109年台綜大轉學考-工程數學D39詳解

 臺灣綜合大學系統109學年度學士班轉學生聯合招生考試

科目名稱：工程數學
類組代碼：D39

解答：$$xf(x)={3\over 2}x^4-3x^2+4 +\int_2^x f(t)\,dt \Rightarrow f(x)+ xf'(x)= 6x^3-6x +f(x)\\ \Rightarrow xf'(x)=6x^3-6x \Rightarrow f'(x)=6x^2-6 \Rightarrow f(x)=2x^3-6x+C \\ 將f(x)=2x^3-6x+C代回原式 \Rightarrow 2x^4-6x^2+Cx ={3\over 2}x^4-3x^2+4 +\int_2^x 2t^3-6t+C\,dt \\ ={3\over 2}x^4-3x^2+4 +({1\over 2}x^4-3x^2 +Cx-8+12-2C) = 2x^4-6x^2+Cx+8-2C \\ \Rightarrow C=4 \Rightarrow \bbox[red,2pt]{f(x)= 2x^3-6x+4}$$

解答：$$y'+2xy = 2x \cos(x^2)，令積分因子I(x)=e^{\int 2x\,dx} =e^{x^2 } \Rightarrow I(x)y'+ 2xyI(x)= 2xI(x)\cos(x^2)\\ \Rightarrow e^{x^2} y'+2x e^{x^2} y= 2xe^{x^2} \cos(x^2) \Rightarrow (e^{x^2} y)'=2xe^{x^2} \cos(x^2)  \Rightarrow e^{x^2} y= \int 2xe^{x^2} \cos(x^2)\,dx \\ = \int e^u \cos u\,du \;(其中u=x^2) ={1\over 2}e^u(\sin u+\cos u) = {1\over 2}e^{x^2}(\sin x^2+ \cos x^2) +C \\ \Rightarrow \bbox[red, 2pt]{y={1\over 2}(\sin x^2+ \cos x^2) +Ce^{-x^2}}$$

解答：$$先求齊次解，即y''+y=0 \Rightarrow 特徵方程式為\lambda^2+1=0 \Rightarrow \lambda=\pm i \Rightarrow y_h=C_1 \cos x+C_2 \sin x\\ 接著令y_p = x(A\cos x+B\sin x) \Rightarrow y_p'= A\cos x+B\sin x+ x(-A\sin x+B\cos x)\\ \Rightarrow y_p'' = -2A\sin x+ 2B\cos x +x(-A\cos x-B\sin x) \Rightarrow y_p''+y_p = -2A\sin x+ 2B\cos x=\cos x\\ \Rightarrow \cases{A=0\\ B=1/2} \Rightarrow y_p={1\over 2}x\sin x \Rightarrow y =y_h+y_p = C_1 \cos x+C_2 \sin x +{1\over 2}x\sin x \\ \Rightarrow y'= -C_1\sin x+ C_2\cos x+{1\over 2}\sin x+{1\over 2}x\cos x \\ 因此\cases{y(0)=1\\ y'(0)=0} \Rightarrow \cases{C_1=1\\ C_2=0} \Rightarrow \bbox[red, 2pt]{y= \cos x+{1\over 2}x\sin x}$$

解答：$$\mathcal L^{-1}\{ {s-2\over s^2+2s+10 }\} = \mathcal L^{-1}\{ {s-2\over (s+1)^2+3^2 }\} = \mathcal L^{-1}\{ {s+1\over (s+1)^2+3^2 } -{3\over (s+1)^2+3^2}\} \\= \bbox[red,2pt]{e^{-t}\cos 3t-e^{-t}\sin 3t}$$

解答：$$\mathcal L\{ \sin t+ t\cos t\} =\mathcal L\{ \sin t \} +\mathcal L\{ t\cos t\} ={1\over s^2+1} +{ s^2-1\over (s^2+1)^2} =\bbox[red, 2pt]{ 2s^2 \over (s^2+1)^2} $$

解答：$$a_0 ={1 \over 2\pi}\int_0^{\pi} x\,dx ={\pi \over 4}\\ a_n={1\over \pi} \int_0^{\pi} x\cos(nx)\,dx ={1\over \pi} \left.\left[{1\over n} x\sin (nx) +{1\over n^2} \cos(nx) \right]\right|_0^\pi ={1\over n^2\pi}((-1)^n-1)\\ b_n = {1\over \pi} \int_0^{\pi} x\sin(nx)\,dx ={1\over \pi} \left.\left[{1\over n^2} \sin (nx) -{1\over n} x\cos(nx) \right]\right|_0^\pi=-{1\over n}(-1)^n\\ \Rightarrow \bbox[red, 2pt]{f(x)= {\pi\over 4}+\sum_{n=1}^\infty \left({1\over n^2\pi}((-1)^n-1) \cos (nx)-{1\over n}(-1)^n \sin (nx) \right)}$$

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