Laplace Transform Example

\[\dot{x}-4x=e^t \; ,\; x(0)=f(o)=1\]

Applying the Laplace transform operator to both sides:

\[\mathop{\mathscr{L}}\{\dot{x}-4x\}=\mathop{\mathscr{L}}\{e^t\}\]

\[\mathop{\mathscr{L}}\{\dot{x}\}-4\mathop{\mathscr{L}}\{x\}=\frac{1}{s-1}\]

\[s\hat{f}(s)-f(0)-4\hat{f}(s)=\frac{1}{s-1}\]

\[(s-4)\hat{f}(s)-1=\frac{1}{s-1}\]

\[\hat{f}(s)=\frac{1}{(s-1)(s-4)}+\frac{1}{s-4}\]

And now for the inverse Laplace transform:

\[x(t)=\mathop{\mathscr{L}^{-1}}\{\hat{f}(s)\}=\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{(s-1)(s-4)}+\frac{1}{s-4}\right\}\]

\[x(t)=\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{(s-1)(s-4)}\right\}+\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{s-4}\right\}\]

The second term is easy:

\[\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{s-4}\right\}=e^{4t}\]

The first term requires partial fraction decomposition or completing the square and involving complex numbers.  I am going to show the latter.

\[(s-1)(s-4)=s^2-5s+4=\left(s-\frac{5}{2}\right)^2-\frac{9}{4}=\left(s-\frac{5}{2}\right)^2+\left(\frac{3i}{2}\right)^2\]

I am doing this with the Laplace transform for  \(\sin(bt)\) , which is  \(\frac{b}{s^2+b^2}\).  In this case  \(b=\frac{3i}{2}\).

\[\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{(s-1)(s-4)}\right\}=\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{\left(s-\frac{5}{2}\right)^2+\left(\frac{3i}{2}\right)^2}\right\}=\mathop{\mathscr{L}^{-1}}\left\{\frac{2}{3i}\cdot\frac{\frac{3i}{2}}{\left(s-\frac{5}{2}\right)^2+\left(\frac{3i}{2}\right)^2}\right\}\]

Since we have  \(s-\frac{5}{2}\)  instead of  \(s\) , we get a multiple of  \(e^{\frac{5}{2}t}\)  in the inverse transform.

\[\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{(s-1)(s-4)}\right\}=\frac{2}{3i}\cdot\sin\left(\frac{3i}{2}t\right)\cdot e^{\frac{5}{2}t}\]

One amazing fact about   \(\sin(ix)\)   is that it is equal to   \(i\cdot\sinh(x)\), the hyperbolic sine, which is equal to \(\frac{i}{2}\left(e^x-e^{-x}\right)\).

Hence

\[\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{(s-1)(s-4)}\right\}=\frac{2}{3i}\cdot i\cdot\sinh\left(\frac{3}{2}t\right)\cdot e^{\frac{5}{2}t}\]

which equals

\[\frac{2}{3}\frac{1}{2}\left(e^{\frac{3t}{2}}-e^{\frac{-3t}{2}}\right)\cdot e^{\frac{5t}{2}}=\frac{1}{3}e^{\frac{3t}{2}+\frac{5t}{2}}-\frac{1}{3}e^{\frac{-3t}{2}+\frac{5t}{2}}=\frac{1}{3}e^{4t}-\frac{1}{3}e^{t}\]

Crazy!  Now add on the \(e^{4t}\) and we get

\[x(t)=\frac{4}{3} e^{4t}-\frac{1}{3} e^{t}\]

Partial fractions will lead to the same solution and is actually a bit easier.  However, this was a wild ride, and well worth the trouble in my opinion.

If you don't have much time to screw around with this, then you could also look up \(\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{(s-1)(s-4)}\right\}\) in a table of Laplace transforms.

One Response to “Laplace Transform Example”

  1. gary says:

    Interesting
    \(e^{i\theta}=\cos(\theta)+i\sin(\theta)\) , so, \(e^{i(ix)}=\cos(ix)+i\sin(ix)\), but \(e^{i(ix)}=e^{-x}\)
    \begin{equation}e^{-x}=\cos(ix)+i\sin(ix)\end{equation}
    and
    \(e^{x}=\cos(-ix)+i\sin(-ix)\)
    so
    \begin{equation}e^{x}=\cos(ix)-i\sin(ix)\end{equation}

    If we subtract equation \((1)\) from equation \((2)\) we get

    \[e^{x}-e^{-x}=-2i\sin(ix)\]


    Solving for \(\sin(ix)\) we get

    \[\sin(ix)=i\frac{e^{x}-e^{-x}}{2}=i\sinh(x)\]

    Beautiful.

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