Q16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the Laplace transformation table and the linearity of the Laplace transform to determine the following transforms.

\(L\left\{ {{{\bf{t}}^{\bf{2}}}{\bf{ - 3t - 2}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin3t}}} \right\}\)

 

Step-by-Step Solution

Answer

The Laplace transform of the given function is \(\frac{2}{{{s^3}}} - \frac{3}{{{s^2}}} - \frac{6}{{{{\left( {s + 1} \right)}^2} + 9}}\).

Step 1: Given data

The given function is \(f\left( t \right) = {{\bf{t}}^{\bf{2}}}{\bf{ - 3t - 2}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin3t}}\).

 

The objective is to find the Laplace transformation.

Step 2: Formulas of Laplace transform

Some important formulas of Laplace transform:

 

\(\begin{array}{l}\left\{ {{t^n}} \right\}\left( s \right) = \frac{{n!}}{{{s^{n + 1}}}}\\\left\{ {{e^{ - at}}\sin bt} \right\}\left( s \right) = \frac{b}{{{{\left( {s + a} \right)}^2} + {b^2}}}\end{array}\)

Step 3: Find the Laplace transform of the given function.

 

Recall the linearity of Laplace transform:

\(L\left( {{f_1} + {f_2}} \right) = L\left( {{f_1}} \right) + L\left( {{f_2}} \right)\) and \(L\left( {{\rm{c}}f} \right) = {\rm{c}}L\left( f \right)\)

 

Now, apply the Laplace transform in the given function as follows:

 

\(\begin{array}{c}\left\{ f \right\}\left( s \right) = \left\{ {{t^2} - 3t - 2{e^{ - t}}\sin 3t} \right\}\left( s \right)\\ = \left\{ {{t^2}} \right\}\left( s \right) - 3\left\{ t \right\}\left( s \right) - 2\left\{ {{e^{ - t}}\sin 3t} \right\}\left( s \right)\;\;\;\;\;\;\;\;\;\;\;...\left( 1 \right)\end{array}\)

 

Consider,

\(\begin{array}{c}\left\{ {{t^2}} \right\}\left( s \right) = \frac{{2!}}{{{s^{2 + 1}}}}{\rm{ }}\left( {{\rm{By Laplace transform table}}} \right)\\ = \frac{2}{{{s^3}}}\end{array}\)

Also,

\(\begin{array}{c}\left\{ t \right\}\left( s \right) = \frac{{1!}}{{{s^{1 + 1}}}}\\ = \frac{1}{{{s^2}}}\end{array}\)

And,

 

 

 

Substitute value of \(\left\{ {{t^2}} \right\}\left( s \right)\), \(\left\{ t \right\}\left( s \right)\) and \(\left\{ {{e^{ - t}}\sin 3t} \right\}\left( s \right)\)in equation (1) and simplify.

\(\begin{array}{c}\left\{ {{t^2} - 3t - 2{e^{ - t}}\sin 3t} \right\}\left( s \right) = \frac{2}{{{s^3}}} - 3\left( {\frac{1}{{{s^2}}}} \right) - 2\left( {\frac{3}{{{{\left( {s + 1} \right)}^2} + 9}}} \right)\\ = \frac{2}{{{s^3}}} - \frac{3}{{{s^2}}} - \frac{6}{{{{\left( {s + 1} \right)}^2} + 9}}\end{array}\)

 

Hence, the Laplace transform is \(\left\{ {{t^2} + {e^t}\sin 2t} \right\}\left( s \right) = \frac{2}{{{s^3}}} - \frac{3}{{{s^2}}} - \frac{6}{{{{\left( {s + 1} \right)}^2} + 9}}\).