Question

Use the Laplace transform to solve the given initial- value problem.

$\mathbf{}\mathbf{2}\mathbf{.}\mathbf{}{\mathbf{y}}^{\mathbf{r}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{\delta }\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$

Short answer

The solution for the initial value problem is$y\left(t\right)=e^{-(t-1)}u(t-1)+2e^{-t}$

Step-by-step solution

Define Laplace transform:

The conversion of the function f (x) to the function $\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\int }\mathbf{\infty }\mathbf{0}\mathit{e}\mathbf{-}\mathit{x}\mathit{t}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}$ is particularly useful in reducing the solution of a standard dividing line equation with constant coefficients in the polynomial equation solution

Find y(t):

Using Laplace transform we get the differential equation subject to indicated initial conditions

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

Use Laplace transform on both side

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

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

Use Theorem 7.5.1, Transform of the Dirae Delta function

For$t_0>0$,$\mathcal{L}\left\{\delta \left(t-t_0\right)\right\}=e^{-2t_0}.$in above equation

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

$\left[sY\right(s)-y(0\left)\right]+Y\left(s\right)=e^{-s}$

Apply initial condition,

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

Use$y\left(t\right)$utilize inverse form of second translation theorem

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

Thus, the solution is

$y\left(t\right)=e^{-(t-1)}u(t-1)+2e^{-t}$