國立臺北科技大學110學年度碩士班招生考試
系所組別:能源與冷凍空調工程
科目:工程數學
解答:$$y'={1\over x}y^2+ {1\over x}y-{2\over x} \Rightarrow xy'=y^2+y-2 \Rightarrow {1 \over y^2+y-2}dy={1\over x}\,dx \\ \Rightarrow {1\over 3}\int \left({1\over y-1}-{1\over y+2} \right) \,dy =\int {1\over x}\,dx \Rightarrow {1\over 3}\left(\ln(y-1)-\ln(y+2) \right)=\ln x+c_1 \\ \Rightarrow {1\over 3}\ln{y-1\over y+2} =\ln (c_2x) \Rightarrow {y-1\over y+2}=c_3x^3 \Rightarrow 1-c_3x^3={3\over y+2} \Rightarrow \bbox[red, 2pt]{y={3\over 1-c_3x^3}-2}$$
解答:$$y''+y'+3y=0 \Rightarrow \lambda^2+\lambda+3=0 \Rightarrow \lambda ={-1\pm \sqrt{11}i \over 2} \\\Rightarrow y_h=e^{-x/2}( c_1\cos({\sqrt{11}\over 2}x) +c_2 \sin{\sqrt{11}\over 2}x) \\ \text{Applying undetermined coefficients, }y_p=A\cos(2x)+ B\sin(2x) \\\Rightarrow y_p'=-2A\sin(2x)+ 2B\cos(2x) \Rightarrow y_p''=-4A\cos(2x)-4B\sin(2x)\\ \Rightarrow y_p''+y_p'+3y_p= (-A+2B)\cos(2x)+(-2A-B)\sin(2x)=5\sin(2x) \\ \Rightarrow \cases{-A+2B=0 \\ -2A-B=5} \Rightarrow \cases{A=-2\\B=-1} \Rightarrow y_p= -2\cos(2x)-\sin(2x) \\ y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=e^{-x/2}( c_1\cos({\sqrt{11}\over 2}x) +c_2 \sin{\sqrt{11}\over 2}x)-2\cos(2x)-\sin(2x)}$$
解答:$$L\{f(t)\} =L\{2t^2\} +L\left\{\int_0^t f(t-\tau)e^{-\tau}\,d \tau \right\} \Rightarrow F(s)={4\over s^3} +L\{f(t)\} L\{e^{-t}\}={4\over s^3}+F(s)\cdot {1\over s+1} \\ \Rightarrow {s\over s+1}F(s)={4\over s^3} \Rightarrow F(s)={4(s+1)\over s^4}={4\over s^3}+{4\over s^4} \Rightarrow f(t)=L^{-1}\{{4\over s^3}+{4\over s^4}\} \\ \Rightarrow \bbox[red, 2pt]{f(t)=2t^2+{2\over 3}t^3}$$
解答:$$L\{y''+2y'+2y \} =L\{\delta(t-3)\} \Rightarrow s^2Y(s)+2sY(s)+2Y(s)= e^{-3s} \\ \Rightarrow Y(s)={1\over s^2+2s+2}e^{-3s} \Rightarrow y(t)=L^{-1}\{ Y(s)\} =L^{-1}\left\{ {1\over s^2+2s+2}e^{-3s} \right\} \\= L^{-1}\left\{ {1\over (s+1)^2 +1}e^{-3s} \right\} \Rightarrow \bbox[red, 2pt]{y(t) =u(t-3)e^{-(t-3)} \sin(t-3)}$$
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