Questions: Take the Laplace transform of the following initial value problem and solve for Y(s) = Ly(t) y'' + 6y' + 17y = T(t), y(0)=0, y'(0)=0 where T is a periodic function defined by T(t) = t, for 0 ≤ t < 1/2 1-t, for 1/2 ≤ t < 1 and T(t+1) = T(t) for all t ≥ 0 Y(s) =

Take the Laplace transform of the following initial value problem and solve for Y(s) = Ly(t)
y'' + 6y' + 17y = T(t), y(0)=0, y'(0)=0
where T is a periodic function defined by
T(t) =
t, for 0 ≤ t < 1/2
1-t, for 1/2 ≤ t < 1
and T(t+1) = T(t) for all t ≥ 0
Y(s) =

Transcript text: Take the Laplace transform of the following initial value problem and solve for $Y(s)=\mathcal{L}\{y(t)\}$ \[ y^{\prime \prime}+6 y^{\prime}+17 y=T(t) \quad y(0)=0, y^{\prime}(0)=0 \] where $T$ is a periodic function defined by \[ T(t)=\left\{\begin{array}{ll} t, & 0 \leq t<1 / 2 \\ 1-t, & 1 / 2 \leq t<1 \end{array} \text { and } T(t+1)=T(t) \text { for all } t \geq 0\right. \] \[ Y(s)= \]

Solution

Solution Steps

To solve the given initial value problem using the Laplace transform, follow these steps:

Take the Laplace transform of both sides of the differential equation: Use the properties of the Laplace transform for derivatives and the given initial conditions.
Transform the periodic function $ T(t) $: Use the Laplace transform for periodic functions.
Solve for $ Y(s) $: Isolate $ Y(s) $ in the transformed equation.

Step 1: Take the Laplace Transform of the Differential Equation

Given the differential equation: \[ y'' + 6y' + 17y = T(t) \] with initial conditions $ y(0) = 0 $ and $ y'(0) = 0 $, we take the Laplace transform of both sides. Using the properties of the Laplace transform for derivatives, we get: \[ s^2 Y(s) + 6s Y(s) + 17 Y(s) = \mathcal{L}\{T(t)\} \]

Step 2: Transform the Periodic Function $ T(t) $

The periodic function $ T(t) $ is defined as: \[ T(t) = \begin{cases} t, & 0 \leq t < \frac{1}{2} \\ 1 - t, & \frac{1}{2} \leq t < 1 \end{cases} \] and $ T(t+1) = T(t) $ for all $ t \geq 0 $.

Using the Laplace transform for periodic functions, we have: \[ \mathcal{L}\{T(t)\} = \frac{1}{1 - e^{-s}} \int_0^1 e^{-st} T(t) \, dt \]

Step 3: Compute the Laplace Transform of $ T(t) $

We compute the integral: \[ \int_0^1 e^{-st} T(t) \, dt = \int_0^{1/2} e^{-st} t \, dt + \int_{1/2}^1 e^{-st} (1 - t) \, dt \]

Evaluating these integrals: \[ \int_0^{1/2} e^{-st} t \, dt = \left[ \frac{t e^{-st}}{s} + \frac{e^{-st}}{s^2} \right]_0^{1/2} = \frac{1}{s^2} \left( 1 - e^{-s/2} \right) \] \[ \int_{1/2}^1 e^{-st} (1 - t) \, dt = \left[ \frac{e^{-st}}{s} - \frac{t e^{-st}}{s} - \frac{e^{-st}}{s^2} \right]_{1/2}^1 = \frac{1}{s^2} \left( e^{-s/2} - e^{-s} \right) \]

Combining these results: \[ \mathcal{L}\{T(t)\} = \frac{1}{1 - e^{-s}} \left( \frac{1}{s^2} \left( 1 - e^{-s/2} \right) + \frac{1}{s^2} \left( e^{-s/2} - e^{-s} \right) \right) = \frac{1}{s^2 (1 - e^{-s})} \]

Step 4: Solve for $ Y(s) $

Substitute $ \mathcal{L}\{T(t)\} $ into the transformed differential equation: \[ s^2 Y(s) + 6s Y(s) + 17 Y(s) = \frac{1}{s^2 (1 - e^{-s})} \]

Isolate $ Y(s) $: \[ Y(s) = \frac{1}{s^2 (1 - e^{-s}) (s^2 + 6s + 17)} \]