108年國立台北大學轉學考-工程數學詳解

國立臺北大學108學年度日間學士班轉學生招生

學制系級：電機三年級
科目：工程數學

解答： $$y^2 \to \bbox[red,2pt]{\text{nonlinear}}$$

解答： $$\mathbf{(1)}\;(1+\ln{x} +{y\over x})dx +(\ln x-1)dy=0 \Rightarrow \cases{M(x,y)=1+\ln{x} +{y\over x} \\ N(x,y)= \ln x-1} \Rightarrow \cases{{\partial M\over \partial y} ={1\over x} \\{\partial N\over \partial x} ={1\over x}} \\ \Rightarrow {\partial M\over \partial y} ={\partial N\over \partial x} \Rightarrow 該微分方程為恰當，\bbox[red,2pt]{故得證} \\ \mathbf{(2)}\; \int M(x,y)\,dx = \int N(x,y)\,dy \Rightarrow \int 1+\ln{x} +{y\over x}\,dx = \int \ln x-1\,dy \\ \Rightarrow   x\ln x +y\ln x+ \phi(y)= y\ln x-y+\varphi(x)+C \Rightarrow \cases{\phi(y)= -y\\ \varphi(x)= x\ln x} \\ \Rightarrow 其解為\bbox[red, 2pt]{y\ln x+x\ln x-y+C=0,C為常數}$$

解答： $$令u(x)=y(x)-x  \Rightarrow u'=y'-1，則原式dy-(1+e^{y-x+5})dx=0 \Rightarrow y'=1+e^{y-x+5} \\ \Rightarrow u'+1=1+e^{u+5} \Rightarrow u'=e^{u+5} \Rightarrow \int {1\over e^u}du = \int e^5 \,dx \Rightarrow -e^{-u}=e^5x +c\\ \Rightarrow -u = \ln(-e^5x-c) \Rightarrow -y+x= \ln(-e^5x-c) \Rightarrow \bbox[red,2pt]{y=x-\ln(-e^5x-c), c\text{為常數}}$$

解答： $$y_c''-4y_c'+4y_c=0 \Rightarrow \lambda^2-4\lambda +4=0 \Rightarrow (\lambda-2)^2=0 \Rightarrow \bbox[red, 2pt]{y_c= C_1e^{2x} +C_2xe^{2x}}\\ y_p =a+ bx+ cx^2+ dx^2e^{2x} \Rightarrow y_p'=b+2cx+2dxe^{2x}+ 2dx^2e^{2x}\\ \Rightarrow y_p''=2c + 2de^{2x}+ 8dxe^{2x} + 4dx^2e^{2x}\\ \Rightarrow y_p''- 4y_p'+ 4y_p = (2c+4a-4b)+ (4b-8c )x +4cx^2 +2d e^{2x} = 12+8x^2-16e^{2x}\\ \Rightarrow \cases{2c+4a-4b=12\\ 4b=8c\\ 4c=8\\ 2d=-16} \Rightarrow \cases{a=6 \\b=4\\ c=2\\ d=-8} \Rightarrow \bbox[red,2pt]{y_p= 6+4x+ 2x^2-8x^2e^{2x}}$$

解答： $$\mathcal{L}\{ 2+2t^4 +e^{2t}+ \sin 2t\} =2\mathcal{L}\{1\} +2\mathcal{L}\{t^4 \}+ \mathcal{L}\{e^{2t}\} +\mathcal{L}\{ \sin 2t\} \\ = {2\over s} +2\cdot {4!\over s^5} +{1\over s-2} +{2\over s^2+2^2} =\bbox[red,2pt]{ {2\over s} +  {48\over s^5} +{1\over s-2} +{2\over s^2+4}}$$

解答： $$\mathcal{L^{-1}}\left\{ {2s-4} \over (s^2+s)(s^2+1)\right\} = \mathcal{L^{-1}}\left\{  {-4\over s} +{3\over s+1} +{ s+3 \over s^2+1}\right\}\\ = -4\mathcal{L^{-1}}\left\{ {1\over s}\right\}+ 3\mathcal{L^{-1}}\left\{{1\over s+1} \right\}+ \mathcal{L^{-1}}\left\{{s \over s^2+1} \right\} +3\mathcal{L^{-1}}\left\{{1\over s^2+1} \right\} \\ = \bbox[red, 2pt]{-4+3e^{-t}+ \cos t+ 3\sin t}$$

解答： $$\mathbf F=4xy \mathbf i +(2x^2+ 2yz)\mathbf j+ (3z^2+ y^2)\mathbf k = F_1\mathbf i+F_2 \mathbf j+ F_3\mathbf k \Rightarrow \cases{F_1= 4xy\\ F_2=2x^2+ 2yz \\ F_3=3z^2 +y^2} \\ \Rightarrow \text{curl }\mathbf F= \begin{vmatrix} \mathbf i& \mathbf j& \mathbf k  \\{\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ F_1 & F_2 & F_3 \end{vmatrix} = ( {\partial \over \partial y}F_3-{\partial \over \partial z}F_2)\mathbf i+( {\partial \over \partial z}F_1 -{\partial \over \partial x}F_3) \mathbf j+  ({\partial \over \partial x}F_2-{\partial \over \partial y}F_1) \mathbf k =\mathbf 0\\ \Rightarrow \text{div }\mathbf F= {\partial \over \partial x}F_1 + {\partial \over \partial y}F_2 +{\partial \over \partial z}F_3 = 4y+2z+ 6z = 4y+8z \\ \Rightarrow \bbox[red, 2pt]{\cases{\text{curl }\mathbf F= 0\mathbf i+ 0\mathbf j+ 0\mathbf k\\ \text{div }\mathbf F= 4y+8z} }$$

解答： $$\mathbf{(1)}\;\int_c\vec F\cdot d\vec r = \int_c 2xydx+ x^2dy \Rightarrow \cases{{\partial \over \partial y} 2xy=2x\\ {\partial \over \partial x} x^2 =2x} \Rightarrow 兩者相等(exact)，因此與積分路徑無關 \\\mathbf{(2)}\;因此取\phi(x,y)滿足\cases{\phi_x= 2xy\\ \phi_y= x^2} \Rightarrow \phi(x,y)=\int 2xy\,dx = \int x^2\,dy \Rightarrow \bbox[red,2pt]{\phi(x,y) =x^2y}\\\mathbf{(3)}\;\int_{(1,1)}^{(2,4)} 2xydx +x^2dy =\phi(2,4)-\phi(1,1) = 16-1=\bbox[red, 2pt]{15}$$   

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