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Section 1.2 Some Basic Examples

Subsection 1.2.1 Solve \(x''(t) - 4x(t) = e^{t}\)

First, we solve the homogeneous version \(x'' - 4x = 0\text{.}\) It turns out that the general solution of this ODE is

\begin{gather*} x(t) = c_1 e^{2t} + c_2 e^{-2t}. \end{gather*}

For 2nd order linear ODEs (basically ODEs of the form \(a x'' + bx' + cx = 0\) for constants \(a,b,c\)), remember that we always start off by plugging in \(x = e^{rt}\text{.}\)

If \(x = e^{rt}\text{,}\) then \(x' = re^{rt}\) and \(x'' = r^2e^{rt}\text{.}\) Substituting these into \(x'' - 4x = 0\) shows that

\begin{gather*} r^2e^{rt} - 4e^{rt} = 0. \end{gather*}

Dividing out by \(e^{rt} > 0\) gives us the characteristic polynomial \(r^2 - 4 = 0\text{.}\) Solving this has solutions \(r = \pm 2\text{.}\) This means that \(e^{2t}\) and \(e^{-2t}\) are solutions of \(x'' - 4x = 0\text{.}\)

So, we get that \(x(t) = c_1 e^{2t} + c_2 e^{-2t}\) is the general solution of \(x'' - 4x = 0\text{.}\)

This makes it clear that we should take \(y_1(t) = e^{2t}\) and \(y_2(t) = e^{-2t}\text{.}\) The Wronskian \(W[y_1,y_2](t) = -4\text{,}\) so these are fundamental solutions.

All we need to find is a particular solution of \(x'' + 4x = e^t\text{.}\) This is quite easy if we've memorized the solution variation of parameters gives us.

Great! Then, this should be a simple task of plugging in stuff (although the simplification might be a little messy).

In this exercise, we have \(g(t) = e^t\) and \(p(t) = 1\text{.}\) Let's also take \(t_0 = 0\text{,}\) just to hopefully make the integration a little easier. So, directly from the formula, the particular solution \(Y(t)\) that we want is

\begin{align*} Y(t) \amp = \int_{t_0}^t \frac{y_1(s) y_2(t) - y_1(t) y_2(s)}{W[y_1,y_2](s)} \cdot \frac{g(s)}{p(s)} \ ds, \\ \amp = \int_0^t \frac{e^{2s} \cdot e^{-2t} - e^{2t} \cdot e^{-2s}}{e^{2s} \cdot (-2 e^{-2s}) - (2 e^{2s}) \cdot e^{-2s} } \cdot e^s \ ds. \end{align*}

Simplifying this will be a bit of a hassle, but the integration shouldn't be too bad:

\begin{align*} Y(t) \amp = \int_0^t \frac{e^{2s - 2t} - e^{2t - 2s}}{(-4)} \cdot e^s \ ds, \\ \amp = -\frac{1}{4} \int_0^t \Big[ e^{3s - 2t} - e^{2t - s} \Big] \ ds, \\ \amp = -\frac{1}{4} \Big[ \frac{1}{3} e^{3s - 2t} + e^{2t - s} \Big]_{s=0}^{s=t}, \\ \amp = -\frac{1}{3}e^t + \frac{1}{12} e^{-2t} + \frac{1}{4} e^{2t}. \end{align*}

Although not necessary, with a little careful observation, one may note that \(e^{-2t}/12 + e^{2t}/4\) solves \(x'' - 4x = 0\text{,}\) so we may instead take \(Y(t) = -e^t/3\) to be our particular solution of choice.

This is okay! The solution that variation of parameters gives is long, and misremembering one small detail could waste a lot of time. A lot of the work to find this solution the “long way” is slightly tedious, so I'll skip over some more of the computational details.

Remember that the basic principle of variation of parameters is that we set \(x = Y(t) = u_1(t) y_1(t) + u_2(t) y_2(t)\text{.}\) In this case, we set \(Y(t) = u_1(t) e^{2t} + u_2(t) e^{-2t}\text{.}\) We want to solve for \(u_1\) and \(u_2\text{.}\)

Now, calculate \(Y'(t)\text{.}\)

\begin{gather*} Y'(t) = u_1'(t)e^{2t} + u_2'(t) e^{-2t} + 2 u_1(t) e^{2t} - 2 u_2(t) e^{-2t}. \end{gather*}

We set \(u_1'(t)e^{2t} + u_2'(t) e^{-2t} = 0\text{,}\) and keep this in the back of our heads. Next up, calculate \(Y''(t)\text{:}\)

\begin{gather*} Y''(t) = 2u_1'(t) e^{2t} - 2u_2'(t) e^{-2t} + 4u_1(t) e^{2t} + 4u_2(t) e^{-2t}. \end{gather*}

Substitute \(x = Y(t)\) into \(x'' - 4x = e^t\) to get

\begin{align*} e^t \amp = \Big( 2u_1'(t) e^{2t} - 2u_2'(t) e^{-2t} + 4u_1(t) e^{2t} + 4u_2(t) e^{-2t} \Big) - \Big( 4u_1(t) e^{2t} + 4u_2(t) e^{-2t} \Big), \\ \amp = 2u_1'(t) e^{2t} - 2u_2'(t) e^{-2t} \end{align*}

So in all, we get a system of equations (which is what we should expect).

\begin{align*} 0 \amp = u_1'(t) e^{2t} + u_2'(t) e^{-2t} \\ e^t \amp = 2u_1'(t) e^{2t} - 2u_2'(t) e^{-2t} \end{align*}

Turn this into a matrix equation to get

\begin{align*} \begin{bmatrix} u_1' \\ u_2' \end{bmatrix} \amp = \begin{bmatrix} e^{2t} \amp e^{-2t} \\ 2e^{2t} \amp -2e^{-2t} \end{bmatrix} \begin{bmatrix} 0 \\ e^t \end{bmatrix}\\ \amp = \frac{1}{4} \begin{bmatrix} e^{-t} \\ -e^{3t} \end{bmatrix} \end{align*}

Now, we have \(u_1'(t) = e^{-t}/4\) and \(u_2'(t) = -e^{3t}/4\text{.}\) We can choose any \(u_1, u_2\) with these derivatives, so for sake of simplicity, choose \(u_1(t) = -e^{-t}/4\) and \(u_2(t) = -e^{3t}/12\text{.}\)

Substitute these into \(Y(t) = u_1(t) e^{2t} + u_2(t) e^{-2t}\text{:}\)

\begin{gather*} Y(t) = -\frac{1}{4} e^{-t} e^{2t} - \frac{1}{12} e^{3t} e^{-2t} = -\frac{1}{3} e^t. \end{gather*}

In both cases, we found that \(Y(t) = -e^t/3\) is a particular solution to \(x'' - 4x = e^t\text{.}\) So, the general solution to \(x'' - 4x = e^t\) is

\begin{gather*} \boxed{x(t) = c_1 e^{2t} + c_2 e^{-2t} - \frac{1}{3} e^t.} \end{gather*}

Subsection 1.2.2 Solve \(x'' - 3x' + 2x = 1\)

Try it yourself! First, solve \(x'' - 3x' + 2x = 0\text{.}\)

Set \(x = e^{rt}\text{.}\) Substituting this into \(x'' - 3x' + 2x = 0\) shows that

\begin{gather*} r^2 - 3r + 2 = 0. \end{gather*}

This occurs when \(r = 1,2\text{,}\) so the general solution to \(x'' - 3x' + 2x = 0\) is

\begin{gather*} x(t) = c_1e^t + c_2e^{2t}. \end{gather*}

Next, we need a particular solution. Use variation of parameters to find some particular solution \(Y(t)\) of \(x'' - 3x' + 2x = 1\text{.}\)

Set \(Y(t) = u_1(t) y_1(t) + u_2(t) y_2(t)\text{.}\) Then, find \(Y'\) and \(Y''\text{,}\) using the special condition that we've set before (it's to avoid terms like \(u_1''\) and \(u_2''\))

Substituting these into \(x'' - 3x' + 2x = 1\text{,}\) along with the special condition, gives us a system of equations that we can solve!

Hint

Alternatively, one might remember that \(Y(t)\) takes the form

\begin{gather*} Y(t) = \int_{t_0}^t \frac{y_1(s) y_2(t) - y_1(t) y_2(s)}{W[y_1,y_2](s)} \cdot \frac{g(s)}{p(s)} \ ds. \end{gather*}

What's \(g(t)\) and \(p(t)\) in the ODE \(x'' - 3x' + 2x = 1\text{?}\)

Finally, the general solution to \(x'' - 3x' + 2x = 1\) is of the form

\begin{gather*} x(t) = c_1 y_1(t) + c_2y_2(t) + Y(t). \end{gather*}
Answer

Here's one possible general solution (there are multiple, based on the \(Y(t)\) chosen)

\begin{gather*} \boxed{x(t) = c_1 e^t + c_2 e^{2t} + \frac{1}{2}.} \end{gather*}