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2024年2月6日 星期二

111年成大土木碩士班-工程數學詳解

國立成功大學111學年度碩士班招生考試

系所: 土木工程系
科目: 工程數學



解答:(1)div(fu)=(f)u+f(u)=(4xi+2yj+2zk)(xzi+yzk)+(2x2+y2+z2)(z+y)=(4x2z+2yz2)+(2x2z+2x2y+y2z+y3+z3+yz2)=6x2z+3yz2+2x2y+y2z+y3+z3(2)2f=4+2+2=8(2)curl(grad f)=curl(4xi+2yj+2zk)=|ijkxyz4x2y2z|=0(4)div(curl u)=div(|ijkxyzxz0yz|)=div(zi+xj)=0(5)grad(uu)=grad(x2z2+y2z2)=2xz2i+2yz2j+2z(x2+y2)k



解答:(1)x+xt=2cos(t)tx+x=2tcos(t)(tx)=2tcos(t)tx=2tcos(t)dttx=2tsin(t)+2cos(t)+c1x(t)=2sin(t)+2tcos(t)+c1t(2)dydx=9y2x41y2dy=9x4dx1y=95x5+c1=9x5+c25y(x)=59x5+c2y(2)=5288+c2=6c2=17336y(x)=30173354x5


解答:(1)a0=12ππ0sin(2t)dt=12π[12cos(2t)]|π0=0an=1ππ0sin(2t)cos(nt)dt=2(n24)π((1)n1)bn=1ππ0sin(2t)sin(nt)dt=0f(x)=2πn=11n24((1)n1)cos(nx))(2)A(ω)=f(x)cos(ωx)dx=π0sin(2x)cos(ωx)dx=2ω24(cos(ωπ)1)B(ω)=f(x)sin(ωx)dx=π0sin(2x)sin(ωx)dx=2sin(ωπ)ω24f(x)=1π0(2ω24(cos(ωπ)1)cos(ωx)+2sin(ωπ)ω24sin(ωx))dω


解答:det
解答:\textbf{(1)}\; u_x-72u_t-10u=0 \Rightarrow \cases{a=1\\ b=-72} \\ 令\cases{i=x\\ s=-bx+at=72x+t} \Rightarrow \cases{u_x=u_ii_x +u_ss_x =u_i+72u_s\\ u_t=u_ii_t+ u_ss_t =0+u_s} 代回原式\\\Rightarrow u_i+72u_s=72u_s+10u \Rightarrow u_i-10u=0 \Rightarrow \text{integration factor: }e^{-10i} \\ \Rightarrow e^{-10i}u_i-10e^{-10i}u=0 \Rightarrow \left( e^{-10i}u\right)'=0 \Rightarrow e^{-10i}u=\rho(s) \Rightarrow u(i,s)=e^{10i}\rho(s) \\ \Rightarrow u(x, t)=e^{10x} \rho(72x+t) \Rightarrow u_x=10e^{10x} \rho(72x+t) +72e^{10x} \rho'(72x+t) \\ \Rightarrow u_x(t=0)=10e^{10x} \rho(72x) +72e^{10x} \rho'(72x) =6e^{-2x} \Rightarrow \left( e^{10x} \rho(72x) \right)'=6e^{-2x} \\ \Rightarrow e^{10x} \rho(72x) =-3e^{-2x} +c_1 \Rightarrow \rho(\omega)=-3e^{-\omega/ 6}+ c_1e^{-5\omega/36} \\ \Rightarrow u(x, t)=e^{10x} \rho(72x+t)= u(x, t)=e^{10x} \left( -3e^{-12x-t/6 +c_1 e^{-10x-5t./36}}\right) \\ \Rightarrow \bbox[red, 2pt]{u(x,t)=-3e^{-2x-t/6}+ c_1e^{-5t/36}}\\ \textbf{(2)}\; \bbox[cyan,2pt]{題目有誤}, u(0,y)=5e^{2/x}不應該有變數x!!
 

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解題僅供參考, 其他 歷年試題及詳解






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