Q7.3-5E - 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.]

$\mathbf{2}{\mathbf{t}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{-}\mathbf{t}\mathbf{+}\mathbf{cos}\mathbf{4}\mathbf{t}$.

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{4}{{(s+1)}^3}-\frac{1}{s^2}+\frac{s}{s^2+16}\text{ for }s>0$.

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 $2t^2{e}^{-t}-t+\cos 4t$

Find the Laplace transform of the given function $2t^2{e}^{-t}-t+\cos 4t$ using the Laplace formula $\mathcal{L}\left\{a{f}_1+b{f}_2\right\}=a\mathcal{L}\left\{f_1\right\}+b\mathcal{L}\left\{f_2\right\}$, $\mathcal{L}\left\{t^n\right\}=\frac{n!}{s^{n+1}}$, $\mathcal{L}\left\{cosat\right\}=\frac{s}{s^2+a^2}$and $\mathcal{L}\left\{t^n{e}^{at}\right\}=\frac{n!}{{(s-a)}^{n+1}}$ as follows:

$\begin{array}{rcl}\mathcal{L}\left\{2t^2{e}^{-t}-t+\cos 4t\right\}& =& 2\mathcal{L}\left\{t^2{e}^{-t}\right\}-\mathcal{L}\left\{t\right\}+\mathcal{L}\left\{\cos 4t\right\}\\ & =& 2·\frac{2!}{{(s-(-1\left)\right)}^{2+1}}-\frac{1!}{s^{1+1}}+\frac{s}{s^2+4^2}\\ & =& \frac{4}{{(s+1)}^3}-\frac{1}{s^2}+\frac{s}{s^2+16}\text{ for }s>0\end{array}$

Therefore, the Laplace transform for the given equation is 

$\frac{4}{{(s+1)}^3}-\frac{1}{s^2}+\frac{s}{s^2+16}\text{ for }s>0$