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7. IVP formula

Let’s recapitulate the solution strategy for \(\opA[x]=f\) from start to finish when given the initial condition \(x(t_0)=x_0\). Recall that we have the option of defining the homogeneous solution using a definite integral,

\[ \tilde{q}(t) = \int_{t_0}^t a(r)\, dr. \]

This form has the property that \(\tilde{q}(t_0)=0\), which turns out to be convenient. Then we define

\[ \tilde{g}(t) = \exp\bigl[ \tilde{q}(t) \bigr], \]

so that \(\tilde{g}(t_0)=1\). The VoP formula has another integral that we can also write in definite form,

\[ \tilde{k}(t) = \int_{t_0}^t \frac{f(s)}{\tilde{g}(s)}\, ds, \]

such that \(\tilde{k}(t_0)=0\). These definitions combine to produce a solution to the IVP with no unknown constants:

(7.2)\[x(t) = \tilde{g}(t) x_0 + \tilde{g}(t) \tilde{k}(t).\]

This works because it’s a combination of homogeneous and particular solutions that clearly satisfies \(x(t_0)=x_0\).

Equation (7.2) invites an interesting interpretation. The first term, \(\tilde{g}(t) x_0\), represents the solution of \(\opA[x]=0\) with initial value \(x(t_0)=x_0\). This can be called the free response, meaning free from external forcing. The other term is the forced response of the system, starting from a zero initial value.

7.1. Exponential form

It’s worth replacing the \(\tilde{g}\) references above with their exponential definitions. The particular solution part becomes

\[ x_p(t) = \exp \bigl[ \tilde{q}(t) \bigr] \int_{t_0}^t \exp \bigl[ -\tilde{q}(s) \bigr]\, f(s) \, ds. \]

The exponential in front is constant as far as the integration is concerned (it does not depend on \(s\)). Thus

\[ x_p(t) = \int_{t_0}^t \exp \bigl[ \tilde{q}(t) -\tilde{q}(s) \bigr]\, f(s) \, ds. \]

By elementary properties of integrals,

\[\begin{align*} \tilde{q}(t) -\tilde{q}(s) & = \int_{t_0}^t a(r)\, dr - \int_{t_0}^s a(r)\, dr \\ &= \int_{t_0}^t a(r)\, dr + \int_s^{t_0} a(r)\, dr \\ &= \int_{s}^t a(r)\, dr. \end{align*}\]

Here is a key new observation. The integrand of \(x_p\),

\[ X(s) = \exp \left[ \tilde{q}(t) -\tilde{q}(s) \right]\, f(s) = \exp \left[ \int_{s}^t a(r)\, dr \right]\, f(s), \]

has the exact same form as

\[ \tilde{g}(t) x_0 = \exp \left[ \int_{t_0}^t a(r)\, dr \right]\, x_0, \]

except that it starts from \(x(s)=f(s)\) rather than \(x(t_0)=x_0\). Therefore \(X(s)\) is the free response of the system at time \(t\) due to the condition \(x(s)=f(s)\), and the particular solution

\[ x_p(t) = \int_{t_0}^t X(s) \, ds \]

integrates (i.e., sums), over all the instants \(s\) between \(t_0\) and \(t\), the free responses of the system due to \(f(s)\). In fact, it’s not much of a stretch to consider this formula a linear combination of free responses, just using a continuous coefficient function \(f(s)\) rather than discrete coefficients! Finally, the complete solution to the IVP just adds the free response of the actual initial condition to the summed force responses.

7.2. The nature of linear solutions

The following summarizes the formulas above.

Formula 7.2 (Solution of 1st order, linear IVP)

The solution of \(x'-a(t)x=f(t)\), \(x(t_0)=x_0\), is

\[ x(t) = \exp \left[ \int_{t_0}^t a(r)\, dr \right]\, x_0 + \int_{t_0}^t \exp \left[ \int_{s}^t a(r)\, dr \right]\, f(s)\, ds. \]

The formula becomes easier to read and parse in the case of a constant coefficient function \(a\):

\[ x(t) = e^{a(t-t_0)} x_0 + \int_{t_0}^t e^{a(t-s)} f(s)\, ds. \]

As solution formulas, these results are ultimately not harder nor easier to apply to a specific example than the steps we used to solve earlier examples. The real value here is in what is revealed about how linear systems actually work: you can decompose them into individual contributions, then add the results together.

6. Variation of parameters 8. Integrating factor