115年台大碩士班-工程數學J詳解

國立臺灣大學 115 學年度碩士班招生考試試題

科目: 工程數學(J)

解答:$$A= \begin{bmatrix}-1&1\\0&1 \end{bmatrix} = \begin{bmatrix}1/2& 1\\1&0 \end{bmatrix} \begin{bmatrix}1&0\\0&-1 \end{bmatrix} \begin{bmatrix}0& 1\\1&-1/2 \end{bmatrix} \\ \Rightarrow e^A = \begin{bmatrix}1/2& 1\\1&0 \end{bmatrix} \begin{bmatrix}e&0\\0&1/e \end{bmatrix} \begin{bmatrix}0& 1\\1&-1/2 \end{bmatrix} = \begin{bmatrix}e/2& 1/e\\ e&0 \end{bmatrix}  \begin{bmatrix}0& 1\\1&-1/2 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}{1\over e}& {e^2-1\over 2e} \\ 0& e\end{bmatrix}}$$

解答:$$y''-6y'+9y=0 \Rightarrow r^2-6r+9=0 \Rightarrow (r-3)^2=0 \Rightarrow y_h= c_1e^{3t}+ c_2 te^{3t}\\ \Rightarrow y_h'= (3c_1+c_2)e^{3t}+ 3c_2te^{3t} \Rightarrow \cases{y(0)=c_1 =2\\ y'(0)=3c_1+c_2=17} \Rightarrow c_2=11 \\\Rightarrow \text{ zero-input response: }y_{zi}(t)=(2+11t)e^{3t} \\ y_p=(At^2+Bt+C)t^2e^{3t} =(At^4+Bt^3+Ct^2)e^{3t} \\ \Rightarrow y_p'=(3At^4+ (4A+3B)t^3 +(3B+3C)t^2 +2Ct)e^{3t} \\ \Rightarrow y_p''= (9At^4+(24A+9B) t^3+(12A+18B+9C)t^2 +(6B+12C)t+ 2C)e^{3t} \\ \Rightarrow y_p''-6y_p'+9y_p=t^2e^{3t} \Rightarrow \cases{A=1/12\\ B=C=0} \Rightarrow y_p={1\over 12}t^4e^{3t} \\ \Rightarrow y_{zs}=y_p+y_h= {1\over 12}t^4e^{3t}+ d_1e^{3t}+d_2te^{3t} \Rightarrow y_{zs}'={1\over 3}t^3e^{3t}+{1\over 4}t^4e^{3t}+ 3d_1e^{3t}+d_2e^{3t}+3d_2te^{3t} \\ \Rightarrow \cases{y_{zs}(0)=d_1=0\\ y'_{zs}(0)=3d_1+d_2=0} \Rightarrow d_1=d_2=0 \Rightarrow \bbox[red, 2pt]{\text{ zero state response: } y={1\over 12}t^4e^{3t}}$$

解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\\Rightarrow x^2y''-3xy'+3y= [m(m-1)-3m+3]x^m= (m^2-4m+3) x^m=0 \Rightarrow (m-3)(m-1)=0 \\ \Rightarrow m=1,3 \Rightarrow y_h=c_1x+c_2x^3 \Rightarrow \cases{y_1=x\\ y_2=x^3} \Rightarrow W= \begin{vmatrix} x&x^3\\1& 3x^2\end{vmatrix}=2x^3 \\ \text{Using variation of parameters, } y_p =-x \int x^2e^x \,dx +x^3 \int  e^x\,dx = -xe^x (x^2-2x+2)+x^3e^x \\=e^x(2x^2-x) \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y= c_1x+c_2x^3 +e^x(2x^2-2x)}$$

解答:$$\textbf{(1) } (f,f)= \int_{-\infty}^\infty 1\cdot 1\cdot e^{-x^2} \,dx =\sqrt{\pi} \Rightarrow \lVert f\rVert =\sqrt{(f,f)} =\sqrt {\sqrt \pi} = \bbox[red, 2pt]{\pi^{1/4}} \\\textbf{(2) }(g_1, f)= \int_{-\infty}^\infty ax \cdot 1\cdot e^{-x^2} \,dx =a \int_{-\infty}^\infty xe^{-x^2}\,dx =0, \text{ for all }a\in \mathbb R\\ \quad (g_2,f) =\int_{-\infty}^\infty (bx^2-2)\cdot 1\cdot e^{-x^2}\,dx =b\cdot {\sqrt \pi\over 2}-2\sqrt \pi =0 \Rightarrow b=4 \\ \Rightarrow \bbox[red, 2pt]{\cases{a=任意實數\\b=4}}$$

解答:$$利用\text{ Superposition Principle }來求解, 假設u(x,y)=v(x) + w(x,y)\\ v(x)滿足\cases{v''(x)=-h\\ v(0)=0\\ v(\pi)=1} \Rightarrow v'=-hx+c_1 \Rightarrow v=-{1\over 2}hx^2+c_1x+ c_2 \Rightarrow \cases{v(0)=c_2=0\\ v(\pi)=-{1\over 2}h\pi^2+c_1\pi=1} \\ \Rightarrow c_1={1\over \pi}+{1\over 2}h\pi \Rightarrow v(x)={x\over \pi}+{h\over 2}x(\pi-x) \\ \cases{u_{xx}+u_{yy} =v''+w_{xx}+w_{yy} =-h+w_{xx}+w_{yy}=-h \Rightarrow w_{xx}+w_{yy}=0 \\ u(0,y)=v(0)+w(0,y)=0+w(0,y)=0 \Rightarrow w(0,y)=0\\ u(\pi,y)=v(\pi)+w(\pi,y)=1+w(\pi,y)=1 \Rightarrow w(\pi,y)=0 \\ u(x,0)=v(x)+w(x,0)=0 \Rightarrow w(x,0)=-v(x) \\ u(x,\infty) =v(x)+ w(x,\infty)\lt \infty  \Rightarrow w(x,\infty) \lt \infty} \Rightarrow 可用分離變數求w(x,y) \\ w(x,y) =X(x)Y(y) \Rightarrow X''Y+XY''=0 \Rightarrow -{Y''\over Y}={X''\over X} =-k^2 \gt 0\\ \Rightarrow X= d_1\cos kx+d_2\sin kx \Rightarrow X(0)=d_1=0 \Rightarrow X(\pi)=d_2\sin k\pi=0 \Rightarrow k\pi =n\pi \Rightarrow X_n=\sin nx \\ \Rightarrow Y= e_1e^{ny} +e_2e^{-ny} \Rightarrow e_1=0 (\because Y(\infty)\lt \infty) \Rightarrow Y_n=e^{-ny},n=1,2,\dots \\ \Rightarrow w(x,y) = \sum_{n=1}^\infty A_n e^{-ny} \sin(nx) \Rightarrow w(x,0) = \sum_{n=1}^\infty A_n   \sin(nx) =-v(x) \\ \Rightarrow A_n={2\over \pi} \int_0^{\pi} (1-v(x))\sin(nx) \,dx = {2\over n}(-1)^n-{2h\over n^3\pi} (1-(-1)^n) \Rightarrow u(x,y) =v(x)+ w(x,y) \\ \Rightarrow \bbox[red, 2pt]{u(x,y)= {x\over \pi}+{h\over 2}x(\pi-x)+ \sum_{n=1}^\infty \left({2\over n}(-1)^n-{2h\over n^3\pi} (1-(-1)^n)  \right) e^{-ny} \sin(nx)}$$

解答:$$\cases{x(t)=t\\ y(t)=t^2\\ z(t)=t^3} \Rightarrow \cases{x'(t)=1\\ y'(t)=2t\\ z'(t)=3t^2} \Rightarrow \int_C \mathbf F\cdot d\mathbf r =\int_0^1 (t^2 \sin(\pi t^3), t^2e^{t^2}, 3t^6) \cdot (1,2t,3t^2)\,dt \\= \int_0^1 \left( t^2\sin(\pi t^3)+2t^3e^{t^3}+ 9t^8 \right) \,dt ={2\over 3\pi}+1+1 =\bbox[red, 2pt]{2+{2\over 3\pi}}$$

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解題僅供參考，碩士班歷年試題及詳解