Q5RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 3–10, determine the Laplace transform of the given function.

${\mathit{e}}^{\mathbf{2}\mathbf{t}}\mathbf{-}{\mathit{t}}^{\mathbf{3}}\mathbf{+}{\mathit{t}}^{\mathbf{2}}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathbf{5}\mathit{t}$

Step-by-Step Solution

Answer

$L\{e^{2t}-t^3+t^2-sin5t\}=\frac{1}{s-2}-\frac{6}{s^4}+\frac{2}{s^3}-\frac{5}{s^2+25}$

Step 1:Given Information

 

The given function is .$e^{2t}-t^3+t^2-sin5t$

Step 2: Determining the Laplace transform:

Using the following Laplace transform definition, to find Laplace of given function.

$\begin{array}{c}L\left\{e^{at}\right\}\left(s\right)=\frac{1}{s-a}\\ L\left\{t^a\right\}\left(s\right)=\frac{a!}{s^{a+1}}\\ L\left\{sinat\right\}\left(s\right)=\frac{a}{s^2+a^2}\end{array}$

Considering the Laplace transform's linearity, we get

$\begin{array}{c}L\{e^{2t}-t^3+t^2-sin5t\}\left(s\right)=L\left\{e^{2t}\right\}\left(s\right)-L\left\{t^3\right\}\left(s\right)+L\left\{t^2\right\}\left(s\right)-L\left\{sin5t\right\}\left(s\right)\\ =\frac{1}{s-2}-\frac{3!}{s^4}+\frac{2!}{s^3}-\frac{5}{s^2+5^2}\\ =\frac{1}{s-2}-\frac{0}{s^4}+\frac{z}{s^3}-\frac{5}{s^2+25}\end{array}$

Step 3: Determining the Result

Thus, the required Laplace transform is .$L\{e^{2t}-t^3+t^2-sin5t\}=\frac{1}{s-2}-\frac{6}{s^4}+\frac{2}{s^3}-\frac{5}{s^2+25}$