Q7.3 - 4E - 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.]

$\mathbf{3}{\mathbf{t}}^{\mathbf{4}}\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{72}{s^5}-\frac{4}{s^3}+\frac{1}{s}\text{ for }s>0$.

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, $3t^4-2t^2+1$

Find the Laplace transform of the given function $3t^4-2t^2+1$ using the Laplace formula

$\mathcal{L}\left\{{\mathrm{af}}_1+{\mathrm{bf}}_2\right\}=\mathrm{a}\mathcal{L}\left\{{\mathrm{f}}_1\right\}+\mathrm{b}\mathcal{L}\left\{{\mathrm{f}}_2\right\}$, $\mathcal{L}\left\{t^n\right\}=\frac{n!}{s^{n+1}}$ and $\mathcal{L}\left\{1\right\}=\frac{1}{s}$ as follows:

$\begin{array}{rcl}\mathcal{L}\left\{3t^4-2t^2+1\right\}& =& 3\mathcal{L}\left\{t^4\right\}-2\mathcal{L}\left\{t^2\right\}+\mathcal{L}\left\{1\right\}\\ & =& 3\times \frac{4!}{s^{4+1}}-2\times \frac{2!}{s^{2+1}}+\frac{1}{s}\\ & =& 3\times \frac{24}{s^5}-2\times \frac{2}{s^3}+\frac{1}{s}\\ & =& \frac{72}{s^5}-\frac{4}{s^3}+\frac{1}{s}\text{ for }s>0\end{array}$

Therefore, the Laplace transform for the given equation is $\frac{72}{s^5}-\frac{4}{s^3}+\frac{1}{s}\text{ for }s>0$.