解答:\先求齊次解,y″+2y′+y=0⇒λ2+2λ+1=0⇒(λ+1)2=0⇒λ=−1⇒yh=c1e−x+c2xe−x再利用 varation of parameter求特解,{y1=e−xy2=xe−xr(x)=xe−x⇒W=|y1y2y′1y′2|=e−2x⇒yp=−y1∫y2rWdx+y2∫y1rWdx=−e−x∫x2dx+xe−x∫xdx=16x3e−x⇒y=yh+yp⇒y=c1e−x+c2xe−x+16x3e−x
解答:(a)L{(t+2)2}=L{t2+4t+4}=2s3+4s2+4s(b)L−1{3s+3}+L−1{3ss2+5}=3e−3t+3cos(√5t)


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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