101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:$$y=x^{ m }\Rightarrow y+xy'=x^{ m }+mx^{ m }=x^{ m }(1+m)=0\Rightarrow m=-1\\ \Rightarrow y+xy'=0的通解為y=Cx^{-1}\Rightarrow xy=C,故選:\bbox[red,2pt]{(A)}$$
解:
一階線性微分方程的標準式為\(y'+P(x)y=Q(x)\),故選\(\bbox[red,2pt]{(B)}\)
解:$$y^{ 2 }-6xy+\left( 3xy-6x^{ 2 } \right) y'=0\Rightarrow \frac { 1 }{ xy } \left( y^{ 2 }-6xy+\left( 3xy-6x^{ 2 } \right) y' \right) =0\\ \Rightarrow \frac { y }{ x } -6+\left( 3-6\frac { x }{ y } \right) y'=0\cdots (1)\\ 令z=\frac { y }{ x } \Rightarrow y=xz\Rightarrow y'=z+xz'代入式(1)\Rightarrow z-6+\left( 3-\frac { 6 }{ z } \right) \left( z+xz' \right) =0\\ \Rightarrow \frac { z^{ 2 }-6z }{ 3z-6 } +z+xz'=0\Rightarrow \frac { 4z^{ 2 }-12z }{ 3z-6 } +xz'=0\Rightarrow xz'=-\frac { 4z^{ 2 }-12z }{ 3z-6 } \\ \Rightarrow -\frac { 3z-6 }{ 4z^{ 2 }-12z } dz=\frac { 1 }{ x } dx\Rightarrow -\left( \frac { 2 }{ 4z } +\frac { 1 }{ 4 } \cdot \frac { 1 }{ z-3 } \right) dz=\frac { 1 }{ x } dx\\ \Rightarrow -\left( \frac { 1 }{ 2 } \ln { \left| z \right| } +\frac { 1 }{ 4 } \ln { \left| z-3 \right| } \right) =\ln { \left| x \right| +C } \Rightarrow -\ln { \left( \sqrt { z } \cdot \sqrt [ 4 ]{ z-3 } \right) } =\ln { \left| x \right| +C } \\ \Rightarrow x\cdot \sqrt { z } \cdot \sqrt [ 4 ]{ z-3 } =C\Rightarrow x\cdot \sqrt { \frac { y }{ x } } \cdot \sqrt [ 4 ]{ \frac { y }{ x } -3 } =C\Rightarrow x^{ 4 }\cdot { \left( \frac { y }{ x } \right) }^{ 2 }\cdot \left( \frac { y }{ x } -3 \right) =C\\ \Rightarrow x^{ 2 }y^{ 2 }\left( \frac { y }{ x } -3 \right) =C\Rightarrow xy^{ 3 }-3x^{ 2 }y^{ 2 }=C, 故選\bbox[red,2pt]{(C)}$$
解:$$y'''-2y''-y'+2y=0\Rightarrow \lambda ^{ 3 }-2\lambda ^{ 2 }-\lambda +2=0\Rightarrow \Rightarrow \left( \lambda +1 \right) \left( \lambda -1 \right) \left( \lambda -2 \right) =0\\ \Rightarrow \lambda =-1,1,2\Rightarrow y=C_1e^{-x}+C_2e^x+C_3e^{2x}, 故選\bbox[red,2pt]{(D)}$$
解:$$y=x^{ m }\Rightarrow y'=mx^{ m-1 }\Rightarrow y''=m(m-1)x^{ m-2 }\\ \Rightarrow x^{ 2 }y''-3xy'+4y=0\Rightarrow m(m-1)x^{ m }-3mx^{ m }+4x^{ m }=x^{ m }\left( m^{ 2 }-4m+4 \right) =0\\ \Rightarrow x^{ m }\left( m-2 \right) ^{ 2 }=0\Rightarrow m=2為二重根\Rightarrow y=(C_{ 1 }+C_{ 2 }\ln { x } )x^{ 2 }=C_{ 1 }x^{ 2 }+C_{ 2 }x^{ 2 }\ln { x } , 故選\bbox[red,2pt]{(C)}$$
解:$$y''-6y'+5y=25x\Rightarrow \lambda ^{ 2 }-6\lambda +5=0\Rightarrow (\lambda -5)(\lambda -1)=0\Rightarrow \lambda =5,1\\ \Rightarrow y_{ h }=C_{ 1 }e^{ x }+C_{ 2 }e^{ 5 x}\\ y_{ p }=ax+b\Rightarrow y'_{ p }=a\Rightarrow y''_{ p }=0\Rightarrow y''-6y'+5y=0-6a+5(ax+b)\\ =5ax+5b-6a=25x\Rightarrow \begin{cases} 5a=25 \\ 5b-6a=0 \end{cases}\Rightarrow \begin{cases} a=5 \\ b=6 \end{cases}\Rightarrow y_{ p }=5x+6\\ \Rightarrow y=y_{ h }+y_{ p }=C_{ 1 }e^{ x }+C_{ 2 }e^{ 5x }+5x+6,故選\bbox[red,2pt]{(A)}$$
解:$$L\left\{ \cos { 3t } \right\} =\frac { s }{ s^{ 2 }+3^{ 2 } } \Rightarrow L\left\{ e^{ 2t }\cos { 3t } \right\} =\frac { s-2 }{ (s-2)^{ 2 }+3^{ 2 } } , 故選\bbox[red,2pt]{(D)}$$
解:$$f\left( t+4 \right) =f\left( t \right) \Rightarrow L\left\{ f\left( t \right) \right\} =\frac { \int _{ 0 }^{ 4 }{ e^{ -st }f\left( t \right) dt } }{ 1-e^{ -4s } } =\frac { 1 }{ 1-e^{ -4s } } \left( \int _{ 0 }^{ 2 }{ 3e^{ -st }dt } -\int _{ 2 }^{ 4 }{ 3e^{ -st }dt } \right) \\ =\frac { 3 }{ 1-e^{ -4s } } \left( \int _{ 0 }^{ 2 }{ e^{ -st }dt } -\int _{ 2 }^{ 4 }{ e^{ -st }dt } \right) =\frac { 3 }{ 1-e^{ -4s } } \left( \left. \left[ \frac { -1 }{ s } e^{ -st } \right] \right| _{ 0 }^{ 2 }-\left. \left[ \frac { -1 }{ s } e^{ -st } \right] \right| _{ 2 }^{ 4 } \right) \\ =\frac { 3 }{ 1-e^{ -4s } } \left( \frac { -1 }{ s } \left( e^{ -2s }-1 \right) +\frac { 1 }{ s } \left( e^{ -4s }-e^{ -2s } \right) \right) =\frac { 3 }{ s\left( 1-e^{ -4s } \right) } \left( 1-2e^{ -2s }+e^{ -4s } \right) \\ =\frac { 3\left( 1-e^{ -2s } \right) ^{ 2 } }{ s\left( 1-e^{ -2s } \right) \left( 1+e^{ -2s } \right) } =\frac { 3\left( 1-e^{ -2s } \right) }{ s\left( 1+e^{ -2s } \right) } ,故選\bbox[red,2pt]{(B)} $$
解:$$L^{ -1 }\left\{ \frac { 1 }{ \left( s^{ 2 }+1 \right) ^{ 2 } } \right\} =L^{ -1 }\left\{ \frac { 1 }{ 2 } \left( \frac { 1 }{ s^{ 2 }+1 } -\frac { s^{ 2 }-1 }{ \left( s^{ 2 }+1 \right) ^{ 2 } } \right) \right\} =\frac { 1 }{ 2 } \left( L^{ -1 }\left\{ \frac { 1 }{ s^{ 2 }+1 } \right\} -L^{ -1 }\left\{ \frac { s^{ 2 }-1 }{ \left( s^{ 2 }+1 \right) ^{ 2 } } \right\} \right) \\ =\frac { 1 }{ 2 } \left( \sin { t } -t\cos { t } \right) ,故選\bbox[red,2pt]{(D)} $$
解:$$L^{ -1 }\left\{ \frac { 8 }{ s^{ 2 }\left( s^{ 2 }+4 \right) } \right\} =L^{ -1 }\left\{ \frac { 2 }{ s^{ 2 } } -\frac { 2 }{ s^{ 2 }+4 } \right\} =2L^{ -1 }\left\{ \frac { 1 }{ s^{ 2 } } \right\} -L^{ -1 }\left\{ \frac { 2 }{ s^{ 2 }+4 } \right\} \\ =2t-\sin { 2t } ,故選\bbox[red,2pt]{(A)}$$
解:$$L^{ -1 }\left\{ \frac { 2s+1 }{ s^{ 2 }+9 } \right\} =L^{ -1 }\left\{ \frac { 2s }{ s^{ 2 }+9 } +\frac { 1 }{ s^{ 2 }+9 } \right\} =L^{ -1 }\left\{ 2\cdot \frac { s }{ s^{ 2 }+9 } +\frac { 1 }{ 3 } \cdot \frac { 3 }{ s^{ 2 }+9 } \right\} \\ =2L^{ -1 }\left\{ \frac { s }{ s^{ 2 }+9 } \right\} +\frac { 1 }{ 3 } L^{ -1 }\left\{ \frac { 3 }{ s^{ 2 }+9 } \right\} =2\cos { 3t } +\frac { 1 }{ 3 } \sin { 3t } ,故選\bbox[red,2pt]{(B)}$$
解:$$a_{ 5 }=\frac { 1 }{ 2 } \int _{ -2 }^{ 2 }{ f\left( x \right) \cos { \frac { 5\pi x }{ 2 } dx } } =\frac { 1 }{ 2 } \left( \int _{ 0 }^{ 1 }{ x\cos { \frac { 5\pi x }{ 2 } dx } } +\int _{ 1 }^{ 2 }{ \cos { \frac { 5\pi x }{ 2 } dx } } \right) \\ =\frac { 1 }{ 2 } \left( \left. \left[ \frac { 2x }{ 5\pi } \sin { \frac { 5\pi x }{ 2 } } +\frac { 4 }{ 25\pi ^{ 2 } } \cos { \frac { 5\pi x }{ 2 } } \right] \right| _{ 0 }^{ 1 }+\left. \left[ \frac { 2 }{ 5\pi } \sin { \frac { 5\pi x }{ 2 } } \right] \right| _{ 1 }^{ 2 } \right) \\ =\frac { 1 }{ 2 } \left( \left( \frac { 2 }{ 5\pi } -\frac { 4 }{ 25\pi ^{ 2 } } \right) +\left( 0-\frac { 2 }{ 5\pi } \right) \right) =\frac { 1 }{ 2 } \left( \left( \frac { 2 }{ 5\pi } -\frac { 4 }{ 25\pi ^{ 2 } } \right) +\left( 0-\frac { 2 }{ 5\pi } \right) \right) \\ =\frac { 1 }{ 2 } \times \frac { -4 }{ 25\pi ^{ 2 } } =\frac { -2 }{ 25\pi ^{ 2 } } ,故選\bbox[red,2pt]{(C)}$$
解:$$\mho _{ 3 }=3\mho _{ 1 }+4\mho _{ 2 }\Rightarrow \left( a,b,c \right) =3\left( 3,2,4 \right) +4\left( 5,4-9 \right) =\left( 9,6,12 \right) +\left( 20,16,-36 \right) \\ =\left( 29,22,-24 \right) \Rightarrow b=22,故選\bbox[red,2pt]{(以上皆非)}$$
解:$$\left[ \begin{matrix} 1 & 2 & -1 & 1 \\ 2 & 4 & -3 & 0 \\ 1 & 2 & 1 & 5 \end{matrix} \right] \xrightarrow { -r_{ 1 }+r_{ 3 },-2r_{ 1 }+r_{ 2 } } \left[ \begin{matrix} 1 & 2 & -1 & 1 \\ 0 & 0 & -1 & -2 \\ 0 & 0 & 2 & 4 \end{matrix} \right] \xrightarrow { 2r_{ 2 }+r_{ 3 } } \left[ \begin{matrix} 1 & 2 & -1 & 1 \\ 0 & 0 & -1 & -2 \\ 0 & 0 & 0 & 0 \end{matrix} \right] ,故選\bbox[red,2pt]{(B)}$$
解:$$\left[\begin{array}{ccc|ccc}2&1&2&1&0&0\\3&2&2&0&1&0\\1&2&3&0&0&1\end{array}\right] \xrightarrow{(-3/2)r_1+r_2,(-1/2)r_1+r_3} \left[\begin{array}{ccc|ccc}2&1&2&1&0&0\\0&1/2&-1&-3/2&1&0\\0&3/2&2&-1/2&0&1\end{array}\right] \xrightarrow{-3r_2+r_3}\\ \left[\begin{array}{ccc|ccc}2&1&2&1&0&0\\0&1/2&-1&-3/2&1&0\\0&0&5&4&-3&1\end{array}\right] \xrightarrow{r_1/2,2r_2,r_3/5} \left[\begin{array}{ccc|ccc}1&1/2&1&1/2&0&0\\ 0&1&-2&-3&2&0\\ 0&0&1&4/5&-3/5&1/5 \end{array}\right]\\ \xrightarrow{(-1/2)r_2+r_1} \left[\begin{array}{ccc|ccc}1&0&2&2&-1&0\\ 0&1&-2&-3&2&0\\ 0&0&1&4/5&-3/5&1/5 \end{array}\right] \xrightarrow{(-2)r_3+r_1,2r_3+r_2}\left[\begin{array}{ccc|ccc}1&0&0&2/5&1/5&-2/5\\ 0&1&0&-7/5&4/5&2/5\\ 0&0&1&4/5&-3/5&1/5 \end{array}\right]\\ \Rightarrow A^{-1}= \left[\begin{matrix}2/5&1/5&-2/5\\-7/5&4/5&2/5\\4/5&-3/5&1/5\end{matrix}\right]=\frac{1}{5}\left[\begin{matrix}2&1&-2\\-7&4&2\\4&-3&1\end{matrix}\right]
,故選\bbox[red,2pt]{(C)}$$
解:$$c_{ 12 }=\left( 1,2,-1 \right) \cdot \left( -1,7,2 \right) =-1+14-2=11 ,故選\bbox[red,2pt]{(A)}$$
解:$$det\left( A-\lambda I \right) =0\Rightarrow \left| \begin{matrix} -\lambda & 0 & -2 \\ 1 & 2-\lambda & 1 \\ 1 & 0 & 3-\lambda \end{matrix} \right| =0\Rightarrow -\lambda \left( \lambda -2 \right) \left( \lambda -3 \right) +2\left( 2-\lambda \right) =0\\ \Rightarrow -\lambda ^{ 3 }+5\lambda ^{ 2 }-6\lambda +4-2\lambda =0\Rightarrow \lambda ^{ 3 }-5\lambda ^{ 2 }+8\lambda -4=0\Rightarrow \left( \lambda -1 \right) \left( \lambda -2 \right) ^2=0\\\Rightarrow\lambda=1,2,故選\bbox[red,2pt]{(D)}$$
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