Chapter 7: Problem 26
26\. $$y^{\prime \prime}+2 y^{\prime}-15 y=g(t) ; \quad y(0)=0, \quad y^{\prime}(0)=8$$
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Chapter 7: Problem 26
26\. $$y^{\prime \prime}+2 y^{\prime}-15 y=g(t) ; \quad y(0)=0, \quad y^{\prime}(0)=8$$
These are the key concepts you need to understand to accurately answer the question.
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\(z^{\prime \prime}+3 z^{\prime}+2 z=e^{-3 t} u(t-2)\) \(z(0)=2, \quad z^{\prime}(0)=-3\)
$$\begin{array}{l}{y^{\prime \prime}-2 y^{\prime}-3 y=2 \delta(t-1)-\delta(t-3)} \\ {y(0)=2, \quad y^{\prime}(0)=2}\end{array}$$
35\. Use the convolution theorem to show that \(\quad \mathscr{L}^{-1}\left\\{\frac{F(s)}{s}\right\\}(t)=\int_{0}^{t} f(v) d v\), where \(F(s)=\mathscr{L}\\{f\\}(s)\).
$$\begin{array}{l}{y^{\prime \prime}+y=-\delta(t-\pi)+\delta(t-2 \pi)} \\\ {y(0)=0, \quad y^{\prime}(0)=1}\end{array}$$
$$y^{\prime \prime}+5 y^{\prime}+6 y=g(t)$$ $$y(0)=0, \quad \quad y^{\prime}(0)=2$$ $$y^{\prime \prime}+5 y^{\prime}+6 y=g(t)$$ $$y(0)=0, \quad \quad y^{\prime}(0)=2$$ where $$g(t)=\left\\{\begin{array}{ll}{0,} & {0 \leq t< 1} \\ {t,} & {1< t <5} \\ {1,} & {5< t}\end{array}\right.$$
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