Question

Consider the initial value problem $$ y' + 5y = \begin{cases} 0 & \text{if } 0 < t < 3 \\ 9 & \text{if } 3 \le t < 5 \\ 0 & \text{if } 5 \le t < \infty \end{cases} $$ $y(0) = 5$. a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of $y(t)$ by $Y(s)$. Do not move any terms from one side of the equation to the other (until you get to part (b) below). $sY(s) + 5Y(s) - 5 = \frac{9e^{-3s}}{s} - \frac{9e^{-5s}}{s}$ help (formulas) b. Solve your equation for $Y(s)$. $Y(s) = \mathcal{L}\{y(t)\} = \frac{1}{s+5} \left( \frac{9e^{-3s}}{s} - \frac{9e^{-5s}}{s} + 5 \right)$ c. Take the inverse Laplace transform of both sides of the previous equation to solve for $y(t)$. $y(t) = e^{-5s} + 9[\dots]$

          Consider the initial value problem
$$
y' + 5y = \begin{cases}
0 & \text{if } 0 < t < 3 \\
9 & \text{if } 3 \le t < 5 \\
0 & \text{if } 5 \le t < \infty
\end{cases}
$$
$y(0) = 5$.

a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of $y(t)$ by $Y(s)$. Do not move any terms from one side of the equation to the other (until you get to part (b) below).

$sY(s) + 5Y(s) - 5 = \frac{9e^{-3s}}{s} - \frac{9e^{-5s}}{s}$ help (formulas)

b. Solve your equation for $Y(s)$.

$Y(s) = \mathcal{L}\{y(t)\} = \frac{1}{s+5} \left( \frac{9e^{-3s}}{s} - \frac{9e^{-5s}}{s} + 5 \right)$

c. Take the inverse Laplace transform of both sides of the previous equation to solve for $y(t)$.

$y(t) = e^{-5s} + 9[\dots]$

consider the initial value problem y 5y begincases 0 textif 0 t 3 9 textif 3 le t 5 0 textif 5 le t infty endcases y0 5 a take the laplace transform of both sides of the given differential e 23171

Added by Manuela S.

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