Question

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

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

Short answer

The solution for the given problem using Laplace transform is,$y\left(t\right)=\frac{1}{2}+\frac{1}{2}U(t-1)-\frac{1}{2}{e}^{-2t}-\frac{1}{2}U(t-1)e^{-2(t-1)}$

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

$F\left(s\right)={\int }_0^{+\infty }f\left(t\right)·e^{-s·t}·dt$

Where, F(s)= Laplace Transform

S = complex number

t= real number >=0

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

Apply Laplace transform to solve the given initial value:

Formula to solve is,

$L\left[\delta \left(t-t_0\right)\right]=e^{-s{t}_0}$

$$\begin{gathered} L\left[y\text{'}\text{'}\right]=s^{2}L\left[y\right]-sy\left(0\right)-y\text{'}\left(0\right) \\ L\left[y\text{'}\right]=sL\left[y\right]-y\left(0\right) \end{gathered}$$

Take Laplace transform on both sides Using Linearity of Laplace transform we have,

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

$L\left[y\text{'}\text{'}+2y\text{'}\right]=L\left[\delta \right(t-1\left)\right]$

$L\left[y\text{'}\text{'}\right]+2L\left[y\text{'}\right]=L\left[\delta \right(t-1\left)\right]$

$s^{2}L\left[y\right]-sy\left(0\right)-y\text{'}\left(0\right)+sL\left[y\right]-y\left(0\right)=e^{-s}$

$$\begin{gathered} L\left[y\right]\left(s^2+2s\right)=e^{-s}+1 \\ L\left[y\right]=\frac{e^{-s}+1}{s(s+2)} \end{gathered}$$

Use of Techniques:

Techniques for partial fractions we have,

$\frac{e^{-s}+1}{s(s+2)}=\frac{A}{s}+\frac{B}{s+2}$

$e^{-s}+1=A(s+2)+Bs$

$\to A+B=0\to B=-A$