115年中山大學材光碩士班-工程數學詳解

國立中山大學 115學年度碩士班考試入學招生考試試題

科目名稱:工程數學【材光系碩士班選考、材料前瞻應材碩士班選考、材光聯合碩士班選考】

解答:$$y'+(\tan x)y= \sin 2x \Rightarrow {1\over \cos x} y'+{\sin x\over \cos^2 x}y ={\sin 2x\over \cos x} \Rightarrow \left( {1\over \cos x}y \right)'=2\sin x \\ \Rightarrow {1\over \cos x}y= \int 2\sin x\,dx =-2\cos x+c_1 \Rightarrow y=-2\cos^2 x+c_1\cos x \Rightarrow y(0)=-2+c_1=2 \\ \Rightarrow c_1=4 \Rightarrow \bbox[red, 2pt]{y=-2\cos^2 x+4\cos x}$$

解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow y'''=m(m-1)(m-2) x^{m-3} \\ \Rightarrow x^3y'''-x^2y''-2xy'+6y=[m(m-1)(m-2)-m(m-1)-2m+6]x^m \\=(m-3)(m-2)(m+1)x^m=0 \Rightarrow m=-1,2,3 \Rightarrow y_h= c_1x^{-1}+c_2x^2 +c_3 x^3\\ y_p=Ax^{-2} \Rightarrow y_p'=-2Ax^{-3} \Rightarrow y_p''=6Ax^{-4} \Rightarrow y_p'''=-24Ax^{-5} \\ \Rightarrow x^3y_p'''-x^2y_p''-2xy_p'+6y_p =-24Ax^{-2}-6Ax^{-2}+4Ax^{-2}+6Ax^{-2}=-20Ax^{-2}=7x^{-2} \\ \Rightarrow -20A=7 \Rightarrow A=-{7\over 20} \Rightarrow y_p=-{7\over 20}x^{-2} \Rightarrow y =y_h+ y_p \\ \Rightarrow \bbox[red, 2pt]{y= c_1x^{-1}+c_2x^2 +c_3 x^3-{7\over 20}x^{-2}}$$

解答:$$\cases{M(x,y)= e^{x+y}+ye^y \\N(x,y)=xe^y-1} \Rightarrow M_y\ne N_x \Rightarrow \text{NOT exact}\\ {N_x-M_y\over M}=-1 \Rightarrow \text{ integrating factor }\mu(y)=e^{-y} \Rightarrow \cases{\mu M =e^x+y\\ \mu N=x-e^{-y}} \Rightarrow (\mu M)_y=1=(\mu N)_x \\ \Rightarrow \Phi(x,y) =\int (e^x+y)\,d x = \int (x-e^{-y})\,dy \Rightarrow e^x +xy+ \phi(y)= xy+e^{-y} + \rho(x) \\ \Rightarrow \Phi(x,y)=xy+e^x+e^{-y}=C \Rightarrow \Phi(0,1) =0+1+e^{-1}=C \Rightarrow \bbox[red, 2pt]{xy+e^x+e^{-y}=1+{1\over e}}$$

解答:$$$$

解答:$$y''-5y'+4y=0 \Rightarrow r^2-5r+4=0 \Rightarrow (r-4)(r-1)=0 \Rightarrow r=1,4 \Rightarrow y_h=c_1e^x +c_2e^{4x} \\ y_p=Axe^x \Rightarrow y_p'=Ae^x +Axe^x \Rightarrow y_p''=2Ae^x+ Axe^x \Rightarrow y_p''-5y_p'+4y_p =-3Ae^x =8e^x \\ \Rightarrow A=-{8\over 3} \Rightarrow y_p=-{8\over 3}xe^x \Rightarrow y=y_h+y_p  \Rightarrow \bbox[red, 2pt]{y =c_1e^x +c_2e^{4x}-{8\over 3}xe^x}$$

解答:$$\text{(a) }L\{e^{3t} \sinh t\} =L\{\sinh t\} (s-3)= \bbox[red, 2pt]{{1\over (s-3)^2-1}} \\\textbf{(b) }L\{(a-bt)^2\} =L\{a^2-2abt +b^2t^2\} =a^2L\{1\} -2ab L\{t\}+b^2 L\{t^2\} =\bbox[red, 2pt]{{a^2\over s}-{2ab\over s^2}+ {2b^2\over s^3}} \\ \textbf{(c) }L^{-1} \left\{ {-s+11\over s^2-2s-3}\right\} = L^{-1}\left\{ -{3\over  s+1} +{2\over s-3}\right\}  =\bbox[red, 2pt]{-3e^{-t}+2e^{3t}} \\ \textbf{(d) }L^{-1}\left\{ {e^{-2s}\over (s-1)^3}\right\} =u(t-2)L^{-1} \left\{ {1\over (s-1)^3}\right\}(t-2) = \bbox[red, 2pt]{u(t-2) \cdot {1\over 2}e^{t-2}(t-2)^2}$$

解答:$$\cases{y_1'=y_2+e^{2t} \Rightarrow y_2=y_1'-e^{2t} \Rightarrow y_2'=y_1''-2e^{2t} \\ y_2'=y_1-3e^{2t}} \Rightarrow y_1''-2e^{2t} =y_1-3e^{2t} \Rightarrow y_1''-y_1=-e^{2t} \\ \Rightarrow y_1''-y_1=0 \Rightarrow y_{1h}=c_1e^t+c_2e^{-t} \Rightarrow y_{1p} =Ae^{2t} \Rightarrow y_{1p}''=4Ae^{2t} \Rightarrow y_{1p}''-y_{1p}=3Ae^{2t}=-e^{2t} \\ \Rightarrow A=-{1\over 3} \Rightarrow y_{1p}=-{1\over 3}e^{2t} \Rightarrow y_1= y_{1h}+y_{1p}=c_1e^t+c_2e^{-t}-{1\over 3}e^{2t} \\ \Rightarrow y_2=y_1'-e^{2t} =c_1e^t-c_2e^{-t}-{2\over 3}e^{2t}-e^{2t}=c_1e^t-c_2e^{-t}-{5\over 3}e^{2t} \\ \Rightarrow \bbox[red, 2pt]{\cases{y_1=c_1e^t+c_2e^{-t}-{1\over 3}e^{2t} \\y_2=c_1e^t-c_2e^{-t}-{5\over 3}e^{2t}}}$$

解答:$$L\{y''\}+ L\{y'\}-6L\{y\}=0 \Rightarrow s^2Y(s)-sy(0)-y'(0)+ sY(s)-y(0)-6Y(s) =0 \\ \Rightarrow s^2Y(s)-s-1+sY(s)-1-6Y(s)=0 \Rightarrow Y(s)={s+2\over s^2+s-6}={4\over 5}\cdot {1\over s-2}+{1\over 5}\cdot {1\over s+3} \\ \Rightarrow y(t) =L^{-1}\{ Y(s)\} ={4\over 5} L^{-1} \left\{ {1\over s-2}\right\}+{1\over 5} L^{-1} \left\{{1\over s+3} \right\} \Rightarrow \bbox[red, 2pt]{y(t)={4\over 5}e^{2t}+{1\over 5}e^{-3t}}$$

解答:$$\tau =t-2 \Rightarrow u(\tau)=y(t)=u(\tau+2) \Rightarrow u'=y' \Rightarrow u''=y'' \\ \Rightarrow u''+2u'-3u=0 \Rightarrow L\{u''\}+2 L\{u'\}-3L\{u\} =0 \Rightarrow s^2U +3s+5+2(sU +3)-3U=0 \\ \Rightarrow U(s)= {-3s-11\over s^2+2s-3} ={1/2\over s+3} -{7/2\over s-1} \Rightarrow u(\tau) =L^{-1}\{U(s)\} ={1\over 2}e^{-3\tau}-{7\over 2}e^{\tau} \\ \Rightarrow \bbox[red, 2pt]{y(t)={1\over 2}e^{-3(t-2)}-{7\over 2}e^{t-2}}$$

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