解答:
(a)$$xy'+3y=2x \Rightarrow y'+{3\over x}y=2 \Rightarrow \text{integrating factor }I(x)=e^{\int (3/x)dx} =x^3\\ \Rightarrow x^3y'+3x^2y=2x^3 \Rightarrow (x^3y)'=2x^3 \Rightarrow x^3y= \int 2x^3\,dx = {1\over 2}x^4+c \\ \Rightarrow \bbox[red,2pt]{y={1\over 2}x+{c\over x^3}}$$(b)$$y''+5y'+6y=0 \Rightarrow \lambda^2+5\lambda+6=0 \Rightarrow (\lambda+3)(\lambda+2)=0 \\ \Rightarrow \lambda=-3,-2 \Rightarrow \bbox[red,2pt]{y=c_1e^{-3x}+c_2 e^{-2x}}$$(c)$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\\Rightarrow x^2y''+1.5xy'-0.5y =(m^2-m)x^m +1.5mx^m-0.5x^m =(m^2+0.5m-0.5)x^m=0\\ \Rightarrow {1\over 2}(2m-1)(m+1)=0 \Rightarrow m={1\over 2},-1 \Rightarrow \bbox[red, 2pt]{y=c_1\sqrt x + {c_2\over x}x}$$
解答:$$\cases{P(x,y)= 3x^2y+ 6xy+{y^2\over 2} \\ Q(x,y)= 3x^2+ y} \Rightarrow \cases{P_y=3x^2+6x+y\\ Q_x=6x} \Rightarrow P_y \ne Q_x \Rightarrow \text{NOT exact} \\ {P_y-Q_x\over Q} =1 \text{ independent of }x \Rightarrow u'=u \Rightarrow \text{ integration factor }u=\bbox[red, 2pt]{e^x} \\ \Rightarrow (e^x P)dx+ (e^xQ)dy=0 \Rightarrow \text{potential }\Phi(x,y) \text{ satisfying} \cases{\Phi_x=e^xP\\ \Phi_y =e^xQ} \\ \Rightarrow \Phi =\int 3x^2ye^x+6xye^x+{y^2\over 2}e^x\,dx = \int3x^2e^x +ye^x\,dy \\ \Rightarrow \Phi=3x^2ye^x+{y^2\over 2}e^x +\phi(y)= 3x^2ye^x+{1\over 2}y^2e^x +\rho(x) \\ \Rightarrow \bbox[red, 2pt]{3x^2ye^x+{y^2\over 2}e^x +c=0}$$解答:$$\先求齊次解，y''+2y'+y=0 \Rightarrow \lambda^2+2\lambda +1=0 \Rightarrow (\lambda+1)^2=0 \Rightarrow \lambda=-1 \Rightarrow y_h=c_1e^{-x}+ c_2xe^{-x}\\再利用\text{ varation of parameter} 求特解, \cases{y_1=e^{-x} \\ y_2=xe^{-x} \\ r(x)=xe^{-x}} \Rightarrow W=\begin{vmatrix} y_1& y_2 \\ y_1'& y_2' \end{vmatrix}=e^{-2x} \\ \Rightarrow y_p=-y_1 \int{y_2 r\over W}\,dx +y_2\int{y_1r\over W}\,dx =-e^{-x}\int x^2\,dx +xe^{-x}\int x\,dx ={1\over 6}x^3e^{-x} \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^{-x}+ c_2xe^{-x} +{1\over 6}x^3e^{-x}}$$
解答:$$\mathbf{(a)}\; L\{(t+2)^2\} =L\{t^2+4t+4\}=\bbox[red, 2pt]{{2\over s^3}+{4\over s^2}+{4\over s}} \\\mathbf{(b)}\;L^{-1}\{{3\over s+3}\}+L^{-1}\{ {3s\over s^2+5}\} = \bbox[red, 2pt]{3e^{-3t}+3\cos(\sqrt 5t)}$$
解答:$$\mathbf{(a)}\;令\cases{u=[x_1,x_2,x_3,x_4]^T\\ v=[y_1,y_2,y_3,y_4]^T} \Rightarrow au+bv=[ax_1 +by_1,ax_2+ by_2, ax_3+by_3, ax_4+by_4]^T \\ \Rightarrow T(au+bv)=[ax_1+by_1,ax_2+by_2, ax_3+by_3, ax_4+by_4]^T\cdot [2,0,2,1]^T\\=2(ax_1+by_1)+ 2(ax_3+by_3)+(ax_4+by_4)\\ 又aT(u)+ bT(v)= a(2x_1+2x_3+x_4)+ b(2y_1+2y_3+y_4)\\=2(ax_1+by_1)+ 2(ax_3+by_3)+(ax_4+by_4)\\ 因此T(au +bv)=aT(u)+bT(v)\Rightarrow T \text{ is a linear transformation} \bbox[red,2pt]{Q.E.D} \\ \mathbf{(b)}\; A=\bbox[red, 2pt]{[2,0,2,1]} \\\mathbf{(c)}\;Ax=0 \Rightarrow 2x_1+2x_3+x_4=0 \Rightarrow ker(A)=\bbox[red,2pt]{\left\{\alpha\begin{pmatrix} 1\\ 0\\ 0\\ -2\end{pmatrix} +\beta \begin{pmatrix} 0\\ 1\\ 0\\ 0\end{pmatrix} +\gamma \begin{pmatrix} 0\\ 0\\ 1\\ -2\end{pmatrix}\mid \alpha,\beta,\gamma \in \mathbb R\right\}}$$
解答:$$\det(A)=-10h-150 \ne 0 \Rightarrow \bbox[red,2pt]{h\ne -10}$$
解答:$$\mathbf{(a)}\; \det(A-\lambda I)=0 \Rightarrow (\lambda-{3\over 5})(\lambda-1)=0 \Rightarrow \lambda=1,{3\over 5} \\ \lambda_1={3\over 5} \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow v=k\begin{pmatrix} -3/2\\ 1\end{pmatrix},取v_1=\begin{pmatrix} -3/2\\ 1\end{pmatrix}\\ \lambda_2=1 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow v=k\begin{pmatrix} -1/2\\ 1\end{pmatrix},取v_2=\begin{pmatrix} -1/2\\ 1\end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{P=\begin{bmatrix} -3/2 & -1/2\\ 1& 1\end{bmatrix}, D=\begin{bmatrix} 3/5 & 0\\ 0 & 1\end{bmatrix}}\\ \mathbf{(b)}\; \lim_{n\to \infty}A^n = \lim_{n\to \infty}(PDP^{-1})^n = \lim_{n\to \infty}PD^nP^{-1} = \begin{bmatrix} -3/2 & -1/2\\ 1& 1\end{bmatrix} \begin{bmatrix} 0 & 0\\ 0 & 1\end{bmatrix} \begin{bmatrix} -1 & -1/2\\ 1& 3/2\end{bmatrix} \\\qquad =\bbox[red, 2pt]{\begin{bmatrix} -1/2 & -3/4\\ 1 & 3/2\end{bmatrix}} \\\mathbf{(c)} \text{eigenvalues of }A^{-1}={1\over \lambda_1},{1\over \lambda_2} =\bbox[red, 2pt]{{5\over 3},1}$$
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