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In the following problems use Laplace transforms to solve for u(r) Keep mind. however. that y() and W() Mst be defined tor al over the semi-infinite dmain ( < 00...

Question

In the following problems use Laplace transforms to solve for u(r) Keep mind. however. that y() and W() Mst be defined tor al over the semi-infinite dmain ( < 00 10 US the Laplace transform method. Hence appropriate moclification to the applied load W() OHL the right hatucl sicle of the abore diflerential eqquation for uls) must he performed as follows,d'u(r) f() 0 <! < 0_ dr'whire f() is the appliedl loaul delinedl over 0 < " < &-Write down f(2). 0 < x 0, if

In the following problems use Laplace transforms to solve for u(r) Keep mind. however. that y() and W() Mst be defined tor al over the semi-infinite dmain ( < 00 10 US the Laplace transform method. Hence appropriate moclification to the applied load W() OHL the right hatucl sicle of the abore diflerential eqquation for uls) must he performed as follows, d'u(r) f() 0 <! < 0_ dr' whire f() is the appliedl loaul delinedl over 0 < " < &- Write down f(2). 0 < x 0, if the bcntn is Idler constant load of Wo for 0 < r < L/2 Write dlown f(2)0 < " 0, il the ba is struck at its free cd _ = L with a impulsive force AId no other force ats tlie bcI , Find ylr) in part (b) given that y(O) = 0, v(0) 0, v"(L) =0 and 4"(L) =0.



Answers

Use the result of Problem 11 to show that
$$ \mathscr{L}^{-1}\left\{\frac{1}{\left(s^{2}+1\right)\left(1-e^{-\pi s}\right)}\right\}(t)=g(t) $$
where $g(t)$ is periodic with period 2$\pi$ and
$$ g(t) : \left\{\begin{array}{ll}{\sin t,} & {0 \leq t \leq \pi} \\ {0,} & {\pi \leq t \leq 2 \pi}\end{array}\right. $$
In Problems 13 and $14,$ use the method of Laplace transforms and the results of Problems 9 and 10 to solve the initial value problem.
$$ \begin{array}{l}{y^{\prime \prime}+3 y^{\prime}+2 y=f(t)} \\ {y(0)=0, \quad y^{\prime}(0)=0}\end{array} $$
where $f(t)$ is the periodic function defined in the stated problem.

For this problem, we're gonna use the convolution room showing here to write our function fft as the convolution of function s don t Tilda to be different from our original absentee which is just equal to t and then a genius team Reasonable through the co sign of two teams such that if we look at the convolution of r f, Tilda and G and we write it with those with respect to W, this will be using this here, um, the integral from zero to t of our function F tilda, which is just tea minus town times the co sign of two Childs NGO town detail, which is exactly what we have up here. Just replaced how it w two. Therefore, this is a convolution of f, Tilda and G. If you want to take the applause transform of F which is little loss transform of F. Tilda Congres to the G using this rule for convolution. This will just be the class transform of F. Tilda times of loss, transform of gene. So we're just going to have the laplace transform of f tilda, which was just tea time. So loss transform of the co sign of to change. And we can use the rule that the applause transform of tea, raises some manager and is in factorial over s to the end plus one and the class transform of the co sign of BT. Is he going to s over a squared plus B squared. Using these two rules will get that this is the posturings from a tea is just tea to the one. So is one factorial which is fun over Esther. The one plus one which is to times R B is too. So this will be s over s squared. Plus two squared, just four. We could just simplify this a little bit. We're gonna get a factor of s that cancels since be won over s times this denominator which would be s cubed plus four s.


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