國立臺北科技大學 l12學年度碩士班招生考試
系所組別 :1201製造科技研究所
第一節 微分方程 試題 (選考)
解答：$$\lambda^2+0.4\lambda+9.04=0 \Rightarrow \lambda=-0.2\pm 3i \Rightarrow y=e^{-0.2x}(A\cos (3x) +B \sin(3x)) \\ \Rightarrow y'=-0.2e^{-0.2x}(A\cos (3x) +B \sin(3x))+ e^{-0.2x}(-3A\sin (3x) +3B \cos(3x)) \\ 初始值\cases{y(0)=0\\ y'(0)=3} \Rightarrow \cases{A=0\\ B=1} \Rightarrow \bbox[red,2pt]{y=e^{-0.2x}\sin(3x)}$$
解答：$$\mathbf{(1)}\;x^2y'+3xy-x+1=0 \Rightarrow y'+{3\over x}y={1\over x}-{1\over x^2}\\ 積分因子I(x)=e^{\int (3/x)dx} =x^3 \Rightarrow x^3y'+3x^2y=x^2-x \Rightarrow (x^3y)'=x^2-x\\ \Rightarrow x^3y={1\over 3}x^3-{1\over 2}x^2+C \Rightarrow y={1\over 3}-{1\over 2x}+{C\over x^3}\\初始值y(2)=0 \Rightarrow 0={1\over 3}-{1\over 4}+{C\over 8} \Rightarrow C=-{2\over 3} \Rightarrow \bbox[red,2pt]{y={1\over 3}-{1\over 2x}-{2\over 3x^3}}\\ \mathbf{(2)}\;先求齊次解，y''+4y=0 \Rightarrow \lambda^2+4=0 \Rightarrow \lambda=\pm 2i \Rightarrow y_h=A\cos(2x)+B\sin(2x)\\ 再令特解y_p=C\cos(3x)+D\sin(3x) \Rightarrow y_p'=-3C\sin(3x)+3D\cos(3x) \\ \Rightarrow y_p''=-9C\cos(3x)-9D\sin(3x) \Rightarrow y_p''+4y_p=-5C\cos(3x)-5D\sin(3x)=2\cos(3x)+3\sin(3x) \\ \Rightarrow \cases{C=-2/5\\ D=-3/5} \Rightarrow y_p=-{2\over 5}\cos(3x)-{3\over 5}\sin(3x) \Rightarrow y=y_h+y_p \\ \Rightarrow y=A\cos(2x)+B\sin(2x)-{2\over 5}\cos(3x)-{3\over 5}\sin(3x)\\ \Rightarrow y'=-2A\sin(2x)+2B \cos(2x)+{6\over 5}\sin(3x)-{9\over 5}\cos(3x)\\ 又\cases{y(0)=3\\ y'(0)=2} \Rightarrow \cases{A-{2\over 5}=3\\ 2B-{9\over 5}=2} \Rightarrow \cases{A=17/5\\ B=19/10}\\ \Rightarrow \bbox[red, 2pt]{y={17\over 5}\cos(2x)+{19\over 10}\sin(2x)-{2\over 5}\cos(3x)-{3\over 5}\sin(3x)}$$
解答：$$令u=y',則原式4x^2u''+12xu'+3u=0\\再令u=x^m \Rightarrow u'=mx^{m-1} \Rightarrow u''=m(m-1)x^{m-2} \Rightarrow 4m(m-1)x^m+12mx^m+3x^m=0\\ \Rightarrow (4m^2+8m+3)x^m=0 \Rightarrow 4m^2+8m+3=0 \Rightarrow (2m+3)(2m+1)＝0 \\ \Rightarrow m=-1/2,-3/2 \Rightarrow u=y'=c_1x^{-1/2} +c_2x^{-3/2} \Rightarrow y=2c_1x^{1/2}-2c_2x^{-1/2}+c_3 \\ \Rightarrow y''=-{1\over 2}c_1x^{-3/2}-{3\over 2}c_2x^{-5/2},因此\cases{y(1)=0\\ y'(1)=1.5\\ y''(1)=-1.75} \Rightarrow \cases{2c_1-2c_2+c_3=0 \\ c_1+c_2=1.5\\ -{1\over 2}c_1-{3\over 2}c_2= -1.75} \\ \Rightarrow \cases{c_1=0.5\\ c_2=1\\ c_3=1} \Rightarrow \bbox[red,2pt]{y=x^{1/2}-2x^{-1/2}+1}$$
解答：$$\mathbf{(1)}\;xy'+y-e^x=0 \Rightarrow y'+{y\over x}={e^x\over x} \Rightarrow 積分因子I(x)=e^{\int (1/x)dx}=x \\ \Rightarrow xy'+y=e^x \Rightarrow (xy)'=e^x \Rightarrow xy=e^x+c_1 \Rightarrow y={e^x\over x}+{c_1\over x}\\ y(1)=e \Rightarrow e+c_1=e \Rightarrow c_1=0 \Rightarrow \bbox[red, 2pt]{y={e^x\over x}}\\ \mathbf{(2)}\; x^2y'+2xy-x+1=0 \Rightarrow y'+{2\over x}y={1\over x}-{1\over x^2} \Rightarrow 積分因子I(x)=e^{\int (2/x)dx}=x^2 \\ \Rightarrow x^2y'+2xy=x-1 \Rightarrow (x^2y)'=x-1 \Rightarrow x^2y={1\over 2}x^2-x+c_1 \Rightarrow y={1\over 2}-{1\over x}+{c_1\over x^2}\\ 又y(1)=0 \Rightarrow {1\over 2}-1+c_1=0 \Rightarrow c_1={1\over 2} \Rightarrow \bbox[red,2pt]{y={1\over 2}-{1\over x}+{1\over 2x^2}}$$
解答：$$\mathbf{(1)}\;先求y_1,再利用降階的方式求y_2\\假設y_1={1\over 1-x}(註：題目有x,x-1,3x-1,因此考慮1/(1-x)) \Rightarrow \cases{y'=1/(1-x)^2\\ y''=2/(1-x)^3} \\ \Rightarrow x(x-1)\cdot {2\over (1-x)^3}+(3x-1)\cdot {1\over (1-x)^2}+{1\over 1-x}={-2x+3x-1+1-x\over (1-x)^2}=0\\ \Rightarrow 確定y_1={1\over 1-x},因此令y_2=uy_1={u\over 1-x} \Rightarrow \cases{y_2'=u/(1-x)+u/(1-x)^2\\ y_2''=u''(1-x)+ 2u'/(1-x)^2+ 2u/(1-x)^3} \\ \Rightarrow x(x-1)y''+(3x-1)y'+y=x(x-1)\left( {u''\over 1-x} +{2u'\over (1-x)^2}+ {2u\over (1-x)^3}\right)\\\qquad + (3x-1)\left({u\over 1-x}+ {u\over (1-x)^2} \right)+{u\over 1-x} =xu''+u'=0 \Rightarrow u=\ln x+c\\ \Rightarrow y_2={\ln x+c\over 1-x} \Rightarrow y=c_1y_1+c_2y_2\Rightarrow \bbox[red, 2pt]{ y= {k_1\over 1-x}+{k_2\ln x\over 1-x},k_1,k_2為常數}$$ $$\mathbf{(2)}\;y''-4xy'+4x^2y=xe^{x^2} \Rightarrow y''-2xy'=2xy'-4x^2y+xe^{x^2}\\ 積分因子I(x)=e^{\int -2x\,dx} =e^{-x^2} \Rightarrow e^{-x^2}y''-2xe^{-x^2}y'=2xe^{-x^2}y'-4x^2e^{-x^2}y+x \\ \Rightarrow (e^{-x^2}y')'=2x(e^{-x^2}y'-2xe^{-x^2}y)+x =2x(e^{-x^2}y)'+x \Rightarrow (e^{-x^2}y')'-2x(e^{-x^2}y)'=x \cdots(1)\\ 考慮(e^{-x^2}y)''=(-2xe^{-x^2}y+ e^{-x^2}y')' =(-2xe^{-x^2}y)'+( e^{-x^2}y')'=-2e^{-x^2}y-2x(e^{-x^2}y)'+( e^{-x^2}y')'\\ \Rightarrow (e^{-x^2}y)''=-2(e^{-x^2}y)-2x(e^{-x^2}y)'+( e^{-x^2}y')'\cdots(2)\\ 將(1)代入(2)\Rightarrow (e^{-x^2}y)''=-2(e^{-x^2}y)+x \Rightarrow u''+2u=x \Rightarrow u=e^{-x^2}y=A\cos \sqrt 2x+B\sin \sqrt 2x+{1\over 2}x\\ \Rightarrow \bbox[red, 2pt]{y=e^{x^2}(A\cos \sqrt 2x+B\sin \sqrt 2x+{1\over 2}x)}$$
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