Section D.1 Orphaned Content

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Additional Narrative.

So far, we have seen three examples of how to find a Laplace transform of a given function, but we have not discussed why this idea is relevant to differential equations. In the next section, we will tie these two concepts together and illustrate how Laplace transform can be used to solve a differential equation.

Additional Examples.

Example D.11.

Find the Laplace transform of the function \(y(t) = 15 - 4e^{9t} + 11t^3.\)

Solution. Solution

We will use properties in the table as follows.

\begin{align*} Y(s) =\amp\ \lap{ y(t) }\\ =\amp\ \lap{ 15 - 4e^{9t} + 11t^3 }\\ =\amp\ 15\lap{ 1 } - 4\lap{ e^{9t} } + 11\lap{ t^3 } \qquad (\knowl{./knowl/xref/LT-Table-L9.html}{\text{\(L9\)}})\\ =\amp\ {\color{blue}\us{s \gt 0}{{\ub{{\color{black}15\cdot \frac{1}{s}}}}}\ {\color{black}-\ } \us{s \gt 9}{\ub{{\color{black}4\cdot \frac{1}{s - 9}}}}\ {\color{black}+\ } \us{s \gt 0}{\ub{{\color{black}11 \cdot \frac{3!}{s^{3 + 1}}}}}} \qquad ( \knowl{./knowl/xref/LT-Table-L1.html}{\text{\(L1\)}}, \knowl{./knowl/xref/LT-Table-L2.html}{\text{\(L2\)}}, \knowl{./knowl/xref/LT-Table-L3.html}{\text{\(L3\)}})\\ =\amp\ \frac{15}{s} - \frac{4}{s-9} + \frac{66}{s^4}, \hspace{0.5cm} s \gt 9 \end{align*}

Example D.12.

Find \(\lap{ e^{3t}\sin(6t) - t^3e^{-5t}}.\)

Solution. Solution

We will use properties in the table as follows.

\begin{align*} =\amp\ \lap{ e^{3t}\sin(6t) - t^3e^{-5t} }\\ =\amp\ \lap{ e^{3t}\sin(6t) } - \lap{ t^2e^{-5t} } \qquad (\knowl{./knowl/xref/LT-Table-L9.html}{\text{\(L9\)}}) \\ =\amp\ \ub{\frac{6}{(s-3)^2 + 6^2}}_{s \gt 3} - \ub{\frac{2!}{\Big(s - (-5)\Big)^{2+1}}}_{s >-5} \qquad (\knowl{./knowl/xref/LT-Table-L7.html}{\text{\(L7\)}}, \knowl{./knowl/xref/LT-Table-L6.html}{\text{\(L6\)}}) \\ =\amp\ \frac{6}{(s-3)^2 + 36} - \frac{2}{(s+5)^3}, \hspace{0.5cm} s \gt 3 \end{align*}

Example D.13.

Find the Laplace transform of the function \(g(t) = t^2 \cos(8t).\)

Solution. Solution

Before we begin, we note that it’s very tempting to think that because we know the Laplace transforms of both \(t^2\) and \(\cos(8t),\) we can simply multiply those together to get the desired Laplace transform. However, this is not the case, just as similar statements were not true for finding the derivatives and integrals of the products of functions. Rather, we will need to use property \(L13\), with \(n = 2\) and \(f(t) = \cos(8t).\)

\begin{align*} G(s) =\amp\ \lap{ g(t) }\\ =\amp\ \lap{ t^2 \cos(8t) }\\ =\amp\ \lap{ t^2 f(t) }\\ =\amp\ (-1)^2\cdot \frac{d^2}{ds^2}\big( F(s) \big) \end{align*}

We need to know what \(F(s)\) is before we can proceed. Let’s go back to the naming system we have instituted. If we have a capital \(F(s),\) that is the Laplace transform of a function lower case \(f(t).\) We identified that function previously: \(f(t) = \cos(8t).\) We use \(L5\) to find its Laplace transform.

\begin{equation*} F(s) = \frac{s}{s^2 + 64}, s >0 \end{equation*}

Then we continue finding \(G(s)\) by taking two derivatives (using the quotient rule for derivatives; details are omitted here).

\begin{align*} G(s) =\amp\ (-1)^2\cdot \frac{d^2}{ds^2}\big( F(s) \big)\\ =\amp\ 1 \cdot \frac{d^2}{ds^2}\left( \frac{s}{s^2 + 64} \right)\\ =\amp\ \frac{d}{ds}\left( \frac{-s^2 + 64}{(s^2 + 64)^2} \right)\\ =\amp\ \frac{2s^3 + 128s}{(s^2 + 64)^3} \end{align*}

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