Q7.3 - 13E - 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.]

${\mathbf{sin}}^{\mathbf{2}}\mathbf{t}$.

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{2}{s^3+4s}$.

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 ${\sin }^{2}t$,

Find the Laplace transform of ${\sin }^{2}t$ using ${\sin }^{2}a=\frac{1}{2}(1-\cos 2a)$, $\mathcal{L}\left\{af\right(x)\pm bg(x\left)\right\}=a\mathcal{L}\left\{f\right\}\pm b\mathcal{L}\left\{g\right(t\left)\right\}$, $\mathcal{L}\left\{1\right\}=\frac{1}{s}$ and $\mathcal{L}\left\{\cos bt\right\}=\frac{s}{s^2+b^2}$ as:

$\begin{array}{rcl}\mathcal{L}\left\{{\sin }^{2}t\right\}& =& \mathcal{L}\left\{\frac{1}{2}(1-\cos 2t)\right\}\\ & =& \frac{1}{2}\left(\mathcal{L}\left\{1\right\}-\mathcal{L}\left\{\cos 2t\right\}\right)\\ & =& \frac{1}{2}\left(\frac{1}{s}-\frac{s}{s^2+4}\right)\\ & =& \frac{1}{2}\left(\frac{s^2+4-s^2}{s\left(s^2+4\right)}\right)\\ & & \end{array}$

Simplify the equation as follows:

$\begin{array}{rcl}\mathcal{L}\left\{{\sin }^{2}t\right\}& =& \frac{1}{2}\left(\frac{4}{s^3+4s}\right)\\ & =& \frac{2}{s^3+4s}\end{array}$

Therefore, the Laplace transform for the given equation is $\frac{2}{s^3+4s}$.