Chapter 7: Problem 29
\(s F(s)+2 F(s)=\frac{10 s^{2}+12 s+14}{s^{2}-2 s+2}\)
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Chapter 7: Problem 29
\(s F(s)+2 F(s)=\frac{10 s^{2}+12 s+14}{s^{2}-2 s+2}\)
These are the key concepts you need to understand to accurately answer the question.
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\(\begin{array}{ll}{x^{\prime}-3 x+2 y=\sin t ;} & {x(0)=0} \\ {4 x-y^{\prime}-y=\cos t ;} &{y(0)=0}\end{array}\)
$$\begin{array}{l}{y^{\prime \prime}+y=-\delta(t-\pi)+\delta(t-2 \pi)} \\\ {y(0)=0, \quad y^{\prime}(0)=1}\end{array}$$
Thanks to Euler's formula (page 166) and the algebraic properties of complex numbers, several of the entries of Table 7.1 can be derived from a single formula; namely, (6) $$\quad \mathscr{L}\left\\{e^{(a+i b) t}\right\\}(s)=\frac{s-a+i b}{(s-a)^{2}+b^{2}}, \quad s>a$$ (a) By computing the integral in the definition of theLaplace transform on page 353 with \(f(t)=e^{(a+i b) t}\)show that $$\mathscr{L}\left\\{e^{(a+i b) t}\right\\}(s)=\frac{1}{s-(a+i b)}, \quad s>a$$ (b) Deduce (6) from part (a) by showing that$$\frac{1}{s-(a+i b)}=\frac{s-a+i b}{(s-a)^{2}+b^{2}}$$ (c) By equating the real and imaginary parts in formula (6), deduce the last two entries in Table 7.1.
\(\frac{e^{-y}}{s^{2}+4}\)
\(\begin{array}{ll}{x^{\prime}=3 x+y-2 z ;} & {x(0)=-6} \\ {y^{\prime}=-x+2 y+z ;} & {y(0)=2} \\ {z^{\prime}=4 x+y-3 z ;} & {z(0)=-12}\end{array}\)
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