Q7.3-2E - 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.]

$\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{6}{s^3}+\frac{1}{s-2}\text{ for }s>2$.

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 $3t^2-e^{2t}$.

Find the Laplace transform of the given function $3t^2-e^{2t}$ using the Laplace formula

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

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

Therefore, the Laplace transform for the given equation is $\frac{6}{s^3}+\frac{1}{s-2}\text{ for }s>2$