111年台綜大轉學考-工程數學D09詳解

臺灣綜合大學系統111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：D09

解答： $$y'+y =y^2 \Rightarrow \int {1\over y^2-y}dy =\int 1\,dx \Rightarrow \int \left({1\over y-1}-{1\over y} \right)\,dy = x+C_1 \Rightarrow \ln |y-1|-\ln |y|=x +C_1 \\ \Rightarrow \ln\left|{y-1\over y} \right|=\ln \left|1-{1\over y}\right| =x +C_1 \Rightarrow 1-{1\over y} = e^{x+C_1} =C_2e^x \Rightarrow {1\over y}=1-{C_2e^x} \\ \Rightarrow\bbox[red,2pt]{ y={1\over 1-C_2e^x},C_2 為常數}$$

解答： $$2xy\,dy = (x^2+y^2)dx \Rightarrow {dy\over dx }={x\over 2y}+{y\over 2x}，因此假設u={y\over x} \Rightarrow y=ux \Rightarrow y'=u'x +u 代回原式\\ \Rightarrow u'x+u ={1\over 2u}+ {u\over 2} \Rightarrow u'x ={1\over 2u}-{u\over 2} ={1-u^2\over 2u} \Rightarrow \int {2u\over 1-u^2}\,du =\int {1\over x}\,dx\\\Rightarrow -\ln (1-u^2)= \ln x+C_1 =\ln C_2x \Rightarrow {1\over 1-u^2} =C_2x \Rightarrow u^2={ y^2\over x^2} =1-{1\over C_2 x}\\ \Rightarrow y^2 = x^2-C_3x \Rightarrow \bbox[red,2pt]{x^2+y^2 =Cx,C為常數}$$

解答： $$\cos(\pi x) \cos(2\pi y)\,dx = 2\sin(\pi x)\sin(2\pi y)\,dy \Rightarrow {\cos (\pi x)\over \sin(\pi x)} \,dx =2{\sin(2\pi y)\over \cos(2\pi y)}\,dy \\ \Rightarrow \int {\cos (\pi x)\over \sin(\pi x)} \,dx =2 \int {\sin(2\pi y)\over \cos(2\pi y)}\,dy \Rightarrow {1\over \pi}\ln (\sin(\pi x)) =-{1\over \pi} \ln (\cos(2\pi y)) +C_1 \\={1\over \pi}\ln {1\over C_2\cos (2\pi y)}  \Rightarrow \sin (\pi x)= {1\over C_2\cos(2\pi y)} \Rightarrow C_2\sin( \pi x)\cos( 2\pi y)=1\\ 將初始值y({3\over 2})={1\over 2}代入上式\Rightarrow  C_2\sin{3\pi \over 2} \cos(\pi)=1 \Rightarrow C_2=1 \Rightarrow \bbox[red, 2pt]{\sin(\pi x)\cos (2\pi y)=1}$$

解答： $$a\cos x+ b\sin x+c =0 \Rightarrow \cases{x=0 \Rightarrow a+c=0\\ x=\pi/2 \Rightarrow b+c=0\\ x=\pi \Rightarrow -a+c=0} \Rightarrow a=b=c=0 \\ \Rightarrow \cos x,\sin x,1為\bbox[red, 2pt]{線性獨立}$$

解答： $$y'''-y'=10 \cos(2x) \Rightarrow \lambda^3-\lambda=0 \Rightarrow \lambda=0,\pm 1 \Rightarrow y_h=C_1+C_2e^x+ C_3e^{-x}\\ 令y_p =A\cos(2x)+ B\sin(2x) \Rightarrow y_p'=-2A\sin(2x) +2B\cos(2x) \\\Rightarrow y_p''= -4A\cos(2x)-4B\sin(2x) \Rightarrow y_p'''= 8A\sin(2x)-8B\cos(2x)\\ \Rightarrow y_p'''-y_p' = 10A\sin(2x)-10B\cos(2x)= 10\cos(2x) \Rightarrow \cases{A=0\\ B=-1} \Rightarrow y_p=-\sin(2x)\\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=C_1 +C_2 e^x +C_3e^{-x} -\sin(2x)}$$

解答： $$y(t)=1+\int_0^t y(\tau)\,d\tau \Rightarrow \mathcal L\{y(t)\} =\mathcal L\{ 1+\int_0^t y(\tau)\,d\tau \} \Rightarrow Y(s)= {1\over s} +{Y(s)\over s} \\ \Rightarrow ({s-1\over s})Y(s)= {1\over s} \Rightarrow Y(s)={1\over s-1} \Rightarrow y(t) =\mathcal L^{-1} \{ {1\over s-1}\} =e^t \Rightarrow \bbox[red, 2pt]{y(t)= e^t}$$

解答： $$A=\begin{bmatrix} 0 & 1\\ 1& 0\end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow \lambda^2-1=0 \Rightarrow \lambda=\pm 1\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)\mathbf x=0 \Rightarrow \begin{bmatrix} -1& 1\\ 1 & -1 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\end{bmatrix} =0 \Rightarrow x_1=x_2, 取v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix} \\ \lambda_2= -1 \Rightarrow (A-\lambda_2 I)\mathbf x=0 \Rightarrow \begin{bmatrix} 1& 1\\ 1 & 1 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\end{bmatrix} =0 \Rightarrow x_1=-x_2, 取v_2= \begin{bmatrix} 1\\ -1 \end{bmatrix} \\ 因此\bbox[red,2pt]{特徵值為1,-1，相對應的特徵向量為\begin{bmatrix} 1\\ 1 \end{bmatrix}及\begin{bmatrix} 1\\ -1 \end{bmatrix} }$$

解答： $$\mathbf{(1)}\;\vec V= yz\vec i+ zx\vec j+xy\vec k \Rightarrow div \vec F={\partial \over \partial x}yz +{\partial \over \partial y}zx + {\partial \over \partial z}xy =\bbox[red, 2pt]0 \\ \mathbf{(2)}\; \text{curl }\vec F =\begin{vmatrix} \vec i & \vec j & \vec k\\ {\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ yz & zx & xy \end{vmatrix} =x\vec i+y\vec j+z\vec k -x\vec i-y\vec j-z\vec k = \bbox[red,2pt]{\vec 0}$$

解答： $$E=\{(x,y,z)\mid x^2+y^2\le 4, x\ge 0,y \ge 0, |z| \le 1\} \Rightarrow \iint_S \mathbf F\cdot \vec n\,dA = \iiint_E \text{div }\mathbf F\,dV\\ =\iint_S \mathbf F\cdot \vec n\,dA = \iiint_E x^2\,dV\\ = \int_0^{\pi/2} \int_0^2 \int_{-1}^1 r^2\cos^2\theta r\,dzdrd\theta = \int_0^{\pi/2} \int_0^2 2r^3\cos^2\theta  \, drd\theta = \int_0^{\pi/2} 8\cos^2 \theta\, d\theta \\ =4\int_0^{\pi /2} \cos(2\theta)+1\,d\theta =\bbox[red, 2pt]{2\pi}$$

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