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        Paul's Online Notes
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        Home / Differential Equations / Higher Order Differential Equations / Laplace Transforms
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        Section 7-5 : Laplace Transforms

        There really isn’t all that much to this section. All we’re going to do here is work a quick example using Laplace transforms for a 3rd order differential equation so we can say that we worked at least one problem for a differential equation whose order was larger than 2.

        Everything that we know from the Laplace Transforms chapter is still valid. The only new bit that we’ll need here is the Laplace transform of the third derivative. We can get this from the general formula that we gave when we first started looking at solving IVP’s with Laplace transforms. Here is that formula,

        \[\mathcal{L}\left\{ {y'''} \right\} = {s^3}Y\left( s \right) - {s^2}y\left( 0 \right) - sy'\left( 0 \right) - y''\left( 0 \right)\]

        Here’s the example for this section.

        Example 1 Solve the following IVP. \[y''' - 4y'' = 4t + 3{u_6}\left( t \right){{\bf{e}}^{30 - 5t}},\hspace{0.25in}y\left( 0 \right) = - 3\,\,\,y'\left( 0 \right) = 1\,\,\,\,\,y''\left( 0 \right) = 4\]
        Show Solution

        As always, we first need to make sure the function multiplied by the Heaviside function has been properly shifted.

        \[y''' - 4y'' = 4t + 3{u_6}\left( t \right){{\bf{e}}^{ - 5\left( {t - 6} \right)}}\]

        It has been properly shifted and we can see that we’re shifting \({{\bf{e}}^{ - 5t}}\). All we need to do now is take the Laplace transform of everything, plug in the initial conditions and solve for \(Y\left( s \right)\). Doing all of this gives,

        \[\begin{align*}{s^3}Y\left( s \right) - {s^2}y\left( 0 \right) - sy'\left( 0 \right) - y''\left( 0 \right) - 4\left( {{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)} \right) & = \frac{4}{{{s^2}}} + \frac{{3{{\bf{e}}^{ - 6s}}}}{{s + 5}}\\ \hspace{0.25in}\hspace{0.25in}\left( {{s^3} - 4{s^2}} \right)Y\left( s \right) + 3{s^2} - 13s & = \frac{4}{{{s^2}}} + \frac{{3{{\bf{e}}^{ - 6s}}}}{{s + 5}}\\ \hspace{0.25in}\hspace{0.25in}\left( {{s^3} - 4{s^2}} \right)Y\left( s \right) & = \frac{4}{{{s^2}}} - 3{s^2} + 13s + \frac{{3{{\bf{e}}^{ - 6s}}}}{{s + 5}}\\ \hspace{0.25in}\hspace{0.25in}\left( {{s^3} - 4{s^2}} \right)Y\left( s \right) & = \frac{{4 - 3{s^4} + 13{s^3}}}{{{s^2}}} + \frac{{3{{\bf{e}}^{ - 6s}}}}{{s + 5}}\\ \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\,\,\,\,\,\,\,\,Y\left( s \right) & = \frac{{4 - 3{s^4} + 13{s^3}}}{{{s^4}\left( {s - 4} \right)}} + \frac{{3{{\bf{e}}^{ - 6s}}}}{{{s^2}\left( {s - 4} \right)\left( {s + 5} \right)}}\\ \hspace{0.25in}\hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\,Y\left( s \right) & = F\left( s \right) + 3{{\bf{e}}^{ - 6s}}G\left( s \right)\end{align*}\]

        Now we need to partial fraction and inverse transform \(F(s)\) and \(G(s)\). We’ll leave it to you to verify the details.

        \[\begin{align*}F\left( s \right) & = \frac{{4 - 3{s^4} + 13{s^3}}}{{{s^4}\left( {s - 4} \right)}} = \frac{{\frac{{17}}{{64}}}}{{s - 4}} - \frac{{\frac{{209}}{{64}}}}{s} - \frac{{\frac{1}{{16}}}}{{{s^2}}} - \frac{{\frac{1}{4}\left( {\frac{{2!}}{{2!}}} \right)}}{{{s^3}}} - \frac{{1\left( {\frac{{3!}}{{3!}}} \right)}}{{{s^4}}}\\ f\left( t \right) & = \frac{{17}}{{64}}{{\bf{e}}^{4t}} - \frac{{209}}{{64}} - \frac{1}{{16}}t - \frac{1}{8}{t^2} - \frac{1}{6}{t^3}\end{align*}\]

        \[\begin{align*}G\left( s \right) & = \frac{1}{{{s^2}\left( {s - 4} \right)\left( {s + 5} \right)}} = \frac{{\frac{1}{{144}}}}{{s - 4}} - \frac{{\frac{1}{{225}}}}{{s + 5}} - \frac{{\frac{1}{{400}}}}{s} - \frac{{\frac{1}{{20}}}}{{{s^2}}}\\ g\left( t \right) & = \frac{1}{{144}}{{\bf{e}}^{4t}} - \frac{1}{{225}}{{\bf{e}}^{ - 5t}} - \frac{1}{{400}} - \frac{1}{{20}}t\end{align*}\]

        Okay, we can now get the solution to the differential equation. Starting with the transform we get,

        \[Y\left( s \right) = F\left( s \right) + 3{{\bf{e}}^{ - 6s}}G\left( s \right)\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}y\left( t \right) = f\left( t \right) + 3{u_6}\left( t \right)g\left( {t - 6} \right)\]

        where \(f(t)\) and \(g(t)\) are the functions shown above.

        Okay, there is the one Laplace transform example with a differential equation with order greater than 2. As you can see the work in identical except for the fact that the partial fraction work (which we didn’t show here) is liable to be messier and more complicated.