• <blockquote id="a6oys"></blockquote>
  • <tt id="a6oys"></tt>
  • Paul's Online Notes
    Paul's Online Notes
    Home / Differential Equations / Systems of DE's / 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.

    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 5-11 : Laplace Transforms

    There’s not too much to this section. We’re just going to work an example to illustrate how Laplace transforms can be used to solve systems of differential equations.

    Example 1 Solve the following system. \[\begin{align*}{{x'}_1} & = 3{x_1} - 3{x_2} + 2 & \hspace{0.25in}{x_1}\left( 0 \right) & = 1\\ {{x'}_2} & = - 6{x_1} - t & \hspace{0.25in}{x_2}\left( 0 \right) & = - 1\end{align*}\]
    Show Solution

    First notice that the system is not given in matrix form. This is because the system won’t be solved in matrix form. Also note that the system is nonhomogeneous.

    We start just as we did when we used Laplace transforms to solve single differential equations. We take the transform of both differential equations.

    \[\begin{align*}s{X_1}\left( s \right) - {x_1}\left( 0 \right) & = 3{X_1}\left( s \right) - 3{X_2}\left( s \right) + \frac{2}{s}\\ s{X_2}\left( s \right) - {x_2}\left( 0 \right) & = - 6{X_1}\left( s \right) - \frac{1}{{{s^2}}}\end{align*}\]

    Now plug in the initial condition and simplify things a little.

    \[\begin{align*}\left( {s - 3} \right){X_1}\left( s \right) + 3{X_2}\left( s \right) & = \frac{2}{s} + 1 = \frac{{2 + s}}{s}\\ 6{X_1}\left( s \right) + s{X_2}\left( s \right) & = - \frac{1}{{{s^2}}} - 1 = - \frac{{{s^2} + 1}}{{{s^2}}}\end{align*}\]

    Now we need to solve this for one of the transforms. We’ll do this by multiplying the top equation by \(s\) and the bottom by -3 and then adding. This gives,

    \[\left( {{s^2} - 3s - 18} \right){X_1}\left( s \right) = 2 + s + \frac{{3{s^2} + 3}}{{{s^2}}}\]

    Solving for \(X_{1}\) gives,

    \[{X_1}\left( s \right) = \frac{{{s^3} + 5{s^2} + 3}}{{{s^2}\left( {s + 3} \right)\left( {s - 6} \right)}}\]

    Partial fractioning gives,

    \[{X_1}\left( s \right) = \frac{1}{{108}}\left( {\frac{{133}}{{s - 6}} - \frac{{28}}{{s + 3}} + \frac{3}{s} - \frac{{18}}{{{s^2}}}} \right)\]

    Taking the inverse transform gives us the first solution,

    \[{x_1}\left( t \right) = \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} - 28{{\bf{e}}^{ - 3t}} + 3 - 18t} \right)\]

    Now to find the second solution we could go back up and eliminate \(X_{1}\) to find the transform for \(X_{2}\) and sometimes we would need to do that. However, in this case notice that the second differential equation is,

    \[{x'_2} = - 6{x_1} - t\hspace{0.25in} \Rightarrow \hspace{0.25in}{x_2} = \int{{ - 6{x_1} - t}}\,dt\]

    So, plugging the first solution in and integrating gives,

    \[\begin{align*}{x_2}\left( t \right) & = - \frac{1}{{18}}\int{{133{{\bf{e}}^{6t}} - 28{{\bf{e}}^{ - 3t}} + 3}}\,dt\\ & = - \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} + 56{{\bf{e}}^{ - 3t}} + 18t} \right) + c\end{align*}\]

    Now, reapplying the second initial condition to get the constant of integration gives

    \[ - 1 = - \frac{1}{{108}}\left( {133 + 56} \right) + c\hspace{0.25in} \Rightarrow \hspace{0.25in}c = \frac{3}{4}\]

    The second solution is then,

    \[{x_2}\left( t \right) = - \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} + 56{{\bf{e}}^{ - 3t}} + 18t - 81} \right)\]

    So, putting all this together gives the solution to the system as,

    \[\begin{align*}{x_1}\left( t \right) & = \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} - 28{{\bf{e}}^{ - 3t}} + 3 - 18t} \right)\\ {x_2}\left( t \right) & = - \frac{1}{{108}}\left( {133{{\bf{e}}^{6t}} + 56{{\bf{e}}^{ - 3t}} + 18t - 81} \right)\end{align*}\]

    Compared to the last section the work here wasn’t too bad. That won’t always be the case of course, but you can see that using Laplace transforms to solve systems isn’t too bad in at least some cases.

    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>