Question

In Problems 35–42 use the Laplace transform to solve the given

equation $\mathbf{f}\left(t\right)\underset{\mathbf{k}\mathbf{=}\mathbf{0}}{\overset{\mathbf{=}}{\mathbf{=}\mathbf{\sum }\left(2k+1-t\right)}}\mathbf{\left[}\mathbf{9}\mathbf{\right(}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{k}\mathbf{)}\mathbf{-}\mathbf{4}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{k}\mathbf{-}\mathbf{1}\mathbf{\left)}\mathbf{\right]}$

Short answer

Therefore, the solution is:

$y\left(t\right)=\frac{1}{2!}{t}^2{e}^t+5t{e}^t$

Step-by-step solution

Given Information.

The given problem is

$f\left(t\right)=\sum _{k=0}^{\infty }(2k+1-t)[9/(t-2k)-4(t-2k-1\left)\right]$

Apply Laplace Transform

Take Laplace transform of both side of the given equation.

$\begin{array}{rcl}\mathcal{L}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}& =& \mathcal{L}\left\{\sum _{\mathrm{k}=0}^{\infty }(2\mathrm{k}+1-\mathrm{t})\left[\mathrm{U}\right(\mathrm{t}-2\mathrm{k})-\mathrm{U}(\mathrm{t}-2\mathrm{k}-1\left)\right]\right\}\\ & =& (1-\mathrm{t})\left[\mathrm{U}\right(\mathrm{t}-0)-\mathrm{U}(\mathrm{t}-1\left)\right]+(3-\mathrm{t})\left[\mathrm{U}\right(\mathrm{t}-2)-\mathrm{U}(\mathrm{t}-3\left)\right]+(5-\mathrm{t})\left[\mathrm{U}\right(\mathrm{t}-4)-\mathrm{U}(\mathrm{t}-5\left)\right]+\cdots \\ & =& -(\mathrm{t}-1)\left[\mathrm{U}\right(\mathrm{t}-0)-\mathrm{U}(\mathrm{t}-1\left)\right]-(\mathrm{t}-3)\left[\mathrm{U}\right(\mathrm{t}-2)-\mathrm{U}(\mathrm{t}-3\left)\right]-(\mathrm{t}-5)\left[\mathrm{U}\right(\mathrm{t}-4)-\mathrm{U}(\mathrm{t}-5\left)\right]+\cdots \\ & =& -0)+(\mathrm{t}-1\left)\mathrm{U}\right(\mathrm{t}-1)-(\mathrm{t}-3\left)\mathrm{U}\right(\mathrm{t}-2)+(\mathrm{t}-3\left)\mathrm{U}\right(\mathrm{t}-3)-(\mathrm{t}-5\left)\mathrm{U}\right(\mathrm{t}-4)+(\mathrm{t}-5\left)\mathrm{U}\right(\mathrm{t}-5)+\cdots \\ & & \end{array}$

$\begin{array}{rcl}d& =& -(t-1)U\left(t\right)+(t-1)U(t-1)-(t-2-1)U(t-2)+(t-3)U(t-3)-(t-4-1)U(t-4)+(t-5)U(t-5)+\cdots \\ & =& -(t-1)U\left(t\right)+(t-1)U(t-1)-(t-2)U(t-2)+U(t-2)+(t-3)U(t-3)\\ & & -(t-4)U(t-4)+U(t-4)+(t-5)U(t-5)+L\\ & =& -\mathcal{L}\left\{tU\right(t\left)\right\}+\mathcal{L}\left\{U\right(t\left)\right\}+\mathcal{L}\left\{\right(t-1\left)U\right(t-1\left)\right\}-\mathcal{L}\left\{\right(t-2\left)U\right(t-2\left)\right\}+\mathcal{L}\left\{\right(t-3\left)U\right(t-3\left)\right\}\\ & & -\mathcal{L}\left\{\right(t-4\left)U\right(t-4\left)\right\}+\mathcal{L}\left\{U\right(t-2\left)\right\}+\mathcal{L}\left\{U\right(t-4\left)\right\}+\cdots \end{array}$

The geometric series

$\frac{1}{1+x}=1-x+x^2-x^3+\cdots$

When $x=e^{-x},s>0$becomes