115年成功大學生醫碩士班-工程數學詳解

國立成功大學115學年度碩士班招生考試試題

系所:生物醫學工程系
科目:工程數學

解答:$$\cases{x'-4x+y'''=6\cos t\cdots(1) \\ x'+2x-2y'''=0 \Rightarrow y'''={1\over 2}x'+x\cdots(2)} \Rightarrow  x'-4x+{1\over 2}x'+x=6\cos \Rightarrow x'-2x=4\cos t \\ \Rightarrow x={4\over 5}(\sin t-2\cos t)+c_1e^{2t} \Rightarrow x(0)=-{8\over 5}+c_1=0 \Rightarrow c_1={8\over 5}\\ \Rightarrow x(t)={4\over 5}(\sin t-2\cos t)+{8\over 5}e^{2t} \Rightarrow x'(t)={4\over 5}(\cos t+2\sin t)+{16\over 5}e^{2t} \\ \Rightarrow y'''={1\over 2}x'+x={16\over 5}e^{2t}-{6\over 5}\cos t+{8\over 5}\sin t \Rightarrow y'' = \int \left( {16\over 5}e^{2t}-{6\over 5}\cos t+{8\over 5}\sin t  \right)\,dt \\ \Rightarrow y''={8\over 5}e^{2t}-{6\over 5}\sin t-{8\over 5}\cos t+c_2 \Rightarrow y''(0)={8\over 5}-{8\over 5}+c_2=0 \Rightarrow c_2=0 \\ \Rightarrow y''={8\over 5}e^{2t}-{6\over 5}\sin t-{8\over 5}\cos t \Rightarrow y'= \int \left( {8\over 5}e^{2t}-{6\over 5}\sin t-{8\over 5}\cos t \right)\,dt \\ \Rightarrow y'={4\over 5}e^{2t}+{6\over 5}\cos t-{8\over 5}\sin t+c_3 \Rightarrow y'(0)={4\over 5}+{6\over 5}+c_3=1\Rightarrow c_3=-1 \\ \Rightarrow y'= {4\over 5}e^{2t}+{6\over 5}\cos t-{8\over 5}\sin t-1 \Rightarrow y= \int \left( {4\over 5}e^{2t}+{6\over 5}\cos t-{8\over 5}\sin t-1 \right)\,dt \\ \Rightarrow y= {2\over 5}e^{2t}+{6\over 5}\sin t+{8\over 5}\cos t-t+c_4 \Rightarrow y(0)={2\over 5}+{8\over 5}+c_4=0 \Rightarrow c_4=-2 \\ \Rightarrow \bbox[red, 2pt]{y(t) ={2\over 5}e^{2t}+{6\over 5}\sin t+{8\over 5}\cos t-t-2}$$

解答:$$\text{Let }  w(x,t)=u(0,t)+{x-0\over 1-0}(u(1,t)-u(0,t)) =\sin t-x\sin t, \\ \text{then let }u(x,t)=v(x,t)+w(x,t)=v(x,t)+(1-x)\sin t \Rightarrow \cases{u_t=v_t+(1-x)\cos t\\ u_{xx}=v_{xx}}\\ \Rightarrow u_{xx}=u_t \Rightarrow v_{xx}=v_t+(1-x)\cos t \Rightarrow v_t-v_{xx}=(x-1)\cos t \\ BCs: \cases{v(0,t)=u(0,t)-w(0,t)= \sin t-\sin t=0\\ v(1,t)=u(1,t)-w(1,t)=0-0=0} \\ \text{Consider } v_t-v_{xx}=0 \text{ with } \cases{v(0,t)=0\\ v(1,t)=0} \\ \text{Assume }v(x,t)= X(x)T(t) \Rightarrow XT'=X''T \Rightarrow {T'\over T}={X''\over X} =-\lambda \\ \text{BCs: } \cases{v(0,t)=0 \\v(1,t)=0} \Rightarrow \cases{X(0)=0\\ X(1)=0}\\ \text{Solving for }X''+\lambda X=0 \Rightarrow \cases{\lambda=0 \Rightarrow X=0\\ \lambda\lt 0 \Rightarrow X=0\\ \lambda=\alpha^2 \gt 0 \Rightarrow  \sin \alpha=0} \Rightarrow X_n= \sin(n\pi x), n=1,2,\dots\\ \text{Homogeneous solution: } v(x,t) = \sum_{n=1}^\infty a_n(t) \sin (n\pi x) \\ \text{Suppose }1-x = \sum_{n=1}^\infty b_n \sin(n\pi x) \Rightarrow b_n= 2\int_0^1 (1-x)\sin(n\pi x)\,dx ={2\over n\pi} \\ \Rightarrow 1-x=\sum_{n=1}^\infty {2\over n\pi} \sin(n\pi x) \\  \\ v_t=v_{xx}-(1-x) \cos t \Rightarrow   \sum_{n=1}^\infty a_n'(t)\sin(n\pi x) = \sum_{n=1}^\infty a_n(t)(-n^2\pi^2) \sin(n\pi x)- \sum_{n=1}^\infty{2\over n\pi }\sin(n\pi x) \cos t \\ \Rightarrow a_n'(t)+n^2\pi^2 a_n(t)=-{2\over n\pi }\cos t \text{, with }a_n(0)=0 \\ \Rightarrow \text{homogeneous sol: }a_n^H(t)=Ce^{-n^2\pi^2t}, \text{ particular sol: }a_n^P(t)= -{2n\pi\over 1+n^4\pi^4} \cos t-{2\over n\pi(1+n^4\pi^4)} \sin t \\ \Rightarrow a_n(t)=a_n^H(t)+a_n^P(t)  \Rightarrow a_n(0)=C-{2n\pi\over 1+n^4\pi^4}=0 \Rightarrow C={2n\pi\over 1+n^4\pi^4} \\ \Rightarrow a_n(t)={2n\pi\over 1+n^4\pi^4} e^{-n^2\pi^2 t}-{2n\pi\over 1+n^4\pi^4} \cos t-{2\over n\pi(1+n^4\pi^4)} \sin t \\\Rightarrow \bbox[red, 2pt]{u(x,t)=(1-x)\sin t + \sum_{n=1}^\infty \left( {2n\pi\over 1+n^4\pi^4} e^{-n^2\pi^2 t}-{2n\pi\over 1+n^4\pi^4} \cos t-{2\over n\pi(1+n^4\pi^4)} \sin t \right) \sin (n\pi x)}$$

解答:$$y(x) ={u(x) \over x} \Rightarrow y'={u'\over x}-{u\over x^2} \Rightarrow y''={u''\over x}-{2u'\over x^2} +{2u\over x^3} \\ \text{Plug }y,y', \text{ and }y'' \text{ back into the original differential equation:} \\ x^2 \left( {u''\over x}-{2u'\over x^2} +{2u\over x^3} \right)+2x \left( {u'\over x}-{u\over x^2} \right) +\beta x^2 \left( {u\over x} \right)=0 \Rightarrow xu''+ \beta xu=0 \Rightarrow u''+\beta u=0 \\\textbf{Case I }\beta \gt 0 \Rightarrow u(x)= c_1 \cos(\sqrt \beta x)+ c_2 \sin (\sqrt \beta x) \Rightarrow y(x) ={1\over x} \left( c_1 \cos(\sqrt \beta x)+ c_2 \sin (\sqrt \beta x) \right)\\ \textbf{Case II }\beta =0 \Rightarrow u''=0 \Rightarrow u(x)=c_1x+c_2 \Rightarrow y(x)=c_1+{c_2\over x} \\ \textbf{Case III }\beta \lt 0 \Rightarrow u(x)=c_1 e^{\sqrt{-\beta}x} +c_2 e^{-\sqrt{-\beta}x} \Rightarrow y(x) = {1\over x}\left( c_1 e^{\sqrt{-\beta}x} +c_2 e^{-\sqrt{-\beta}x} \right) \\ \Rightarrow \bbox[red, 2pt]{y(x) = \begin{cases} \displaystyle {1\over x} \left( c_1 \cos(\sqrt \beta x)+ c_2 \sin (\sqrt \beta x) \right),& \beta \gt 0\\ c_1+\displaystyle {c_2\over x},& \beta=0\\ \displaystyle {1\over x}\left( c_1 e^{\sqrt{-\beta}x} +c_2 e^{-\sqrt{-\beta}x} \right), & \beta \lt 0\end{cases}}$$

========================== END =========================

解題僅供參考，碩士班歷年試題及詳解