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:\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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