112年雲科大碩士班-工程數學(2)詳解

國立雲林科技大學112學年度碩士班招生考試

系所: 電子系
科目: 工程數學(2)

解答: $$\mathbf{(a)}\; xy'+3y=2x \Rightarrow y'+{3\over x}y=2 \Rightarrow 積分因子I(x)=e^{\int {3\over x}\,dx} =x^3 \\ \Rightarrow \quad x^3y'+3x^2y=2x^3 \Rightarrow (x^3y)'=2x^3 \Rightarrow x^3y=\int 2x^3\,dx ={1\over 2}x^4+C \\ \Rightarrow \bbox[red, 2pt]{y={1\over 2}x+{C\over x^3},C為常數} \\\mathbf{(b)}\; y''+5y'+6y=0 \Rightarrow \lambda^2+5\lambda +6=0 \Rightarrow (\lambda+3)(\lambda +2)=0 \Rightarrow \lambda =-2,-3\\ \quad \Rightarrow \bbox[red, 2pt]{y=c_1e^{-2x}+c_2e^{-3x},c_1,c_2皆為常數} \\\mathbf{(c)}\; 令y=x^m  \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2}代回原式 \\\quad \Rightarrow m(m-1)x^m +1.5m x^m-0.5x^m=0 \Rightarrow (m^2+0.5m-0.5)x^m=0 \\ \quad \Rightarrow (m-0.5)(m+1)x^m=0 \Rightarrow (m-0.5)(m+1)=0 \Rightarrow m=0.5,-1 \Rightarrow y=c_1x^{0.5}+c_2x^{-1} \\ \quad \Rightarrow \bbox[red, 2pt]{y=c_1\sqrt x+{c_2\over x},c_1,c_2皆為常數}$$

解答: $$令\cases{P(x,y)=3x^2y +6xy+y^2/2\\ Q(x,y) =3x^2+y} \Rightarrow \cases{P_y= 3x^2+6x+y\\ Q_x= 6x} \Rightarrow {P_y-Q_x\over Q}=1\\ \Rightarrow 取積分因子I(x)=e^{\int (P_y-Q_x)/Q\,dx} =e^{\int 1\,dx} =\bbox[red,2pt]{e^x}\\ \Rightarrow \cases{I(x)P(x,y)=(3x^2y+6xy+y^2/2)e^x\\ I(x)Q(x,y)=(3x^2+y)e^x} \Rightarrow {\partial \over \partial y}(I(x)P(x,y)) ={\partial \over \partial y}(I(x)Q(x,y))=(3x^2+6x+y)e^x\\ \Rightarrow \Phi(x,y)=\int I(x)P(x,y)\,dx +\phi(y)= \int I(x)Q(x,y)\,dy+ \rho(x)\\ 又\cases{\int (3x^2y+6xy+y^2/2)e^x\,dx =3x^2ye^x+{1\over 2}y^2e^x\\ \int (3x^2+y)e^x\,dy =3x^2ye^x+{1\over 2}y^2e^x} \Rightarrow \phi(y)=\rho(x)=c\\ \Rightarrow \bbox[red, 2pt]{3x^2ye^x+{1\over 2}y^2e^x+C=0}$$

解答: $$先求齊次解: y''+2y'+y=0 \Rightarrow \lambda^2+ 2\lambda+1=0 \Rightarrow (\lambda+1)^2=0 \\ \Rightarrow y_h=c_1e^{-x}+ c_2xe^{-x},再利用變數變換法求y_p\\ \cases{y_1=e^{-x}\\ y_2=xe^{-x}\\ r(x)=xe^{-x}} \Rightarrow \cases{y_1'= -e^{-x} \\ y_2'=e^{-x}-xe^{-x}} \Rightarrow W(y_1,y_2)=y_1y_2'-y_1'y_2= e^{-2x} \\ \Rightarrow y_p=-y_1 \int{y_2r\over W}+ y_2\,dx \int {y_1r\over W}\,dx =-e^{-x} \int x^2\,dx +xe^{-x} \int x\,dx ={1\over 6}x^3e^{-x} \\ \Rightarrow y=y_h +y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^{-x}+ c_2xe^{-x} +{1\over 6}x^3e^{-x},c_1及c_2皆為常數}$$

解答: $$\mathbf{(a)}\; f(t)=(t+2)^2 =t^2+4t+4 \Rightarrow L\{f(t)\} =L\{t^2\}+ 4L\{t\} +4L\{1\} \\\quad= \bbox[red, 2pt]{{2\over s^3} +{4\over s^2}+{4\over s}} \\ \mathbf{(b)}\; L^{-1}\{F(S)\} =3 L^{-1}\{{1\over S+3}\} +3 L^{-1}\{{S\over S^2+(\sqrt 5)^2}\} =\bbox[red, 2pt]{3e^{-3t}+3\cos(\sqrt 5 t)}$$

解答: $$\mathbf{(a)}\; T\left(\begin{bmatrix}x_1 \\x_2 \\x_3 \\x_4\end{bmatrix} \right) =\begin{bmatrix}x_1 \\x_2 \\x_3 \\x_4\end{bmatrix}\cdot \begin{bmatrix}2 \\0 \\2 \\1\end{bmatrix}=2x_1+2x_3+x_4\\ 假設u=\begin{bmatrix}u_1 \\u_2 \\u_3 \\u_4\end{bmatrix}, v=\begin{bmatrix}v_1 \\v_2 \\v_3 \\v_4\end{bmatrix},則au+bv=\begin{bmatrix}au_1+bv_1 \\au_2+ bv_2 \\au_3+ bv_3 \\ au_4 +bv_4\end{bmatrix}\\ \Rightarrow T(au+bv)=2(au_1+bv_1)+ 2(au_3+bv_3) +au_4+bv_4 \\ 又aT(u)+bT(v)=a(2u_1 +2u_3+u_4) +b(2v_1+ 2v_3+v_4)\\ =2(au_1+bv_1) +2(au_3+bv_3)+ au_4+bv_4 = T(au+bv) \Rightarrow T(au+bv)=aT(u) +bT(v) \\ \Rightarrow T \text{is linear transformation} \\ \mathbf{(b)}\; \begin{bmatrix}2 & 0 & 2 & 1\end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3 \\x_4\end{bmatrix}= 2x_1+2x_3+x_4 =T(x) \Rightarrow \bbox[red, 2pt]{ A=\begin{bmatrix}2 & 0 & 2 & 1\end{bmatrix}} \\ \mathbf{(c)}\; \text{kernel of }A =\{(x_1,x_2,x_3,x_4) \mid 2x_1+2x_3 +x_4=0,x_1,x_2,x_3,x_4 \in \mathbb R\}$$

解答: $$\det(A)=-10h-150 \ne 0 \Rightarrow \bbox[red, 2pt]{h\ne -15}$$

解答: $$\mathbf{(a)}\; \det(A-\lambda I)=(\lambda-0.6)(\lambda-1)=0 \Rightarrow \lambda_1=0.6,\lambda_2=1\\ \lambda_1=0.6 \Rightarrow (A-\lambda_1 I)\mathbf x=0 \Rightarrow x_1+{3\over 2}x_2=0, 取 v_1 =\begin{pmatrix}-3/2 \\1\end{pmatrix} \\ \lambda_2=1  \Rightarrow (A-\lambda_1 I)\mathbf x=0 \Rightarrow x_1+{1\over 2}x_2=0,取v_2=\begin{pmatrix}-1/2 \\1\end{pmatrix}\\ 因此D=\begin{bmatrix} \lambda_1 & 0 \\0 &\lambda_2 \end{bmatrix}\Rightarrow \bbox[red, 2pt]{D=\begin{bmatrix} 0.6 & 0 \\0 &1 \end{bmatrix}}, P=[v_1|v_2] \Rightarrow \bbox[red, 2pt]{P=\begin{bmatrix} -3/2 & -1/2 \\1 &1 \end{bmatrix}} \\\mathbf{(b)}\; \lim_{n\to \infty}A^n =\lim_{n\to \infty}(PDP^{-1})^n =\lim_{n\to \infty}PD^nP^{-1} =P\begin{bmatrix} 0 & 0 \\0 &1 \end{bmatrix}P^{-1} \\=\begin{bmatrix} -3/2 & -1/2 \\1 &1 \end{bmatrix}\begin{bmatrix} 0 & 0 \\0 &1 \end{bmatrix}\begin{bmatrix} -1 & -1/2 \\1& 3/2 \end{bmatrix}= \bbox[red,2pt]{\begin{bmatrix} -1/2 & -3/4 \\1& 3/2 \end{bmatrix}} \\ \mathbf{(c)}\;A^{-1}的特徵值={1\over \lambda_1},{1\over \lambda_2} =\bbox[red, 2pt]{{5\over 3},1}$$

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