Exam 12: Boundary-Value Problems in Rectangular Coordinates

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In the previous three problems, the solution of the original problem is

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In the previous two problems, the product solutions are

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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

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The differential equation 2ux2+2uy2=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial y ^ { 2 } } = u is

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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.

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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

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After separating variables in the previous problem, the eigenvalue problem becomes

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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

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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

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The solution of the eigenvalue problem from the previous problem is

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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.

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In the previous two problems, the product solutions are

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In the previous three problems, the solution of the original problem is

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In the previous problem, the eigenfunction expansion if xetx e ^ { t } is

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The differential equation 2ux2ut=u\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } - \frac { \partial u } { \partial t } = u is Select all that apply.

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In the previous three problems, the solution of the original problem is

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In the previous three problems, the solution of the original problem is

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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

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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.

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In the previous three problems, the solution for u(t)u ( t ) is

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