108年台綜大轉學考-工程數學D39詳解

 臺灣綜合大學系統108學年度學士班轉學生聯合招生考試

科目名稱：工程數學
類組代碼：D39

解答： $$\mathbf{(a)}\;令X=\begin{bmatrix} a & b\\ c& d \end{bmatrix} \Rightarrow \begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix}X+ 2\begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix} =\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix} \begin{bmatrix} a & b\\ c& d \end{bmatrix}+ 2\begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix}\\ =\begin{bmatrix} a+3c & b+3d\\ 2a+4c& 2b+4d \end{bmatrix} +\begin{bmatrix} 4 & 2 \\ -2 & 6 \end{bmatrix}  =\begin{bmatrix} a+3c+4 & b+3d+2 \\ 2a+4c-2 & 2b+4d+6 \end{bmatrix} =\begin{bmatrix} 1 & 5 \\ 2 & -1 \end{bmatrix}\\ \Rightarrow \cases{\cases{a+3c=-3 \\ 2a+4c=4} \Rightarrow \cases{a=12\\ c=-5} \\ \cases{b+3d=3\\ 2b+4d= -7} \Rightarrow \cases{b=-33/2\\ d=13/2}} \Rightarrow \bbox[red, 2pt]{X=\begin{bmatrix} 12 & -33/2\\ -5 & 13/2 \end{bmatrix}} \\\mathbf{(b)}\;(A\mid I_3) = \left(\begin{array}{rrr|rrr}1 & -2 & 3 & 1 & 0 & 0 \\2 & -3 & 3 & 0 & 1 & 0 \\3 & -4 & 2 & 0 & 0 & 1\end{array}\right) \xrightarrow{\matrix{-2r_1+r_2\to r_2 \\-3r_1+r_3\to r_3}} \left(\begin{array}{rrr|rrr}1 & -2 & 3 & 1 & 0 & 0 \\0 & 1 & -3 & -2 & 1 & 0 \\0 & 2 & -7 & -3 & 0 & 1\end{array}\right) \\ \xrightarrow{\matrix{2r_2+r_1\to r_1 \\-2r_2+r_3\to r_3}} \left(\begin{array}{rrr|rrr}1 & 0 & -3 & -3 & 2 & 0 \\0 & 1 & -3 & -2 & 1 & 0 \\0 & 0 & -1 & 1 & -2 & 1\end{array}\right) \xrightarrow{\matrix{-3r_3+r_1\to r_1 \\-3r_3+r_2\to r_2}} \left(\begin{array}{rrr|rrr}1 & 0 & 0 & -6 & 8 & -3 \\0 & 1 & 0 & -5 & 7 & -3 \\0 & 0 & -1 & 1 & -2 & 1\end{array}\right) \\\xrightarrow{ -r_3 }  \left(\begin{array}{rrr|rrr}1 & 0 & 0 & -6 & 8 & -3 \\0 & 1 & 0 & -5 & 7 & -3 \\0 & 0 & 1 & -1 & 2 & -1\end{array}\right) \Rightarrow \bbox[red,2pt]{A^{-1} =\left(\begin{array}{rrr|rrr}  -6 & 8 & -3 \\ -5 & 7 & -3 \\  -1 & 2 & -1\end{array}\right) } \\ X= A^{-1}AX =\left(\begin{array}{rrr|rrr}  -6 & 8 & -3 \\ -5 & 7 & -3 \\  -1 & 2 & -1\end{array}\right)\left(\begin{array}{rrr }  -1 \\ 2 \\ 1\end{array}\right) =\left( \begin{matrix}19 \\16 \\4\end{matrix}\right) \Rightarrow \bbox[red,2pt]{X= \left( \begin{matrix}19 \\16 \\4\end{matrix}\right)} \\ \det(5A) =5^3 \det(A)= 125\times \begin{vmatrix} 1 & -2 & 3\\ 2 & -3 & 3\\ 3& -4 & -2\end{vmatrix} =125\times (-1) =-125 \Rightarrow \bbox[red, 2pt]{\det(5A) =-125}$$

解答： $$\mathbf{(a)}\;x^3y'+ 2y=x^3+2x \Rightarrow y'+ {2\over x^3}y = 1+{2\over x^2},取積分因子I(x)=e^{\int {2\over x^3}\,dx} =e^{-1/x^2} \\ \Rightarrow e^{-{1\over x^2}}y' + {2\over x^3}e^{-{1\over x^2}} y = e^{-{1\over x^2}}+ {2\over x^2}e^{-{1\over x^2}} \Rightarrow \left( e^{-{1\over x^2}}y\right)' =e^{-{1\over x^2}}+ {2\over x^2}e^{-{1\over x^2}}  \\ \Rightarrow e^{-{1\over x^2}}y =\int e^{-{1\over x^2}}+ {2\over x^2}e^{-{1\over x^2}}\,dx = xe^{-{1\over x^2}}+C \Rightarrow y=x+ Ce^{1\over x^2}\\ 將y(1)=e+1代入上式\Rightarrow e+1 =1+Ce \Rightarrow C=1 \Rightarrow \bbox[red, 2pt]{y=x+ e^{1\over x^2}}$$ $$\mathbf{(b)}\;令y=xv(x) \Rightarrow y'=v(x)+xv'(x) \Rightarrow y''=2v'(x)+ xv''(x) \Rightarrow y'''=3v''(x)+ xv'''(x) \\ \Rightarrow x^3(3v'' + xv''' )- 3x^2(2v' + xv'' )+ (6-x^2)(xv +x^2v' )-(6-x^2)xv =0 \\ \Rightarrow x^4v'''-x^4v'=0 \Rightarrow v'''-v'=0 \Rightarrow v(x)= C_1+C_2e^x +C_3e^{-x}\\ \Rightarrow \bbox[red, 2pt]{y = C_1x + C_2xe^x +C_3xe^{-x}}$$ $$\mathbf{(c)}\;先求齊次解，即y''-3y'+2y=0 \Rightarrow \lambda^2-3\lambda+2=0 \Rightarrow \lambda=1,2 \Rightarrow y_h=C_1e^x +C_2e^{2x}\\ 再用參數變換法:令\cases{y_1=e^x\\ y_2=e^{2x}} \Rightarrow W=\begin{vmatrix} y_1& y_2\\ y_1' & y_2'\end{vmatrix} =\begin{vmatrix} e^x & e^{2x}\\ e^x & 2e^{2x}\end{vmatrix} =e^{3x} \\ \Rightarrow y_p =-y_1\int {y_2 \sin(e^{-x}) \over W} \,dx+ y_2 \int{y_1 \sin(e^{-x})\over W}\,dx =-e^x \int {\sin(e^{-x})\over e^x}\,dx + e^{2x}\int {\sin(e^{-x})\over e^{2x}}\,dx \\ =-e^x \cos(e^{-x})+e^{2x}\left( e^{-x}\cos(e^{-x})-\sin(e^{-x})\right) = -e^{2x} \sin(e^{-x})\\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red,2pt]{y=C_1e^x +C_2e^{2x}-e^{2x}\sin(e^{-x})}$$

解答： $$\mathcal L^{-1}\{{s-2\over s^2+2s +10} \}=\mathcal L^{-1}\{ {(s+1)-3\over (s+1)^2+ 9}\} =\mathcal L^{-1}\{ {s+1\over (s+1)^2+3^2}\} -\mathcal L^{-1}\{ {3\over (s+1)^2+3^2}\} \\= \bbox[red, 2pt]{e^{-t}\cos(3t)-e^{-t}\sin(3t)}$$

解答： $$\mathcal F(f(x))= {1\over \sqrt{2\pi}} \int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx\\= {1\over \sqrt{2\pi}} \int_{-\infty}^\infty e^{-x^2/a^2}e^{-i\omega x}\,dx ={1\over \sqrt{2\pi}} \int_{-\infty}^\infty e^{-x^2/a^2}(\cos(\omega x)-i\sin (\omega x))\,dx \\={1\over \sqrt{2\pi}} \int_{-\infty}^\infty e^{-x^2/a^2}\cos(\omega x)\,dx \;(\because f(x)\sin(\omega x)為奇函數，其積分值為0)\\ ={2\over \sqrt{2\pi}} \int_{0}^\infty e^{-x^2/a^2}\cos(\omega x)\,dx =\sqrt{2\over \pi}\int_{0}^\infty e^{-x^2/a^2}\cos(\omega x)\,dx \;(\because f(x)\cos(\omega x)對稱y軸)\\ 取g(\omega)= \int_0^\infty e^{-x^2/a^2}\cos(\omega x)\,dx,其中\omega\ge 0; \Rightarrow g'(\omega)= \int_0^\infty -x e^{-x^2/a^2} \sin(\omega x) \,dx\\ 由\cases{u=\sin(\omega x) \Rightarrow du =\omega \cos (\omega x)\,dx\\ dv=-xe^{-x^2/a^2}\,dx \Rightarrow v= {a^2\over 2}e^{-x^2/a^2}} \\\Rightarrow g'(\omega) = \left.{a^2\over 2}e^{-x^2/a^2}\sin (\omega x) \right|_0^\infty -{\omega a^2\over 2 } \int_{0}^\infty\cos(\omega x)e^{-x^2/a^2}\,dx =0 -{\omega a^2\over 2 }g(\omega) \\ \Rightarrow g'(\omega)=-{\omega a^2\over 2 }g(\omega) \Rightarrow \left(e^{a^2\omega^2/4} g(\omega)\right)'=0 \Rightarrow g(\omega) =C_1e^{-a^2\omega^2/4} \Rightarrow \mathcal F(f(x))= \sqrt{2\over \pi}g(\omega) \\ \Rightarrow \bbox[red, 2pt]{\mathcal F(f(x))=\sqrt{2\over \pi}C_1e^{-a^2\omega^2/4}}$$

解答： $$\cases{A(1,1,1)\\ B(2,2,2)\\ C(3,4,X)} \Rightarrow \cases{\vec u =\overrightarrow{AB}=(1,1,1)\\ \vec v=\overrightarrow{AC} =(2,3,X-1)} \Rightarrow \vec u\times \vec v =(x-4,3-x,1) \\ \Rightarrow \triangle ABC面積={1\over 2} \lVert(x-4,3-x,1) \rVert={1\over 2}\sqrt{2x^2-14x+26} ={1\over 2}\sqrt{2(x-(7/2))^2+3/2} \\ \Rightarrow \triangle ABC面積最小值={1\over 2}\sqrt{3\over 2} =\bbox[red,2pt]{\sqrt 6\over 4}$$

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