Differential Equations - IVP's With Step Functions (2024)

First let’s rewrite the forcing function to make sure that it’s being shifted correctly and to identify the function that is actually being shifted.

\[y'' - y' + 5y = 4 + {u_2}\left( t \right){{\bf{e}}^{ - 2\left( {t - 2} \right)}}\]

So, it is being shifted correctly and the function that is being shifted is \({{\bf{e}}^{ - 2t}}\). Taking the Laplace transform of everything and plugging in the initial conditions gives,

\[\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - \left( {sY\left( s \right) - y\left( 0 \right)} \right) + 5Y\left( s \right) & = \frac{4}{s} + \frac{{{{\bf{e}}^{ - 2s}}}}{{s + 2}}\\ \left( {{s^2} - s + 5} \right)Y\left( s \right) - 2s + 3 & = \frac{4}{s} + \frac{{{{\bf{e}}^{ - 2s}}}}{{s + 2}}\end{align*}\]

Now solve for \(Y(s)\).

\[\begin{align*}\left( {{s^2} - s + 5} \right)Y\left( s \right) & = \frac{4}{s} + \frac{{{{\bf{e}}^{ - 2s}}}}{{s + 2}} + 2s - 3\\ \left( {{s^2} - s + 5} \right)Y\left( s \right) & = \frac{{2{s^2} - 3s + 4}}{s} + \frac{{{{\bf{e}}^{ - 2s}}}}{{s + 2}}\\ Y\left( s \right) & = \frac{{2{s^2} - 3s + 4}}{{s\left( {{s^2} - s + 5} \right)}} + {{\bf{e}}^{ - 2s}}\frac{1}{{\left( {s + 2} \right)\left( {{s^2} - s + 5} \right)}}\\ Y\left( s \right) & = F\left( s \right) + {{\bf{e}}^{ - 2s}}G\left( s \right)\end{align*}\]

Notice that we combined a couple of terms to simplify things a little. Now we need to partial fraction \(F(s)\) and \(G(s)\). We’ll leave it to you to check the details of the partial fractions.

\[\begin{align*}F\left( s \right) & = \frac{{2{s^2} - 3s + 4}}{{s\left( {{s^2} - s + 5} \right)}} = \frac{1}{5}\left( {\frac{4}{s} + \frac{{6s - 11}}{{{s^2} - s + 5}}} \right)\\ G\left( s \right) & = \frac{1}{{\left( {s + 2} \right)\left( {{s^2} - s + 5} \right)}} = \frac{1}{{11}}\left( {\frac{1}{{s + 2}} - \frac{{s - 3}}{{{s^2} - s + 5}}} \right)\end{align*}\]

We now need to do the inverse transforms on each of these. We’ll start with \(F(s)\).

\[\begin{align*}F\left( s \right) & = \frac{1}{5}\left( {\frac{4}{s} + \frac{{6\left( {s - \frac{1}{2} + \frac{1}{2}} \right) - 11}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}}} \right)\\ & = \frac{1}{5}\left( {\frac{4}{s} + \frac{{6\left( {s - \frac{1}{2}} \right)}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}} - \frac{{8\frac{{\sqrt {19} }}{2}\frac{2}{{\sqrt {19} }}}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}}} \right)\\ f\left( t \right) & = \frac{1}{5}\left( {4 + 6{{\bf{e}}^{\frac{t}{2}}}\cos \left( {\frac{{\sqrt {19} }}{2}t} \right) - \frac{{16}}{{\sqrt {19} }}{{\bf{e}}^{\frac{t}{2}}}\sin \left( {\frac{{\sqrt {19} }}{2}t} \right)} \right)\end{align*}\]

Now \(G(s)\).

\[\begin{align*}G\left( s \right) & = \frac{1}{{11}}\left( {\frac{1}{{s + 2}} - \frac{{s - \frac{1}{2} + \frac{1}{2} - 3}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}}} \right)\\ & = \frac{1}{{11}}\left( {\frac{1}{{s + 2}} - \frac{{s - \frac{1}{2}}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}} + \frac{{\frac{5}{2}\frac{{\sqrt {19} }}{{\sqrt {19} }}}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}}} \right)\\ g\left( t \right) & = \frac{1}{{11}}\left( {{{\bf{e}}^{ - 2t}} - {{\bf{e}}^{\frac{t}{2}}}\cos \left( {\frac{{\sqrt {19} }}{2}t} \right) + \frac{5}{{\sqrt {19} }}{{\bf{e}}^{\frac{t}{2}}}\sin \left( {\frac{{\sqrt {19} }}{2}t} \right)} \right)\end{align*}\]

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

\[\begin{align*}Y\left( s \right) & = F\left( s \right) + {{\bf{e}}^{ - 2s}}G\left( s \right)\\ y\left( t \right) & = f\left( t \right) + {u_2}\left( t \right)g\left( {t - 2} \right)\end{align*}\]

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

Let’s rewrite the differential equation so we can identify the function that is actually being shifted.

\[y'' - y' = \cos \left( {2t} \right) + \cos \left( {2\left( {t - 6} \right)} \right){u_6}\left( t \right)\]

So, the function that is being shifted is \(\cos \left( {2t} \right)\) and it is being shifted correctly. Taking the Laplace transform of everything and plugging in the initial conditions gives,

\[\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - \left( {sY\left( s \right) - y\left( 0 \right)} \right) & = \frac{s}{{{s^2} + 4}} + \frac{{s{{\bf{e}}^{ - 6s}}}}{{{s^2} + 4}}\\ \left( {{s^2} - s} \right)Y\left( s \right) + 4s - 4 & = \frac{s}{{{s^2} + 4}} + \frac{{s{{\bf{e}}^{ - 6s}}}}{{{s^2} + 4}}\end{align*}\]

Now solve for \(Y(s)\).

\[\begin{align*}\left( {{s^2} - s} \right)Y\left( s \right) & = \frac{{s + s{{\bf{e}}^{ - 6s}}}}{{{s^2} + 4}} - 4s + 4\\ Y\left( s \right) & = \frac{{s\left( {1 + {{\bf{e}}^{ - 6s}}} \right)}}{{s\left( {s - 1} \right)\left( {{s^2} + 4} \right)}} - 4\frac{{s - 1}}{{s\left( {s - 1} \right)}}\\ & = \frac{{1 + {{\bf{e}}^{ - 6s}}}}{{\left( {s - 1} \right)\left( {{s^2} + 4} \right)}} - \frac{4}{s}\\ Y\left( s \right) & = \left( {1 + {{\bf{e}}^{ - 6s}}} \right)F\left( s \right) - \frac{4}{s}\end{align*}\]

Notice that we combined the first two terms to simplify things a little. Also, there was some canceling going on in this one. Do not expect that to happen on a regular basis. We now need to partial fraction \(F(s)\). We’ll leave the details to you to check.

\[\begin{align*}F\left( s \right) & = \frac{1}{{\left( {s - 1} \right)\left( {{s^2} + 4} \right)}} = \frac{1}{5}\left( {\frac{1}{{s - 1}} - \frac{{s + 1}}{{{s^2} + 4}}} \right)\\ f\left( t \right) & = \frac{1}{5}\left( {{{\bf{e}}^t} - \cos \left( {2t} \right) - \frac{1}{2}\sin \left( {2t} \right)} \right)\end{align*}\]

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

\[\begin{align*}Y\left( s \right) & = F\left( s \right) + F\left( s \right){{\bf{e}}^{ - 6s}} - \frac{4}{s}\\ y\left( t \right) & = f\left( t \right) + {u_6}\left( t \right)f\left( {t - 6} \right) - 4\end{align*}\]

where \(f(t)\) is given above.

Let’s take the Laplace transform of everything and note that in the third term we are shifting 4\(t\).

\[\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - 5\left( {sY\left( s \right) - y\left( 0 \right)} \right) - 14Y\left( s \right) & = \frac{9}{s} + \frac{{{{\bf{e}}^{ - 3s}}}}{s} + 4\frac{{{{\bf{e}}^{ - s}}}}{{{s^2}}}\\ \left( {{s^2} - 5s - 14} \right)Y\left( s \right) - 10 & = \frac{{9 + {{\bf{e}}^{ - 3s}}}}{s} + 4\frac{{{{\bf{e}}^{ - s}}}}{{{s^2}}}\end{align*}\]

Now solve for \(Y(s)\).

\[\begin{align*}\left( {{s^2} - 5s - 14} \right)Y\left( s \right) - 10 & = \frac{{9 + {{\bf{e}}^{ - 3s}}}}{s} + 4\frac{{{{\bf{e}}^{ - s}}}}{{{s^2}}}\\ Y\left( s \right) & = \frac{{9 + {{\bf{e}}^{ - 3s}}}}{{s\left( {s - 7} \right)\left( {s + 2} \right)}} + \frac{{4{{\bf{e}}^{ - s}}}}{{{s^2}\left( {s - 7} \right)\left( {s + 2} \right)}} + \frac{{10}}{{\left( {s - 7} \right)\left( {s + 2} \right)}}\\ Y\left( s \right) & = \left( {9 + {{\bf{e}}^{ - 3s}}} \right)F\left( s \right) + 4{{\bf{e}}^{ - s}}G\left( s \right) + H\left( s \right)\end{align*}\]

So, we have three functions that we’ll need to partial fraction for this problem. I’ll leave it to you to check the details.

\[\begin{align*}F\left( s \right) & = \frac{1}{{s\left( {s - 7} \right)\left( {s + 2} \right)}} = - \frac{1}{{14}}\frac{1}{s} + \frac{1}{{63}}\frac{1}{{s - 7}} + \frac{1}{{18}}\frac{1}{{s + 2}}\\ f\left( t \right) & = - \frac{1}{{14}} + \frac{1}{{63}}{{\bf{e}}^{7t}} + \frac{1}{{18}}{{\bf{e}}^{ - 2t}}\end{align*}\] \[\begin{align*}G\left( s \right) & = \frac{1}{{{s^2}\left( {s - 7} \right)\left( {s + 2} \right)}} = \frac{5}{{196}}\frac{1}{s} - \frac{1}{{14}}\frac{1}{{{s^2}}} + \frac{1}{{441}}\frac{1}{{s - 7}} - \frac{1}{{36}}\frac{1}{{s + 2}}\\ g\left( t \right) & = \frac{5}{{196}} - \frac{1}{{14}}t + \frac{1}{{441}}{{\bf{e}}^{7t}} - \frac{1}{{36}}{{\bf{e}}^{ - 2t}}\end{align*}\] \[\begin{align*}H\left( s \right) & = \frac{{10}}{{\left( {s - 7} \right)\left( {s + 2} \right)}} = \frac{{10}}{9}\frac{1}{{s - 7}} - \frac{{10}}{9}\frac{1}{{s + 2}}\\ h\left( t \right) & = \frac{{10}}{9}{{\bf{e}}^{7t}} - \frac{{10}}{9}{{\bf{e}}^{ - 2t}}\end{align*}\]

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

\[\begin{align*}Y\left( s \right) & = 9F\left( s \right) + {{\bf{e}}^{ - 3s}}F\left( s \right) + 4{{\bf{e}}^{ - s}}G\left( s \right) + H\left( s \right)\\ y\left( t \right) & = 9f\left( t \right) + {u_3}\left( t \right)f\left( {t - 3} \right) + 4{u_1}\left( t \right)g\left( {t - 1} \right) + h\left( t \right)\end{align*}\]

where \(f(t)\), \(g(t)\) and \(h(t)\) are given above.

The first step is to get \(g(t)\) written in terms of Heaviside functions so that we can take the transform.

\[g\left( t \right) = 2 + \left( {t - 2} \right){u_6}\left( t \right) + \left( {4 - t} \right){u_{10}}\left( t \right)\]

Now, while this is \(g(t)\) written in terms of Heaviside functions it is not yet in proper form for us to take the transform. Remember that each function must be shifted by a proper amount. So, getting things set up for the proper shifts gives us,

\[\begin{align*}g\left( t \right) & = 2 + \left( {t - 6 + 6 - 2} \right){u_6}\left( t \right) + \left( {4 - \left( {t - 10 + 10} \right)} \right){u_{10}}\left( t \right)\\ g\left( t \right) & = 2 + \left( {t - 6 + 4} \right){u_6}\left( t \right) + \left( { - 6 - \left( {t - 10} \right)} \right){u_{10}}\left( t \right)\end{align*}\]

So, for the first Heaviside it looks like \(f\left( t \right) = t + 4\) is the function that is being shifted and for the second Heaviside it looks like \(f\left( t \right) = - 6 - t\) is being shifted.

Now take the Laplace transform of everything and plug in the initial conditions.

\[\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) + 3\left( {sY\left( s \right) - y\left( 0 \right)} \right) + 2Y\left( s \right) & = \frac{2}{s} + {{\bf{e}}^{ - 6s}}\left( {\frac{1}{{{s^2}}} + \frac{4}{s}} \right) - {{\bf{e}}^{ - 10s}}\left( {\frac{1}{{{s^2}}} + \frac{6}{s}} \right)\\ \left( {{s^2} + 3s + 2} \right)Y\left( s \right) + 2 & = \frac{2}{s} + {{\bf{e}}^{ - 6s}}\left( {\frac{1}{{{s^2}}} + \frac{4}{s}} \right) - {{\bf{e}}^{ - 10s}}\left( {\frac{1}{{{s^2}}} + \frac{6}{s}} \right)\end{align*}\]

Solve for \(Y(s)\).

\[\begin{align*}\left( {{s^2} + 3s + 2} \right)Y\left( s \right) & = \frac{2}{s} + {{\bf{e}}^{ - 6s}}\left( {\frac{1}{{{s^2}}} + \frac{4}{s}} \right) - {{\bf{e}}^{ - 10s}}\left( {\frac{1}{{{s^2}}} + \frac{6}{s}} \right) - 2\\ \left( {{s^2} + 3s + 2} \right)Y\left( s \right) & = \frac{{2 + 4{{\bf{e}}^{ - 6s}} - 6{{\bf{e}}^{ - 10s}}}}{s} + \frac{{{{\bf{e}}^{ - 6s}} - {{\bf{e}}^{ - 10s}}}}{{{s^2}}} - 2\\ Y\left( s \right) & = \frac{{2 + 4{{\bf{e}}^{ - 6s}} - 6{{\bf{e}}^{ - 10s}}}}{{s\left( {s + 1} \right)\left( {s + 2} \right)}} + \frac{{{{\bf{e}}^{ - 6s}} - {{\bf{e}}^{ - 10s}}}}{{{s^2}\left( {s + 1} \right)\left( {s + 2} \right)}} - \frac{2}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\\ & \\ Y\left( s \right) & = \left( {2 + 4{{\bf{e}}^{ - 6s}} - 6{{\bf{e}}^{ - 10s}}} \right)F\left( s \right) + \left( {{{\bf{e}}^{ - 6s}} - {{\bf{e}}^{ - 10s}}} \right)G\left( s \right) - H\left( s \right)\end{align*}\]

Now, in the solving process we simplified things into as few terms as possible. Even doing this, it looks like we’ll still need to do three partial fractions.

We’ll leave the details of the partial fractioning to you to verify. The partial fraction form and inverse transform of each of these are.

\[\begin{align*}F\left( s \right) & = \frac{1}{{s\left( {s + 1} \right)\left( {s + 2} \right)}} = \frac{{\frac{1}{2}}}{s} - \frac{1}{{s + 1}} + \frac{{\frac{1}{2}}}{{s + 2}}\\ f\left( t \right) & = \frac{1}{2} - {{\bf{e}}^{ - t}} + \frac{1}{2}{{\bf{e}}^{ - 2t}}\end{align*}\] \[\begin{align*}G\left( s \right) & = \frac{1}{{{s^2}\left( {s + 1} \right)\left( {s + 2} \right)}} = - \frac{{\frac{3}{4}}}{s} + \frac{{\frac{1}{2}}}{{{s^2}}} + \frac{1}{{s + 1}} - \frac{{\frac{1}{4}}}{{s + 2}}\\ g\left( t \right) & = - \frac{3}{4} + \frac{1}{2}t + {{\bf{e}}^{ - t}} - \frac{1}{4}{{\bf{e}}^{ - 2t}}\end{align*}\] \[\begin{align*}H\left( s \right) & = \frac{2}{{\left( {s + 1} \right)\left( {s + 2} \right)}} = \frac{2}{{s + 1}} - \frac{2}{{s + 2}}\\ h\left( t \right) & = 2{{\bf{e}}^{ - t}} - 2{{\bf{e}}^{ - 2t}}\end{align*}\]

Putting this all back together is going to be a little messy. First rewrite the transform a little to make the inverse transform process possible.

\[Y\left( s \right) = 2F\left( s \right) + {{\bf{e}}^{ - 6s}}\left( {4F\left( s \right) + G\left( s \right)} \right) - {{\bf{e}}^{ - 10s}}\left( {6F\left( s \right) + G\left( s \right)} \right) - H\left( s \right)\]

Now, taking the inverse transform of all the pieces gives us the final solution to the IVP.

\[y\left( t \right) = 2f\left( t \right) - h\left( t \right) + {u_6}\left( t \right)\left( {4f\left( {t - 6} \right) + g\left( {t - 6} \right)} \right) - {u_{10}}\left( t \right)\left( {6f\left( {t - 10} \right) + g\left( {t - 10} \right)} \right)\]

where \(f(t)\), \(g(t)\), and \(h(t)\) are defined above.