ^ { \prime \prime } + \lambda Y = 0
D) X+λX=0,Y+λY=0X ^ { \prime } + \lambda X = 0 , Y ^ { \prime } + \lambda Y = 0
E) X+λX=0,YλY=0X ^ { \prime } + \lambda X = 0 , Y ^ { \prime } - \lambda Y = 0
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24
The solution of the eigenvalue problem from the previous problem is

A) λ=nπ/L,X(x)=sin(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
B) λ=nπ/L,X(x)=cos(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \cos ( n \pi x / L ) , n = 1,2,3 , \ldots
C) λ=(nπ/L)2,X(x)=sin(nπx/L),n=1,2,3,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
D) λ=(nπ/L)2,X(x)=cos(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
E) λ=(nπ/L),X(x)=sin(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) , X ( x ) = \sin ( n \pi x / L ) , n = 0,1,2 , \ldots
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The solution of the eigenvalue problem from the previous problem is

A) λ=nπ/L,X(x)=sin(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
B) λ=nπ/L,X(x)=cos(nπx/L),n=1,2,3,\lambda = n \pi / L , X ( x ) = \cos ( n \pi x / L ) , n = 1,2,3 , \ldots
C) λ=(nπ/L)2,X(x)=sin(nπx/L),n=1,2,3,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots
D) λ=(nπ/L)2,X(x)=cos(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) ^ { 2 } , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots
E) λ=(nπ/L),X(x)=sin(nπx/L),n=0,1,2,\lambda = ( n \pi / L ) , X ( x ) = \sin ( n \pi x / L ) , n = 0,1,2 , \ldots
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26
The solution of yn2y=0,y(0)=0,y(π)=0,n=1,2,3y ^ { \prime \prime } - n ^ { 2 } y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 , n = 1,2,3 \ldots is

A) y=0y = 0
B) y=csin(nx)y = c \cdot \sin ( n x )
C) y=ccos(nx)y = c \cdot \cos ( n x )
D) y=c(enxenx)y = c \left( e ^ { n x } - e ^ { - n x } \right)
E) y=c(enx+enx)y = c \left( e ^ { n x } + e ^ { - n x } \right)
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27
The differential equation 2ux22uy2=sinu\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } - \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = \sin u is Select all that apply.

A) linear
B) nonlinear
C) hyperbolic
D) elliptic
E) parabolic
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28
In the previous three problems, the solution of the original problem is

A) u(x,t)=n1cnsin(nπx/L)sinh(nπ(yH)/L)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L ) , where cn=0Lf(x)cos(nπx/L)dx/sinh(nπH/L)c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / \sinh ( - n \pi H / L )
B) u(x,t)=n1cnsin(nπx/L)sinh(nπ(yH)/L)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L ) , where cx=20Lf(x)sin(nπx/L)dx/(Lsinh(nπH/L))c _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )
C) u(x,t)=n1cnsin(nπx/L)cosh(nπ(yH)/L)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \cosh ( n \pi ( y - H ) / L ) , where cn=20Lf(x)cos(nπx/L)dx/(Lsinh(nπH/L))c _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )
D) u(x,t)=n1cnsin(nπx/L)cosh((nπ/L)2(yH))u ( x , t ) = \sum _ { n - 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) \cosh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right) , where cx=20Lf(x)sin(nπx/L)dx/(Lsinh(nπH/L))c _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / ( L \sinh ( - n \pi H / L ) )
E) u(x,t)=n=0cncos(nπx/L)sinh((nπ/L)2(yH))u ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) \sinh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right) , where cn=0Lf(x)cos(nπx/L)dx/sinh(nπH/L)c _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / \sinh ( - n \pi H / L )
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In the previous two problems, the product solutions are

A) sin(nπx/L)sinh(nπ(yH)/L)\sin ( n \pi x / L ) \sinh ( n \pi ( y - H ) / L )
B) cos(nπx/L)sinh(nπ(yH)/L)\cos ( n \pi x / L ) \sinh ( - n \pi ( y - H ) / L )
C) cos(nπx/L)cosh((nπ/L)2(yH))\cos ( n \pi x / L ) \cosh \left( ( n \pi / L ) ^ { 2 } ( y - H ) \right)
D) sin(nπx/L)cosh((nπ/L)2(yH))\sin ( n \pi x / L ) \cosh \left( - ( n \pi / L ) ^ { 2 } ( y - H ) \right)
E) cos(nπx/L)cosh(nπ(yH)/L)\cos ( n \pi x / L ) \cosh ( - n \pi ( y - H ) / L )
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30
The differential equation 2ux2+2uy2=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = u is

A) first order, linear, homogeneous
B) first order, linear, non-homogeneous
C) second order, nonlinear
D) second order, linear, homogeneous
E) second order, linear, non-homogeneous
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31
In the problem 2ux2+xet=ut,u(0,t)=0,u(L,t)=0,u(x,0)=0\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + x e ^ { t } = \frac { \partial u } { \partial t } , u ( 0 , t ) = 0 , u ( L , t ) = 0 , u ( x , 0 ) = 0 , the eigenvalues and eigenfunctions of the underlying homogeneous problem are

A) λ=n2,X=sin(nx)\lambda = n ^ { 2 } , X = \sin ( n x )
B) λ=n2,X=cos(nx)\lambda = n ^ { 2 } , X = \cos ( n x )
C) λ=n2π2/L2,X=sin(nπx/L)\lambda = n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } , X = \sin ( n \pi x / L )
D) λ=n2π2/L2,X=cos(nπx/L)\lambda = n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } , X = \cos ( n \pi x / L )
E) λ=n2π2,X=sin(nπx)\lambda = n ^ { 2 } \pi ^ { 2 } , X = \sin ( n \pi x )
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32
The quantity of heat in an element of a rod of mass mm is proportional to Select all that apply.

A) mass
B) thermal conductivity
C) specific heat
D) thermal diffusivity
E) temperature
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33
In the previous two problems, the solution for uu takes the form

A) u(x,t)=n1un(t)sin(nπx/L)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } u _ { n } ( t ) \sin ( n \pi x / L )
B) u(x,t)=n=1un(t)cos(nπx)u ( x , t ) = \sum _ { n = 1 } ^ { \infty } u _ { n } ( t ) \cos ( n \pi x )
C) u(x,t)=n1un(t)sin(n2π2x2/L2)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } u _ { n } ( t ) \sin \left( n ^ { 2 } \pi ^ { 2 } x ^ { 2 } / L ^ { 2 } \right)
D) u(x,t)=n1un(t)cos(n2π2x2/L2)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } u _ { n } ( t ) \cos \left( n ^ { 2 } \pi ^ { 2 } x ^ { 2 } / L ^ { 2 } \right)
E) u(x,t)=n1un(t)sin(nπx)u ( x , t ) = \sum _ { n - 1 } ^ { \infty } u _ { n } ( t ) \sin ( n \pi x )
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34
In the previous two problems, the product solutions are

A) sin(nπx/L)enx/L\sin ( n \pi x / L ) e ^ { n x / L }
B) sin(nπx/L)enx/L\sin ( n \pi x / L ) e ^ { - n x / L }
C) cos(nπx/L)e(nπ/D)2t\cos ( n \pi x / L ) e ^ { ( n \pi / D ) ^ { 2 } t }
D) sin(nπx/L)e(nπ/L)2t\sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t }
E) cos(nπx/L)ent/L\cos ( n \pi x / L ) e ^ { - n t / L }
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35
Consider the equation umut=0u _ { m } - u _ { t } = 0 with conditions u(0,t)=0,u=(L,t)=0,u(x,0)=f(x)u ( 0 , t ) = 0 , u = ( L , t ) = 0 , u ( x , 0 ) = f ( x ) . When separating variables with u(x,t)=X(x)T(t)u ( x , t ) = X ( x ) T ( t ) , the resulting problems for X,TX , T are

A) X+λX=0,X(0)=0,X(L)=0,Tt+λT=0,T(0)=f(0)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { t } + \lambda T = 0 , T ( 0 ) = f ( 0 )
B) X+λX=0,X(0)=0,X(L)=0,Tt+λT=0,T(0)=f(x)X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { t } + \lambda T = 0 , T ( 0 ) = f ( x )
C) XλX=0,X(0)=0,X(L)=0,T+λT=0X ^ { \prime \prime } - \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { \prime } + \lambda T = 0
D) X+λX=0,X(0)=0,X(L)=0,TλT=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { \prime } - \lambda T = 0
E) X+λX=0,X(0)=0,X(L)=0,T+λT=0X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ( L ) = 0 , T ^ { \prime } + \lambda T = 0
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36
In the previous three problems, the solution of the original problem is

A) u(x,t)=n=1cnsin(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cn=0Lf(x)cos(nπx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x
B) u(x,t)=n=1cncos(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cn=0Lf(x)sin(nπx/L)dxc _ { n } = \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x
C) u(x,t)=n=1cnsin(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cn=20Lf(x)cos(nπx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x / L
D) u(x,t)=n=1cnsin(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 1 } ^ { \infty } c _ { n } \sin ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cn=20Lf(x)sin(nπx/L)dx/Lc _ { n } = 2 \int _ { 0 } ^ { L } f ( x ) \sin ( n \pi x / L ) d x / L
E) u(x,t)=n=0cncos(nπx/L)e(nπ/L)2tu ( x , t ) = \sum _ { n = 0 } ^ { \infty } c _ { n } \cos ( n \pi x / L ) e ^ { - ( n \pi / L ) ^ { 2 } t } , where cx=20Lf(x)cos(nπx/L)dxc _ { x } = 2 \int _ { 0 } ^ { L } f ( x ) \cos ( n \pi x / L ) d x
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37
The general solution of y+n2y=0,y(0)=0,y(π)=0,n=1,2,3y ^ { \prime \prime } + n ^ { 2 } y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 , n = 1,2,3 \ldots is

A) y=0y = 0
B) y=csin(nx)y = c \cdot \sin ( n x )
C) y=ccos(nx)y = c \cdot \cos ( n x )
D) y=c(enxenx)y = c \left( e ^ { n x } - e ^ { - n x } \right)
E) y=c(enx+enx)y = c \left( e ^ { n x } + e ^ { - n x } \right)
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38
The solution of y+λy=0,y(0)=0,y(π)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 if λ=0\lambda = 0 is

A) y=ax+by = a x + b
B) y=xπy = x - \pi
C) y=0y = 0
D) y=sinxy = \sin x
E) y=cosxy = \cos x
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39
In the previous three problems, the solution for u(t)u ( t ) is

A) un(t)=(1)n2L(eten2a2t/L2)/(nπ(1+nπ/L))u _ { n } ( t ) = ( - 1 ) ^ { n } 2 L \left( e ^ { t } - e ^ { - n ^ { 2 } a ^ { 2 } t / L ^ { 2 } } \right) / ( n \pi ( 1 + n \pi / L ) )
B) ux(t)=(1)n2L(et+en2x2t/L2)/(nπ(1+n2π2/L2))u _ { x } ( t ) = ( - 1 ) ^ { n } 2 L \left( e ^ { t } + e ^ { - n ^ { 2 } x ^ { 2 } t / L ^ { 2 } } \right) / \left( n \pi \left( 1 + n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } \right) \right)
C) ux(t)=(1)n+12L(et+en2x2t/L2)/(nπ(1+n2π2/L2))u _ { x } ( t ) = ( - 1 ) ^ { n + 1 } 2 L \left( e ^ { t } + e ^ { - n ^ { 2 } x ^ { 2 } t / L ^ { 2 } } \right) / \left( n \pi \left( 1 + n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } \right) \right)
D) ux(t)=(1)n+12L(eten2x2t/L2)/(nπ(1+n2π2/L2))u _ { x } ( t ) = ( - 1 ) ^ { n + 1 } 2 L \left( e ^ { t } - e ^ { - n ^ { 2 } x ^ { 2 } t / L ^ { 2 } } \right) / \left( n \pi \left( 1 + n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } \right) \right)
E) un(t)=2L(eten2x2t/L2)/(nπ(1+n2π2/L2))u _ { n } ( t ) = 2 L \left( e ^ { t } - e ^ { - n ^ { 2 } x ^ { 2 } t / L ^ { 2 } } \right) / \left( n \pi \left( 1 + n ^ { 2 } \pi ^ { 2 } / L ^ { 2 } \right) \right)
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40
The differential equation 2ux2+2uy2=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = u is Select all that apply.

A) linear
B) nonlinear
C) hyperbolic
D) elliptic
E) parabolic
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