Q7.3-3E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1-20, determine the Laplace transform of the given function using Table 7.1 on page 356 and the properties of the transform given in Table 7.2. [Hint: In Problems 12-20, use an appropriate trigonometric identity.]

${\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{cos}\mathbf{3}\mathbf{t}\mathbf{+}{\mathbf{e}}^{\mathbf{6}\mathbf{t}}\mathbf{-}\mathbf{1}$.

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{\mathrm{s}+1}{{(\mathrm{s}+1)}^2+9}+\frac{1}{\mathrm{s}-6}-\frac{1}{\mathrm{s}}\text{ for }\mathrm{s}>6$.

Definition of Laplace transform

• The integral transform of a given derivative function with real variable t into a complex function with variable s is known as the Laplace transform. 
• Let f(t) be supplied for t(0), and assume that the function meets certain constraints that will be presented subsequently.
• The Laplace transform formula defines the Laplace transform of f(t), which is indicated by $\mathcal{L}\left\{f\left(t\right)\right\}$ or F(s).

Determine the Laplace transform for the given equation

Given that $e^{-t}\cos 3t+e^{6t}-1$.

Find the Laplace transform of the given function $e^{-t}\cos 3t+e^{6t}-1$ using the Laplace formula

$\mathcal{L}\left\{a{f}_1+b{f}_2\right\}=a\mathcal{L}\left\{f_1\right\}+b\mathcal{L}\left\{f_2\right\}$, $\mathcal{L}\left\{e^{at}cosbt\right\}=\frac{s-a}{{(s-a)}^2+b^2}$, $\mathcal{L}\left\{1\right\}=\frac{1}{s}$, $\mathcal{L}\left\{t^n\right\}=\frac{n!}{s^{n+1}}$and $\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a}$ as follows:

$\begin{array}{rcl}\mathcal{L}\left\{e^{-t}\cos 3t+e^{6t}-1\right\}& =& \mathcal{L}\left\{e^{-t}\cos 3t\right\}+\mathcal{L}\left\{e^{6t}\right\}-\mathcal{L}\left\{1\right\}\\ & =& \frac{s-(-1)}{{(s-(-1\left)\right)}^2+3^2}+\frac{1}{s-6}-\frac{1}{s}\\ & =& \frac{s+1}{{(s+1)}^2+9}+\frac{1}{s-6}-\frac{1}{s}\text{ for }s>6\end{array}$

Therefore, the Laplace transform for the given equation is 

$\frac{s+1}{{(s+1)}^2+9}+\frac{1}{s-6}-\frac{1}{s}\text{ for }s>6$