110年北科大冷凍空調碩士班-工程數學詳解

 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： $$\text{By method of Frobenius,} x^2y''+5xy'+(x+4)y'=0 \equiv x^2 y''+xp(x) y'+q(x)y=0, \\\text{where }p(x)=5, q(x)=4+x. \text{ indicial equation: }r(r-1)+p_0r +q_0=0 \\ \Rightarrow r(r-1)+5r+4 =r^2-4r+4=0 \Rightarrow r_1=r_2=2 \\ \Rightarrow y_1=\sum_{n=1}^\infty a_nx^{n-2} \text{ and }y_2=y_1\ln(x)+\sum_{n=0}^\infty b_nx^{n-2}\\ x^2y_1''+ 5xy_1+ (x+4)y_1=0 \\ \Rightarrow (a_0+a_1)x^{-1} +(a_1+4a_2)+ (a_2+9a_3)x+\cdots +(a_{n-1}+n^2a_n)x^{n-2}+ \cdots =0 \\ \Rightarrow a_n=-{1\over n^2}a_{n-1} ={1\over n^2(n-1)^2}a_{n-2} =\cdots = (-1)^n {1\over n^2(n-1)^2 \cdots 1^2}a_0 \\ \Rightarrow \bbox[red, 2pt]{y_1=a_0 \sum_{n=0}^\infty {(-1)^n\over (n!)^2}x^{n-2}} \Rightarrow y_2=a_0 \sum_{n=0}^\infty {(-1)^n\over (n!)^2}x^{n-2}\ln x+ \sum_{n=0}^\infty b_nx^{n-2} \\x^2y_2''+ 5xy_2+ (x+4)y_2=0  \Rightarrow \cdots \Rightarrow \bbox[red, 2pt]{y_2=y_1\ln x-b_0x^3(3+{13\over 4}x+ {31\over 18}x^2+ {173\over 288}x^3+\cdots )} \\ \Rightarrow \bbox[red, 2pt]{y=y_1+y_2}$$

解答： $$g(t)=\begin{cases}0 & t\lt 2\\t^2+1 & x \ge 2\end{cases} \Rightarrow g(t)=u(t-2)(t^2+1) \\ \Rightarrow L\{g(t)\} =L\{u(t-2)(t^2+1)\} =e^{-2s} L\{(t+2)^2+1\} = \bbox[red, 2pt]{e^{-2s} \left( {2\over s^3} +{4\over s^2} +{5\over s}\right)}$$

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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