111年北科大能源與冷凍碩士班-工程數學詳解

國立臺北科技大學111學年度碩士班招生考試

系所組別 :1410、 1420 能源與冷凍空調工程系碩士班 甲 、乙組

科目:工程數學

解答:$$\text{integration factor }I(x)=e^{\int \tan x\,dx} = \sec x \Rightarrow  y'\sec x+ y \tan x \sec x=\sec x \sin (2x) \\ \Rightarrow (y \sec x)'=\sec x\sin(2x) \Rightarrow y\sec x = \int \sec x\sin(2x)\,dx = -2\cos x+c_1 \\ \Rightarrow y=-2\cos^2 x+c_1\cos x \Rightarrow y(0)=-2+c_1=1 \Rightarrow c_1=3 \Rightarrow \bbox[red, 2pt]{y=-2\cos^2 x+ 3\cos x }$$

解答:$$y''+4y=0 \Rightarrow y_h= c_1\cos(2x) +c_2 \sin(2x) \\ y_p=ax^2+bx+c \Rightarrow y_p'=2ax+b \Rightarrow y_p''= 2a \Rightarrow y_p''+ 4y_p=4ax^2+4bx+4c+2a= 8x^2 \\ \Rightarrow \cases{4a=8\\ 4b=0\\ 4c+2a=0} \Rightarrow \cases{a=2\\ b=0\\ c=-1} \Rightarrow y_p=2x^2-1 \Rightarrow y=y_p+y_h \\ \Rightarrow \bbox[red, 2pt]{y=c_1\cos(2x) +c_2 \sin(2x) +2x^2-1}$$

解答:$$y=\sum_{n=0}^\infty a_n x^n \Rightarrow y'=\sum_{n=0}^\infty na_n x^{n-1} \Rightarrow \cases{y''= \sum_{n=0}^\infty n(n-1)a_n x^{n-2} \\xy'= \sum_{n=0}^\infty na_n x^{n} }\\ \Rightarrow \cases{xy'-y= \sum_{n=0}^\infty (n-1)a_n x^n \\ x^2y''= \sum_{n=0}^\infty n(n-1)a_n x^n} \Rightarrow (1-x^2)y'' =\sum_{n=0}^\infty((n+2)(n+1)a_{n+2}-n(n-1)a_n)x^n \\ \Rightarrow (1-x^2)y''+(xy'-y) =\sum_{n=0}^\infty((n+2)(n+1)a_{n+2}-(n-1)^2a_n )x^n=0\\ \Rightarrow (n+2)(n+1)a_{n+2}-(n-1)^2a_n=0, n\ge 0 \Rightarrow a_{n+2} ={(n-1)^2 \over (n+2)(n+1)} a_n\\ \Rightarrow a_n={((n-3)(n-5)\cdots 1)^2\over n!}a_0,n\in \mathbb N \Rightarrow \cases{a_2={1\over 2}a_0 \\ a_3= a_5=a_7=\cdots =0\\ a_4={1\over 24}a_0\\  a_6= {1\over 80}a_0}\\ \Rightarrow \bbox[red, 2pt]{y=a_0+ a_1 x +{1\over 2}a_0x^2+ {1\over 24} a_0x^4 + {1\over 80}a_0x^6+ \cdots} $$ 這影片可以參考

解答:$$L\{f(t)\} =L\{t\sin(\omega t)\} =-{d\over ds} L\{\sin(\omega t)\}=-{d\over ds} \left({ \omega\over s^2+ \omega^2} \right) =\bbox[red, 2pt]{{2\omega s\over (s^2+\omega)^2}}$$

解答:$$L\{y''\}-L\{y\}=L\{t\} \Rightarrow s^2Y(s)-sy(0)-y'(0)-Y(s)={1\over s^2} \Rightarrow (s^2-1)Y(s)={1\over s^2} +s+1 \\ \Rightarrow Y(s) ={1\over s^2(s^2-1)}+{1 \over s-1} ={1\over s^2-1}-{1\over s^2}+{1 \over s-1}= {1\over 2}({1\over s-1}-{1\over s+1})-{1\over s^2}+{1 \over s-1} \\={3\over 2}\cdot {1\over s-1} -{1\over 2}\cdot {1\over s+1}-{1\over s^2} \Rightarrow y(t)=L^{-1}\{ Y(s)\} \\ \Rightarrow \bbox[red, 2pt]{y(t)={3\over 2}e^t-{1\over 2}e^{-t}-t}$$

解答:$$L\{y''\}+3L\{y'\}+ 2L\{y\}=L\{\delta(t-1)\} \Rightarrow  s^2Y(s)+ 3sY(s)+2Y(s)= e^{-s} \\ \Rightarrow Y(s)= {1\over s^2+3s+2}e^{-s} =({1\over s+1}-{1\over s+2})e^{-s} \\ \Rightarrow y(t)=L^{-1} \{ Y(s)\} =(e^{-(t-1)}-e^{-2(t-1)})u(t-1) \Rightarrow \bbox[red, 2pt]{y(t)=u(t-1)\left( e^{1-t}-e^{2-2t}\right)}$$
 

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解題僅供參考,其他歷年試題及詳解