pancha3

2021-08-22

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

diskusje5

Step 1

$\frac{dy}{dt}-y=1$
$\mathcal{L}\left\{\frac{dy}{dt}-y\right\}=\mathcal{L}\left\{1\right\}$
$\mathcal{L}\left\{\frac{dy}{dt}\right\}-\mathcal{L}\left\{y\right\}=\frac{1}{s}$
$sY\left(s\right)-y\left(0\right)-Y\left(s\right)=\frac{1}{s}$
$sY\left(s\right)-0-Y\left(s\right)=\frac{1}{s}$
$sY\left(s\right)-Y\left(s\right)=\frac{1}{s}$
$\left(s-1\right)Y\left(s\right)=\frac{1}{s}$
$Y\left(s\right)=\frac{1}{s\left(s-1\right)}$
$y\left(t\right)={\mathcal{L}}^{-1}\left\{Y\left(s\right)\right\}$
$={\mathcal{L}}^{-1}\left\{\frac{1}{s\left(s-1\right)}\right\}$
$={\mathcal{L}}^{-1}\left\{\frac{s-\left(s-1\right)}{s\left(s-1\right)}\right\}$
$={\mathcal{L}}^{-1}\left\{\frac{s}{s\left(s-1\right)}-\frac{s-1}{s\left(s-1\right)}\right\}$
$={\mathcal{L}}^{-1}\left\{\frac{1}{s-1}-\frac{1}{s}\right\}$
$={\mathcal{L}}^{-1}\left\{\frac{1}{s-1}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}$
$={e}^{t}-1$
That is
$y\left(t\right)={e}^{t}-1$

Do you have a similar question?