112年暨大碩士班-工程數學詳解

 國立暨南國際大學112學年度碩士班入學考試

科目: 工程數學(線性代數、微分方程)
適用:電機系

解答: $$\mathbf{(a)}\;2xyy'=y^2-x^2 \Rightarrow y'={y\over 2x}-{x\over 2y} \Rightarrow y'-{y\over 2x}=-{x\over 2y},此為n=-1的\text{Bernoulli Equation}\\ 因此令u=y^2 \Rightarrow u'=2yy'代回原式 \Rightarrow xu'=u-x^2 \Rightarrow u'-{u\over x}=-x \\ 積分因子I(x)=e^{-\int {1\over x}\,dx}={1\over x} \Rightarrow {u'\over x}-{u\over x^2}=-1 \Rightarrow ({u\over x})'=-1 \Rightarrow {u\over x}=-x+C \\\Rightarrow u=y^2=-x^2+Cx \Rightarrow \bbox[red, 2pt]{y=\pm \sqrt{-x^2+Cx},C為常數}　\\\mathbf{(b)}\; y''-4y'+4y=0 \Rightarrow \lambda^2-4\lambda+4=0 \Rightarrow (\lambda-2)^2=0 \Rightarrow \bbox[red, 2pt]{y=c_1e^{2x} +c_2xe^{2x},c_1及c_2為常數}\\\mathbf{(c)}\; y'''+12y''+48y'+64y=0 \Rightarrow \lambda^3+12 \lambda^2+48\lambda +64=0 \Rightarrow (\lambda+4)^3=0 \\ \quad \Rightarrow \bbox[red, 2pt]{y=e^{-4x}(c_1+ c_2x+ c_3x^2),c_1,c_2,c_3皆為常數}$$

解答: $$\mathbf{(a)}\; L\{f(t)\} =F(s)=\int_0^\infty f(t)e^{-st}\,dt \\\qquad f(t)=L^{-1}\{F(s)\} \iff F(s)=L\{f(t)\}\\\mathbf{(b)} \;F(s)=\int_0^\infty f(t)e^{-st}\,dt \Rightarrow {d\over ds}F(s) =\int_0^\infty -tf(t)e^{-st}\,dt =-L\{tf(t)\}\\ \qquad \Rightarrow L\{tf(t)\} =-{d\over ds}F(s) \\ \mathbf{(c)}\; L\{t\cos(kt)\} =-{d\over ds}L\{\cos (kt)\} =-{d\over ds}{s\over s^2+k^2}={s^2-k^2\over (s^2+k^2)^2}\\ \quad L\{t\sin(kt)\} =-{d\over ds}L\{\sin(kt)\}=-{d\over ds}{k\over s^2+k^2} ={2ks\over (s^2+k^2)^2} \\\mathbf{(d)}\;L\{y''\}+16L\{y\} =L\{\cos(4t)\} \Rightarrow s^2Y(s)-sy(0)-y'(0)+16Y(s) ={s\over s^2+16}\\ \Rightarrow Y(s)(s^2+16)-1 ={s\over s^2+16} \Rightarrow Y(s)={s\over (s^2+16)^2}+{1\over s^2+16}\\ \Rightarrow y(t)=L^{-1}\{{s\over (s^2+16)^2}\} +L^{-1}\{{1\over s^2+16}\}\\= L^{-1}\{{1\over 64}\cdot {4\over s^2+4^2}-{1\over 16}\cdot{s^2-4^2\over (s^2+16)^2}\} +L^{-1}\{{1\over 4}\cdot {4\over s^2+4^2}\}\\={1\over 64}\sin(4t)-{1\over 16}t\cos(4t)+ {1\over 4}\sin(4t) ={17\over 64}\sin(4t)-{1\over 16}t\cos(4t) \\ \Rightarrow \bbox[red, 2pt]{y(t)={17\over 64}\sin(4t)-{1\over 16}t\cos(4t) }$$

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