Question

In Problems 1–12 use the Laplace transform to solve the given initial value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathit{\pi }\mathbf{)}\mathbf{+}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{-}\mathbf{4}\mathit{\pi }\mathbf{)}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution is$y\left(t\right)=\sin tu(t-2\pi )+\sin tu(t-4\pi )+\cos t$ .

Step-by-step solution

Define Laplace Transform.

The use of Laplace transformation is to convertdifferential equations into algebraic equations. The formula for Laplace transform is

${\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{̀}\mathbf{o}}_{\mathbf{0}}^{\mathbf{+}\mathbf{\yen }}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\times }{\mathbf{e}}^{\mathbf{-}\mathbf{s}\mathbf{\times }\mathbf{t}}\mathbf{\times }\mathbf{dt}$

Where, F(s) = Laplace Transform

S = complex number

t =real number >=0

t’ = first derivative of the function f(t)

Use Laplace Transform.

To find the solution for given differential equation indicated with initial conditions using Laplace transform.

$y\text{'}\text{'}+y=\delta (t-2\pi )+\delta (t-4\pi )y\left(0\right)=1y\text{'}\left(0\right)=0$

Applying Laplace operator in both sides of expression we get,

$\mathcal{L}\left\{y\text{'}\text{'}\right\}+\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{\delta \right(t-2\pi \left)\right\}+\mathcal{L}\left\{\delta \right(t-2\pi \left)\right\}$

We know,

$\mathcal{L}\left\{f^n\left(t\right)\right\}=s^{n}F\left(s\right)-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-\dots s{f}^{n-2}\left(0\right)-f^{n-1}\left(0\right)$

From property of Dirac delta function,

${\int }_0^{\infty }f\left(t\right)\delta \left(t-t_0\right)=f\left(t_0\right)$

So, the transform became

$\left[s^{2}Y\left(s\right)-sy\left(0\right)-y\left(0\right)\right]+Y\left(s\right)=e^{-2\pi s}+e^{-4\pi s}$

After applying initial Conditions, we get

$Y\left(s\right)=\frac{e^{-2\pi s}}{s^2+1}+\frac{e^{-4\pi s}}{s^2+1}+\frac{s}{s^2+1}$

For evaluating y(t) utilize inverse form of second translation theorem

${\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}={\mathcal{L}}^{-1}\left\{\frac{e^{-2\pi s}}{s^2+1}\right\}+{\mathcal{L}}^{-1}\left\{\frac{e^{-4\pi s}}{s^2+1}\right\}+{\mathcal{L}}^{-1}\left\{\frac{s}{s^2+1}\right\}$

$y\left(t\right)=\sin (t-2\pi )u(t-2\pi )+\sin (t-4\pi )u(t-4\pi )+\cos t$

Using trigonometric properties we get the solution,

.$y\left(t\right)=\sin tu(t-2\pi )+\sin tu(t-4\pi )+\cos t$