Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 4\right) - 2 \, e^{\left(2 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{2}{s - 2} + 3 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -4 \, \delta\left(t - 2\right) - 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} - \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 1\right) + 4 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 4\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{5 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 2} + 3 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 5 \, \delta\left(t - 1\right) + 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 4} + 5 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 3} + 3 \, e^{\left(-3 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 4\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} + \frac{4}{s - 3} + 3 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -4 \, \delta\left(t - 4\right) + 2 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 5 \, \delta\left(t - 1\right) - 2 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{2}{s - 2} + 5 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -3 \, \delta\left(t - 1\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -5 \, \delta\left(t - 5\right) + 2 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 4} - 5 \, e^{\left(-5 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -2 \, \delta\left(t - 1\right) - 5 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 5} - 2 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -5 \, \delta\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 4} - 5 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 4\right) + 4 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 4\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{5 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 2} + 3 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -5 \, \delta\left(t - 1\right) - 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{5}{s - 2} - 5 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -4 \, \delta\left(t - 1\right) + 3 \, e^{\left(4 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{3}{s - 4} - 4 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 1\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s - 2} + 3 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 2 \, \delta\left(t - 5\right) - 5 \, e^{\left(2 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 2} + 2 \, e^{\left(-5 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -5 \, \delta\left(t - 3\right) - 4 \, e^{\left(5 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-3 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 4 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{5}{s - 4} + 4 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -2 \, \delta\left(t - 2\right) - 5 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{5}{s - 3} - 2 \, e^{\left(-2 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -3 \, \delta\left(t - 4\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} - 3 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 1\right) - 3 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 2} + 3 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -5 \, \delta\left(t - 2\right) + 3 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 5} - 5 \, e^{\left(-2 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -5 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 4} - 5 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -3 \, \delta\left(t - 3\right) + 4 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 5} - 3 \, e^{\left(-3 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -2 \, \delta\left(t - 3\right) + 5 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 3} - 2 \, e^{\left(-3 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 4} + 3 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -5 \, \delta\left(t - 4\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{2 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 4} - 5 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 5 \, \delta\left(t - 1\right) + 4 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{4}{s - 2} + 5 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 5\right) - 3 \, e^{\left(3 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-5 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 5 \, \delta\left(t - 4\right) - 4 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 5} + 5 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -4 \, \delta\left(t - 5\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 2} - 4 \, e^{\left(-5 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -3 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{5}{s - 4} - 3 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 4 \, \delta\left(t - 5\right) - 4 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{5 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} + 4 \, e^{\left(-5 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 2 \, \delta\left(t - 2\right) - 5 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 5} + 2 \, e^{\left(-2 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 3 \, \delta\left(t - 3\right) - 5 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 5} + 3 \, e^{\left(-3 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 4 \, \delta\left(t - 1\right) - 5 \, e^{\left(3 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 3} + 4 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 2 \, \delta\left(t - 3\right) - 5 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 5} + 2 \, e^{\left(-3 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -4 \, \delta\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} - 4 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -4 \, \delta\left(t - 2\right) - 4 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 5} - 4 \, e^{\left(-2 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -3 \, \delta\left(t - 1\right) + 3 \, e^{\left(2 \, t\right)} + 3 \, \mathrm{u}\left(t - 3\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{3 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 2} - 3 \, e^{\left(-s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -2 \, \delta\left(t - 3\right) + 3 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 5} - 2 \, e^{\left(-3 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -2 \, \delta\left(t - 2\right) - 2 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 4} - 2 \, e^{\left(-2 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 5 \, \delta\left(t - 5\right) + 3 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 3} + 5 \, e^{\left(-5 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 2 \, \delta\left(t - 2\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 3} + 2 \, e^{\left(-2 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -5 \, \delta\left(t - 3\right) + 4 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} + \frac{4}{s - 4} - 5 \, e^{\left(-3 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} - \frac{2}{s - 3} - 5 \, e^{\left(-4 \, s\right)} \]
Compute the Laplace transform \(\mathcal{L}\{y\}\) of \(y = 5 \, \delta\left(t - 4\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 1\right) \) by using a transform table.
Then show how the integral definition of the Laplace transform to obtains same result.
Answer:
\[ \mathcal{L}\{y\} = -\frac{2 \, e^{\left(-s\right)}}{s} - \frac{2}{s - 5} + 5 \, e^{\left(-4 \, s\right)} \]