105年專科學力鑑定考試--工程數學詳解

105年專科學校畢業程度自學進修學力鑑定考試

專業科目(一)：工程數學 詳解

解：$$L\left\{ \left( f\ast g \right) \left( t \right) \right\} =L\left\{ f\left( t \right) \right\} \cdot L\left\{ g\left( t \right) \right\} \Rightarrow L\left\{ H\left( t \right) \right\} =L\left\{ { e }^{ 3t } \right\} \cdot L\left\{ \cos { t } \right\} \\ =\frac { 1 }{ s-3 } \cdot \frac { s }{ s^{ 2 }+1 } =\frac { s }{ \left( s-3 \right) \left( s^{ 2 }+1 \right) } ，故選：\bbox[red,2pt]{(D)}$$

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解：$$\vec{u}\cdot\vec{v}=\left(3\vec{i}+2\vec{j}- 3\vec{k}\right)\cdot\left(\vec{i}-2\vec{j}+ 3\vec{k}\right)=3\times 1+ 2\times(-2)+(-3)\times3=3-4-9\\=-10 ，故選\bbox[red,2pt]{(A)}$$

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解：
$$\left( A+3B \right) C=\left( \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}+3\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \right) \begin{bmatrix} 5 & 1 \\ 1 & 2 \end{bmatrix}=\left( \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}+\begin{bmatrix} 6 & 3 \\ 3 & 6 \end{bmatrix} \right) \begin{bmatrix} 5 & 1 \\ 1 & 2 \end{bmatrix}\\ =\begin{bmatrix} 7 & 6 \\ 5 & 10 \end{bmatrix}\begin{bmatrix} 5 & 1 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 41 & 19 \\ 35 & 25 \end{bmatrix}， 故選\bbox[red,2pt]{(D)}$$

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解：$$\left| \begin{matrix} 2 & 3 & 5 \\ a & 0 & 2 \\ 3 & 1 & 4 \end{matrix} \right| =7\Rightarrow 0+5a+18-0-12a-4=7\Rightarrow 7a=7\Rightarrow a=1， 故選\bbox[red,2pt]{(B)}$$

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解：$$由y\left( x \right) =\left( A+Bx \right) { e }^{ x }\Rightarrow y'\left( x \right) =\left( A+B \right) { e }^{ x }+Bx{ e }^{ x }\\ 可知\begin{cases} y\left( 0 \right) =1 \\ y'\left( 0 \right) =2 \end{cases}\Rightarrow \begin{cases} A=1 \\ A+B=2 \end{cases}\Rightarrow \begin{cases} A=1 \\ B=1 \end{cases}\Rightarrow A+B=2， 故選\bbox[red,2pt]{(C)}$$

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解：$$\begin{cases} \vec { u } =\vec { i } +2\vec { j } -3\vec { k } =\left< 1,0,0 \right> +2\left< 0,1,0 \right> -3\left< 0,0,1 \right> =\left< 1,2-3 \right>  \\ \vec { v } =2\vec { i } -\vec { j } +\vec { k } =\left< 1,0,0 \right> +2\left< 0,1,0 \right> +\left< 0,0,1 \right> =\left< 2,-1,1 \right>  \\ \vec { w } =-\vec { i } -2\vec { j } +2\vec { k } =-\left< 1,0,0 \right> -2\left< 0,1,0 \right> +2\left< 0,0,1 \right> =\left< -1,-2,2 \right>  \end{cases}\\ \Rightarrow 2\left( \vec { u } +\vec { v }  \right) -\vec { w } =2\left( \left< 1,2-3 \right> +\left< 2,-1,1 \right>  \right) -\left< -1,-2,2 \right> =2\left< 3,1-2 \right> -\left< -1,-2,2 \right> \\ =\left< 6,2-4 \right> -\left< -1,-2,2 \right> =\left< 7,4,-6 \right> =7\vec{i}+4\vec{j}-6\vec{k}， 故選\bbox[red,2pt]{(B)}$$

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解：

可以簡化成\(y'+P(x)y=Q(x)\)就是一階線性微分方程式。

\(x\frac{dy}{dx}+3y=x^2-x\Rightarrow   y'+\frac{3}{x}\cdot   y=x-1\)，故選\(\bbox[red,2pt]{(B)}\)

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解：$$特徵方程式\lambda^2+2\lambda+5=0\Rightarrow \lambda=\frac{-2\pm4i}{2}=-1\pm2i\Rightarrow y=e^{-x}\left(A\cos{2x}+B\sin{2x}\right)，\\故選\bbox[red,2pt]{(A)} $$

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解：$$\begin{cases} 偶函數:f\left( x \right) =-f\left( x \right)  \\ 奇函數:f\left( -x \right) =-f\left( x \right)  \end{cases}\Rightarrow \begin{cases} 偶函數:y_2,y_3 \\ 奇函數:y_1  \end{cases}，故選\bbox[red,2pt]{(D)} $$

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解：$$f\left( t \right) =e^{ 2t }\left( t+1 \right) ^{ 2 }=t^{ 2 }e^{ 2t }+2te^{ 2t }+e^{ 2t }\Rightarrow L\left\{ f\left( t \right)  \right\} =L\left\{ t^{ 2 }e^{ 2t } \right\} +2L\left\{ te^{ 2t } \right\} +L\left\{ e^{ 2t } \right\} \\ =\frac { 2 }{ { \left( s-2 \right)  }^{ 3 } } +\frac { 2 }{ { \left( s-2 \right)  }^{ 2 } } +\frac { 1 }{ s-2 } =\frac { 2+2\left( s-2 \right) +{ \left( s-2 \right)  }^{ 2 } }{ { \left( s-a \right)  }^{ 3 } } =\frac { s^{ 2 }-2s+2 }{ ^{ 2 } } ，故選\bbox[red,2pt]{(B)}$$

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解：$$\lambda ^{ 2 }-6\lambda +9=0\Rightarrow \left( \lambda -3 \right) ^{ 2 }=0\Rightarrow y=\left( A+Bx \right) e^{ 3x }\Rightarrow y'=3Ae^{ 3x }+Be^{ 3x }+3Bxe^{ 3x }\\ \Rightarrow \begin{cases} y\left( 0 \right) =1 \\ y'\left( 0 \right) =2 \end{cases}\Rightarrow \begin{cases} A=1 \\ 3A+B=2 \end{cases}\Rightarrow \begin{cases} A=1 \\ B=-1 \end{cases}\Rightarrow y=\left( 1-x \right) e^{ 3x }\Rightarrow y\left( 1 \right) =0，故選\bbox[red,2pt]{(D)}$$

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解：$$L^{-1}\left\{ \frac { s }{ s^{ 2 }-2s+5 }  \right\} =L^{-1}\left\{ \frac { s }{ \left( s-1 \right) ^{ 2 }+2^{ 2 } }  \right\} =L^{-1}\left\{ \frac { s-1 }{ \left( s-1 \right) ^{ 2 }+2^{ 2 } } +\frac { 1 }{ 2 } \cdot \frac { 2 }{ \left( s-1 \right) ^{ 2 }+2^{ 2 } }  \right\} \\ =L^{-1}\left\{ \frac { s-1 }{ \left( s-1 \right) ^{ 2 }+2^{ 2 } }  \right\} +\frac { 1 }{ 2 } L^{-1}\left\{ \frac { 2 }{ \left( s-1 \right) ^{ 2 }+2^{ 2 } }  \right\} =e^t\cos{\left(2t\right)}+\frac{1}{2}e^t\sin{\left(2t\right)}，故選\bbox[red,2pt]{(C)}$$

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解：$$\left( A \right) \frac { dy }{ dx } =\frac { y^{ 2 }+x }{ 2xy+y } \Rightarrow \begin{cases} \frac { d }{ dx } \left( 2xy+y \right) =2y \\ -\frac { d }{ dy } \left( y^{ 2 }+x \right) =-2y \end{cases}兩者不相等\\ \left( B \right) \frac { dy }{ dx } =\frac { 1-e^{ x }y }{ e^{ x }+xy } \Rightarrow \begin{cases} \frac { d }{ dx } \left( e^{ x }+xy \right) =e^{ x }+y \\ -\frac { d }{ dy } \left( 1-e^{ x }y \right) =e^{ x } \end{cases}兩者不相等\\ \left( C \right) \frac { dy }{ dx } =\frac { \sin { y }  }{ y^{ 2 }-x\cos { y }  } \Rightarrow \begin{cases} \frac { d }{ dx } \left( y^{ 2 }-x\cos { y }  \right) =-\cos { y }  \\ -\frac { d }{ dy } \left( \sin { y }  \right) =-\cos { y }  \end{cases}兩者相等\\ \left( D \right) \frac { dy }{ dx } =\frac { -y\sin { x }  }{ x\cos { y }  } \Rightarrow \begin{cases} \frac { d }{ dx } \left( x\cos { y }  \right) =\cos { y }  \\ -\frac { d }{ dy } \left( -y\sin { x }  \right) =\sin { x }  \end{cases}兩者不相等\\故選\bbox[red,2pt]{(C)}$$

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解：$$L^{ -1 }\left\{ \frac { s+2 }{ s^{ 2 }+3s }  \right\} =L^{ -1 }\left\{ \frac { s }{ s^{ 2 }+3s } +\frac { 2 }{ s^{ 2 }+3s }  \right\} =L^{ -1 }\left\{ \frac { 1 }{ s+3 } +\frac { 2 }{ 3 } \left( \frac { 1 }{ s } -\frac { 1 }{ s+3 }  \right)  \right\} \\ =L^{ -1 }\left\{ \frac { 1 }{ 3 } \cdot \frac { 1 }{ s+3 } +\frac { 2 }{ 3 } \cdot \frac { 1 }{ s }  \right\} =\frac { 1 }{ 3 } L^{ -1 }\left\{ \frac { 1 }{ s+3 }  \right\} +\frac { 2 }{ 3 } L^{ -1 }\left\{ \frac { 1 }{ s }  \right\} =\frac { 1 }{ 3 } e^{ -3t }+\frac { 2 }{ 3 } ，故選\bbox[red,2pt]{(B)}$$

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解：$$\begin{cases} 2x-y+z=3\cdots \left( 1 \right)  \\ 4x-y-z=5\cdots \left( 2 \right)  \\ x+y+2z=12\cdots \left( 3 \right)  \end{cases}\Rightarrow (2)-(1),(2)+(3)\Rightarrow \begin{cases} 2x-2z=2\cdots \left( 4 \right)  \\ 5x+z=17\cdots \left( 5 \right)  \end{cases}\Rightarrow (5)\times 2+(4)\Rightarrow 12x=36\Rightarrow x=3，故選\bbox[red,2pt]{(C)}$$

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解：$$\vec { u } \times \vec { v } =\left( 2,1,3 \right) \times \left( 2,-3,5 \right) =\left( \begin{vmatrix} 1 & 3 \\ -3 & 5 \end{vmatrix},\begin{vmatrix} 3 & 2 \\ 5 & 2 \end{vmatrix},\begin{vmatrix} 2 & 1 \\ 2 & -3 \end{vmatrix} \right) =\left( 14,-4,-8 \right) ，故選\bbox[red,2pt]{(A)}$$

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解：$$\frac { dy }{ dx } =-4x^3y^2\Rightarrow \frac{1}{y^2}dy=-4x^3dx\Rightarrow -\frac{1}{y}=-x^4+C\\\Rightarrow y(1)=2\Rightarrow -\frac{1}{2}=-1+C\Rightarrow C=\frac{1}{2}\Rightarrow y(0)=-\frac{1}{C}=-2，故選\bbox[red,2pt]{(A)}$$

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解：$$特徵方程式\lambda ^{ 2 }+2\lambda -3=0\Rightarrow (\lambda +3)(\lambda -1)=0\Rightarrow \lambda =-3,1\Rightarrow y_{ h }=Ae^{ -3x }+Be^{ x }\\ 令y_{ p }=cx+d\Rightarrow y''_{ p }+2y'_{ p }-3y_{ p }=2c-3cx-3d=-3cx+2c-3d=3x\Rightarrow c=-1,d=-\frac { 2 }{ 3 } \\ 通解y=y_{ h }+y_{ p }=Ae^{ -3x }+Be^{ x }-x-\frac { 2 }{ 3 } ，故選\bbox[red,2pt]{(D)}$$

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解：$$det\left( A-\lambda I \right) =0\Rightarrow \begin{vmatrix} 3-\lambda  & 2 \\ 5 & 6-\lambda  \end{vmatrix}=0\Rightarrow \lambda^2-9\lambda+8=0\Rightarrow (\lambda-8)(\lambda-1)=0\Rightarrow \lambda=8,1\\故選\bbox[red,2pt]{(C)}$$

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解：$$a_{ 0 }=\frac { 1 }{ 2\pi } \int _{ -\pi }^{ \pi }{ f\left( x \right) dx } =\frac { 1 }{ 2\pi } \left[ \int _{ -\pi }^{ 0 }{ f\left( x \right) dx } +\int _{ 0 }^{ \pi }{ f\left( x \right) dx } \right] =\frac { 1 }{ 2\pi } \left[ \int _{ -\pi }^{ 0 }{ \left( -1 \right) dx } +\int _{ 0 }^{ \pi }{ 1dx } \right] =0\\ a_{ 1 }=\frac { 1 }{ \pi } \left[ \int _{ -\pi }^{ 0 }{ \left( -\cos { x } \right) dx } +\int _{ 0 }^{ \pi }{ \cos { x } dx } \right] =\frac { 1 }{ \pi } \left. \left[ -\sin { x } \right] \right| _{ -\pi }^{ 0 }+\frac { 1 }{ \pi } \left. \left[ \sin { x } \right] \right| _{ 0 }^{ \pi }=0\\ b_{ 1 }=\frac { 1 }{ \pi } \left[ \int _{ -\pi }^{ 0 }{ \left( -\sin { x } \right) dx } +\int _{ 0 }^{ \pi }{ \sin { x } dx } \right] =\frac { 1 }{ \pi } \left. \left[ \cos { x } \right] \right| _{ -\pi }^{ 0 }+\frac { 1 }{ \pi } \left. \left[ -\cos { x } \right] \right| _{ 0 }^{ \pi }=\frac { 4 }{ \pi }\\ \Rightarrow  a_0+a_1+b_1=0+0+\frac{4}{\pi}=\frac{4}{\pi} ，故選\bbox[red,2pt]{(A)}$$

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