國立中山大學112學年度碩士班暨碩士在職專班招生考試
科目名稱：工程數學【材光系碩士班選考，材料前瞻應材碩士班選考，材光聯合碩士班選考】
解答：$$積分因子\mu(x)=e^{\int \tan x\,dx} =\sec x \Rightarrow y'\sec x +y\sec x\tan x = \sec x\sin(2x) \Rightarrow (y\sec x)'=2\sin x \\ \Rightarrow y \sec x=\int 2\sin x\,dx =-2\cos x+C \Rightarrow y=-2\cos^2 x+C\cos x\\ 將初始值y(0)=1代入上式\Rightarrow -2+C=1 \Rightarrow C=3 \Rightarrow \bbox[red, 2pt]{y=-2\cos^2 x+3\cos x}$$
解答：$$y'+y=-{x\over y} \Rightarrow 2yy'+2y^2=-2x \Rightarrow v'+2v=-2x, 其中v(x)=y^2(x) \\ 積分因子 \mu=e^{\int 2\,dx} =e^{2x} \Rightarrow v'e^{2x}+2v e^{2x}=-2xe^{2x} \Rightarrow (ve^{2x})'=-2xe^{2x} \\ \Rightarrow ve^{2x} =\int -2xe^{2x}\,dx = {1\over 2}e^{2x}(1-2x)+ C \Rightarrow v=y^2 ={1\over 2}(1-2x)+Ce^{-2x}\\ y(0)=1 \Rightarrow C={1\over 2} \Rightarrow y^2 = {1\over 2}(1-2x+e^{-2x}) \Rightarrow \bbox[red, 2pt]{y=\sqrt{1-2x+e^{-2x} \over 2}}$$解答：$$先求齊次解：xy'+4y=0,取y=x^m \Rightarrow y'=mx^{m-1} 代回原式\Rightarrow (m+4)x^m=0 \\ \Rightarrow m=-4 \Rightarrow y_h=Cx^{-4}\\ 又取y_p=Ax^4 \Rightarrow y_p'=4Ax^3代回原式\Rightarrow 8Ax^4=8x^4 \Rightarrow A=1 \Rightarrow y_p=x^4\\ y=y_h+y_p =Cx^{-4}+x^4,由於y(1)=2 \Rightarrow 2=C+1 \Rightarrow C=1 \Rightarrow \bbox[red,2pt]{y=x^4+{1\over x^4}}$$
解答：$$取y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow y'''=m(m-1)(m-2) x^{m-3}\\ \Rightarrow x^3y'''+3x^2y''-6xy'-6y= (m(m-1)(m-2)+ 3m(m-1)-6m-6)x^m=0 \\ \Rightarrow (m+1)(m+2)(m-3)x^m=0 \Rightarrow m=-1,-2,3 \Rightarrow \bbox[red, 2pt]{y=C_1x^{-1} +C_2x^{-2}+C_3 x^3}$$
解答：$$先求齊次解：y''-3y'+2y=0 \Rightarrow (\lambda-2)(\lambda-1)=0 \Rightarrow \lambda=1,2 \Rightarrow y_h=C_1e^x +C_2e^{2x}\\ 取y_p=Axe^{2x}+Bx^2+Cx+D \Rightarrow y_p'=Ae^{2x}+2Axe^{2x}+ 2Bx+C \Rightarrow y_p''=4Ae^{2x}+4Axe^{2x}+2B \\ \Rightarrow y_p''-3y_p'+2y_p=Ae^{2x}+2Bx^2+(2C-6B)x+2B-3C+2D=3e^{2x}+2x^2-7\\ \Rightarrow \cases{A=3\\ B=1\\ C=3\\ D=0} \Rightarrow y_p=3xe^{2x}+ x^2+3x \Rightarrow y=y_h+y_p =C_1e^x +C_2e^{2x}+3xe^{2x}+ x^2+3x\\ \Rightarrow y'=C_1e^x +(2C_2+3)e^{2x}+ 6xe^{2x}+2x+3 \Rightarrow \cases{y(0)=1=C_1+C_2\\ y'(0)=0=C_1+2C_2+6}\\ \Rightarrow \cases{C_1=8\\ C_2=-7} \Rightarrow \bbox[red,2pt]{y= 8e^x-7e^{2x}+3xe^{2x}+ x^2+3x}$$
解答：$$\cases{y_1'=-y_1-y_2\\ y_2'=y_1-y_2} \Rightarrow y'=\begin{bmatrix}-1 & -1 \\1 & -1 \end{bmatrix}y =Ay\\求矩陣Ａ的特徵值及特徵向量：\\ \det(A-\lambda I)=0 \Rightarrow \lambda^2+2\lambda+2=0 \Rightarrow \lambda=-1\pm i\\ \lambda_1=-1+i \Rightarrow (A-\lambda_1 I)x=0 \Rightarrow x_1=ix_2,取v_1=\begin{bmatrix}i \\1 \end{bmatrix} \\ \lambda_2=-1-i \Rightarrow (A-\lambda_2 I)x=0 \Rightarrow x_1=-ix_2,取v_2=\begin{bmatrix}1 \\i \end{bmatrix}\\ 因此Ａ=PDP^{-1},其中P=\begin{bmatrix}i & 1 \\ 1 &i \end{bmatrix},D=\begin{bmatrix}-1+i &0 \\0 & -1-i \end{bmatrix}\\ \Rightarrow y'=PDP^{-1}y \Rightarrow P^{-1}y'= DP^{-1}y \Rightarrow z'=Dz,其中z=P^{-1}y\\ \Rightarrow z'=\begin{bmatrix}-1+i &0 \\0 & -1-i \end{bmatrix}\begin{bmatrix}z_1 \\z_2 \end{bmatrix}= \begin{bmatrix}(-1+i)z_1 \\(-1-i)z_2 \end{bmatrix} \Rightarrow \cases{z_1=C_1e^{(-1+i)x} \\ z_2= C_2 e^{(-1-i)x}}\\ \Rightarrow y=Pz =\begin{bmatrix}i & 1 \\ 1 &i \end{bmatrix} \begin{bmatrix}C_1e^{(-1+i)x} \\C_2e^{(-1-i)x}\end{bmatrix} =\begin{bmatrix}iC_1e^{(-1+i)x} +C_2e^{(-1-i)x}\\C_1e^{(-1+i)x}+ iC_2e^{(-1-i)x} \end{bmatrix}\\ 初始值\cases{y_1(0)=1\\ y_2(0)=0} \Rightarrow \cases{iC_1+C_2=1\\ C_1+iC_2=0} \Rightarrow \cases{C_1=-i/2\\ C_2=1/2}\\ \Rightarrow \cases{y_1={1\over 2}e^{-x}(e^{ix}+e^{-ix}) =e^{-x} \cos x\\ y_2= {1\over 2}e^{-x}(-ie^{ix}+ie^{-ix}) =e^{-x}\sin x} \Rightarrow \bbox[red,2pt]{\cases{y_1=e^{-x}\cos x\\ y_2=e^{-x}\sin x}}$$
解答：$$\mathbf{(a)}\;L^{-1}\{{2\over s^4} \}-L^{-1}\{{48\over s^6} \} ={1\over 3}L^{-1}\{{3!\over s^4} \}-{2\over 5}L^{-1}\{{5!\over s^6} \} =\bbox[red, 2pt]{{1\over 3}t^4-{2\over 5}t^5} \\\mathbf{(b)}\; L^{-1} \{{6s+7\over 2s^2+4s+10}\} =L^{-1} \{3\cdot {s+1\over (s+1)^2+2^2}+ {1\over 4}\cdot {2 \over (s+1)^2+2^2}\} \\\quad = \bbox[red, 2pt]{3e^{-t}\cos (2t)+{1\over 4}e^{-t}\sin(2t)}$$解答：$$L\{y''\}-{1\over 4}L\{ y\} =s^2Y(s)-sy(0)-y'(0)-{1\over 4}Y(s)=(s^2-{1\over 4})Y(s)-12s=0 \\ \Rightarrow Y(s)={12s\over s^2-1/4}={6\over s+1/2}+{6\over s-1/2}\\ \Rightarrow y(t)=L^{-1}\{Y(s)\}=6L^{-1}\{{1\over s+1/2}\}+ 6L^{-1}\{{1\over s-1/2}\} =6(e^{-t/2} +e^{t/2}) \\ \Rightarrow \bbox[red, 2pt]{y(t)= 6(e^{-t/2} +e^{t/2})}$$
解答：$$取w(t)=y(t+2) \Rightarrow y''+2y'-3y=w''+2w'-3w=0 \\\Rightarrow L\{w''\}+2 L\{ w'\}-3L\{w\} =s^2W(s)-sw(0)-w'(0)+2(sW(s)-w(0))-3W(s)\\ =(s^2+2s-3)W(s)+ 3s+11=0 \Rightarrow W(s)=-{3s+11\over s^2+2s-3}={1\over 2}\cdot {1\over s+3}-{7\over 2}\cdot {1\over s-1} \\ \Rightarrow w(t)=L^{-1}\{W(s)\}={1\over 2} e^{-3t}-{7\over 2}e^t \Rightarrow w(t-2)= \bbox[red, 2pt]{y(t)={1\over 2} e^{-3(t-2)}-{7\over 2}e^{t-2}}$$
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