Chapter 7: Problem 7
\((D-4)[x]+6 y=9 e^{-3 r} ; \quad x(0)=-9$$x-(D-1)[y]=5 e^{-3 t} ; \quad y(0)=4\)
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Chapter 7: Problem 7
\((D-4)[x]+6 y=9 e^{-3 r} ; \quad x(0)=-9$$x-(D-1)[y]=5 e^{-3 t} ; \quad y(0)=4\)
These are the key concepts you need to understand to accurately answer the question.
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$$\begin{array}{l}{y^{\prime \prime}-y=4 \delta(t-2)+t^{2}} \\ {y(0)=0, \quad y^{\prime}(0)=2}\end{array}$$
$$\begin{array}{l}{y^{\prime \prime}+y=-\delta(t-\pi)+\delta(t-2 \pi)} \\\ {y(0)=0, \quad y^{\prime}(0)=1}\end{array}$$
Residue Computation. Let \(P(s) / Q(s)\) be a rational function with deg \(P<\) deg \(Q\) and suppose \(s-r\) is a non- repeated linear factor of \(Q(s) .\) Prove that the portion of the partial fraction expansion of \(P(s) / Q(s)\) corresponding to \(s-r\) is \(\frac{A}{s-r}\) where A ( called the residue) is given by the formula \(A=\lim _{s \rightarrow r} \frac{(s-r) P(s)}{Q(s)}=\frac{P(r)}{Q^{\prime}(r)}\)
\(\int_{-1}^{1}(\cos 2 t) \delta(t) d t\)
$$y^{\prime \prime}+3 y^{\prime}+2 y=g(t)$$ $$y(0)=2, \quad y^{\prime}(0)=-1$$ where $$g(t)=\left\\{\begin{array}{ll}{e^{-t},} & {0 \leq t< 3} \\ {1,} & {3< t}\end{array}\right.$$
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