• <blockquote id="a6oys"></blockquote>
  • <tt id="a6oys"></tt>
  • Paul's Online Notes
    Paul's Online Notes
    Home / Differential Equations / Higher Order Differential Equations / Laplace Transforms
    Show General Notice Show Mobile Notice Show All Notes Hide All Notes
    General Notice

    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    Paul
    August 27, 2020

    Mobile Notice
    You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

    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.

    亚洲欧美中文日韩视频 - 视频 - 在线观看 - 影视资讯 - av网 <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <文本链> <文本链> <文本链> <文本链> <文本链> <文本链>