MHB Inverse Laplace Transform problem

[Dustinsfl]

I can't seem to part of an inverse Laplace transform correct.

\begin{align*}
f(t) &= \frac{6}{5}\mathcal{L}^{-1}\bigg\{\frac{1}{s + 2}\bigg\} +
\frac{3}{5}\mathcal{L}^{-1}\bigg\{\frac{3s - 1}
{s^2 + 5s + 11}\bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{\big(s + \frac{5}{2}\big)^2 + \frac{19}{4}}\Bigg\}
- \frac{6}{5\sqrt{19}}\mathcal{L}^{-1}
\Bigg\{\frac{\frac{\sqrt{19}}{2}}
{\big(s + \frac{5}{2}\big)^2 + \frac{19}{4}}\Bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{s^2\big|_{s\to s + \frac{5}{2}} +
\frac{19}{4}}\Bigg\} - \frac{6}{5\sqrt{19}}\mathcal{L}^{-1}
\Bigg\{\frac{\frac{\sqrt{19}}{2}}
{s^2\big|_{s\to s + \frac{5}{2}} + \frac{19}{4}}\Bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}e^{-5/2t}
\cos\bigg(\frac{\sqrt{19}}{2}t\bigg) - \frac{6}{5\sqrt{19}}
e^{-5/2t}\sin\bigg(\frac{\sqrt{19}}{2}t\bigg)
\end{align*}

When I check this with Mathematica, I should have \(\frac{17}{5\sqrt{19}}e^{-5/2t}\sin\left(\frac{\sqrt{19}}{2}t\right)\).

What am I doing wrong?

 

[Chris L T521]

I can't seem to part of an inverse Laplace transform correct.

\begin{align*}
f(t) &= \frac{6}{5}\mathcal{L}^{-1}\bigg\{\frac{1}{s + 2}\bigg\} +
\frac{3}{5}\mathcal{L}^{-1}\bigg\{\frac{3s - 1}
{s^2 + 5s + 11}\bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{\big(s + \frac{5}{2}\big)^2 + \frac{19}{4}}\Bigg\}
- \frac{6}{5\sqrt{19}}\mathcal{L}^{-1}
\Bigg\{\frac{\frac{\sqrt{19}}{2}}
{\big(s + \frac{5}{2}\big)^2 + \frac{19}{4}}\Bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{s^2\big|_{s\to s + \frac{5}{2}} +
\frac{19}{4}}\Bigg\} - \frac{6}{5\sqrt{19}}\mathcal{L}^{-1}
\Bigg\{\frac{\frac{\sqrt{19}}{2}}
{s^2\big|_{s\to s + \frac{5}{2}} + \frac{19}{4}}\Bigg\}\\
&= \frac{6}{5}e^{-2t} + \frac{9}{5}e^{-5/2t}
\cos\bigg(\frac{\sqrt{19}}{2}t\bigg) - \frac{6}{5\sqrt{19}}
e^{-5/2t}\sin\bigg(\frac{\sqrt{19}}{2}t\bigg)
\end{align*}

When I check this with Mathematica, I should have \(\frac{17}{5\sqrt{19}}e^{-5/2t}\sin\left(\frac{\sqrt{19}}{2}t\right)\).

What am I doing wrong?

Careful!

\[\frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s}{s^2\big|_{s\to s + \frac{5}{2}} +
\frac{19}{4}}\Bigg\} \neq \frac{9}{5} e^{-5/2 t}\cos\Bigg(\frac{\sqrt{19}}{2}t\Bigg)\]

The result that you want is actually

\[\frac{9}{5}\mathcal{L}^{-1}
\Bigg\{\frac{s\color{red}{\big|_{s\to s+\frac{5}{2}}}}{s^2\big|_{s\to s + \frac{5}{2}} +
\frac{19}{4}}\Bigg\} = \frac{9}{5} e^{-5/2 t}\cos\Bigg(\frac{\sqrt{19}}{2}t\Bigg)\]

Let's start over with that second fraction; I would do things this way:

\[\begin{aligned}\frac{3}{5}\mathcal{L}^{-1}\Bigg\{\frac{3s-1}{s^2+5s+11}\Bigg\} &= \frac{9}{5}\mathcal{L}^{-1}\Bigg\{\frac{s-\frac{1}{3}}{\left(s+\frac{5}{2}\right)^2+\frac{19}{4}}\Bigg\}\\ &= \frac{9}{5}\mathcal{L}^{-1}\Bigg\{\frac{s+\frac{5}{2} - \frac{17}{6}}{\left(s+\frac{5}{2} \right)^2 + \frac{19}{4} } \Bigg\}\\ &= \frac{9}{5}\mathcal{L}^{-1}\Bigg\{ \frac{s+\frac{5}{2}}{\left(s+\frac{5}{2}\right)^2 + \frac{19}{4}}\Bigg\} - \frac{51}{10}\mathcal{L}^{-1}\Bigg\{ \frac{1}{\left(s+\frac{5}{2}\right)^2 + \frac{19}{4}} \Bigg\}\\ &= \frac{9}{5}\mathcal{L}^{-1}\Bigg\{ \frac{s\big|_{s\to s+\frac{5}{2}}}{ s^2\big|_{s\to s+\frac{5}{2}} + \frac{19}{4}}\Bigg\} - \frac{51}{5\sqrt{19}}\mathcal{L}^{-1}\Bigg\{ \frac{\frac{\sqrt{19}}{2} }{s^2\big|_{s\to s+ \frac{5}{2}} + \frac{19}{4}} \Bigg\} \\ &= \frac{9}{5} e^{-5/2 t} \cos\Bigg(\frac{\sqrt{19}}{2} t\Bigg) - \frac{51}{5\sqrt{19}}e^{-5/2 t} \sin\Bigg(\frac{\sqrt{19}}{2} t\Bigg) \end{aligned}\]

Putting this together with the inverse Laplace Transform of the first fraction gives us

\[\begin{aligned} f(t) &= \frac{6}{5}e^{-2t} + \frac{9}{5} e^{-5/2 t} \cos\Bigg(\frac{\sqrt{19}}{2} t\Bigg) - \frac{51}{5\sqrt{19}}e^{-5/2 t} \sin\Bigg(\frac{\sqrt{19}}{2} t\Bigg) \\ &= \frac{3}{5}e^{-5/2 t}\Bigg( 2e^{t/2} + 3\cos\Bigg(\frac{\sqrt{19}}{2} t\Bigg) - \frac{17}{\sqrt{19}} \sin\Bigg(\frac{\sqrt{19}}{2} t\Bigg) \Bigg)\\ &= \frac{3}{95}e^{-5/2 t}\Bigg( 38e^{t/2} + 57\cos\Bigg(\frac{\sqrt{19}}{2} t\Bigg) - 17\sqrt{19} \sin\Bigg(\frac{\sqrt{19}}{2} t\Bigg) \Bigg)\end{aligned}\]

Which matches the answer given to me by Mathematica:

I hope this clarifies where you made your mistake! (Smile)