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### Section 4.6 : Nonconstant Coefficient IVP's

In this section we are going to see how Laplace transforms can be used to solve some differential equations that do not have constant coefficients. This is not always an easy thing to do. However, there are some simple cases that can be done.

To do this we will need a quick fact.

#### Fact

If \(f(t)\) is a piecewise continuous function on \(\left[ {0,\infty } \right)\) of exponential order then,

\[\begin{equation}\mathop {\lim }\limits_{s \to \infty } F\left( s \right) = 0\label{eq:eq1}\end{equation}\]A function \(f(t)\) is said to be of exponential order \(\alpha \) if there exists positive constants \(T\) and \(M\) such that

\[\left| {f\left( t \right)} \right| \le M{{\bf{e}}^{\alpha \,t}}\hspace{0.25in}{\mbox{for all }}t \ge T\]Put in other words, a function that is of exponential order will grow no faster than

\[M{{\bf{e}}^{\alpha \,t}}\]for some \(M\) and \(\alpha \) and all sufficiently large \(t\). One way to check whether a function is of exponential order or not is to compute the following limit.

\[\mathop {\lim }\limits_{t \to \infty } \frac{{\left| {f\left( t \right)} \right|}}{{{{\bf{e}}^{\alpha \,t}}}}\]If this limit is finite for some \(\alpha \) then the function will be of exponential order \(\alpha \). Likewise, if the limit is infinite for every \(\alpha\) then the function is not of exponential order.

Almost all of the functions that you are liable to deal with in a first course in differential equations are of exponential order. A good example of a function that is not of exponential order is

\[f\left( t \right) = {{\bf{e}}^{{t^3}}}\]We can check this by computing the above limit.

\[\mathop {\lim }\limits_{t \to \infty } \frac{{{{\bf{e}}^{{t^3}}}}}{{{{\bf{e}}^{\alpha \,t}}}} = \mathop {\lim }\limits_{t \to \infty } {{\bf{e}}^{{t^3} - \alpha t}} = \mathop {\lim }\limits_{t \to \infty } {{\bf{e}}^{t\left( {{t^2} - \alpha } \right)}} = \infty \]This is true for any value of \(\alpha \) and so the function is not of exponential order.

Do not worry too much about this exponential order stuff. This fact is occasionally needed in using Laplace transforms with non constant coefficients.

So, let’s take a look at an example.

So, for this one we will need to recall that #30 in our table of Laplace transforms tells us that,

\[\begin{align*}\mathcal{L}\left\{ {ty'} \right\} & = - \frac{d}{{ds}}\left( {\mathcal{L}\left\{ {y'} \right\}} \right)\\ & = - \frac{d}{{ds}}\left( {sY\left( s \right) - y\left( 0 \right)} \right)\\ & = - sY'\left( s \right) - Y\left( s \right)\end{align*}\]So, upon taking the Laplace transforms of everything and plugging in the initial conditions we get,

\[\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) + 3\left( { - sY'\left( s \right) - Y\left( s \right)} \right) - 6Y\left( s \right) & = \frac{2}{s}\\ - 3sY'\left( s \right) + \left( {{s^2} - 9} \right)Y\left( s \right) & = \frac{2}{s}\\ Y'\left( s \right) + \left( {\frac{3}{s} - \frac{s}{3}} \right)Y\left( s \right) & = - \frac{2}{{3{s^2}}}\end{align*}\]Unlike the examples in the previous section where we ended up with a transform for the solution, here we get a linear first order differential equation that must be solved in order to get a transform for the solution.

The integrating factor for this differential equation is,

\[\mu \left( s \right) = {{\bf{e}}^{\int{{\left( {\frac{3}{s} - \frac{s}{3}} \right)ds}}}} = {{\bf{e}}^{\ln \left( {{s^3}} \right) - \frac{{{s^2}}}{6}}} = {s^3}{{\bf{e}}^{ - \frac{{{s^2}}}{6}}}\]Multiplying through, integrating and solving for \(Y(s)\) gives,

\[\begin{align*}\int{{{{\left( {{s^3}{{\bf{e}}^{ - \frac{{{s^2}}}{6}}}Y\left( s \right)} \right)}^\prime }\,ds}} & = \int{{ - \frac{2}{3}s{{\bf{e}}^{ - \frac{{{s^2}}}{6}}}\,ds}}\\ {s^3}{{\bf{e}}^{ - \frac{{{s^2}}}{6}}}Y\left( s \right) & = 2{{\bf{e}}^{ - \frac{{{s^2}}}{6}}} + c\\ Y\left( s \right) & = \frac{2}{{{s^3}}} + c\frac{{{{\bf{e}}^{\frac{{{s^2}}}{6}}}}}{{{s^3}}}\end{align*}\]Now, we have a transform for the solution. However, that second term looks unlike anything we’ve seen to this point. This is where the fact about the transforms of exponential order functions comes into play. We are going to assume that whatever our solution is, it is of exponential order. This means that

\[\mathop {\lim }\limits_{s \to \infty } \left( {\frac{2}{{{s^3}}} + \frac{{c{{\bf{e}}^{\frac{{{s^2}}}{6}}}}}{{{s^3}}}} \right) = 0\]The first term does go to zero in the limit. The second term however, will only go to zero if \(c = 0\). Therefore, we must have \(c = 0\) in order for this to be the transform of our solution.

So, the transform of our solution, as well as the solution is,

\[Y\left( s \right) = \frac{2}{{{s^3}}}\hspace{0.25in}y\left( t \right) = {t^2}\]We’ll leave it to you to verify that this is in fact a solution if you’d like to.

Now, not all nonconstant differential equations need to use \(\eqref{eq:eq1}\). So, let’s take a look at one more example.

From the first example we have,

\[\mathcal{L}\left\{ {ty'} \right\} = - sY'\left( s \right) - Y\left( s \right)\]We’ll also need,

\[\begin{align*}\mathcal{L}\left\{ {ty''} \right\} & = - \frac{d}{{ds}}\left( {\mathcal{L}\left\{ {y''} \right\}} \right)\\ & = - \frac{d}{{ds}}\left( {{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)} \right)\\ & = - {s^2}Y'\left( s \right) - 2sY\left( s \right) + y\left( 0 \right)\end{align*}\]Taking the Laplace transform of everything and plugging in the initial conditions gives,

\[\begin{align*} - {s^2}Y'\left( s \right) - 2sY\left( s \right) + y\left( 0 \right) - \left( { - sY'\left( s \right) - Y\left( s \right)} \right) + Y\left( s \right) & = \frac{2}{s}\\\left( {s - {s^2}} \right)Y'\left( s \right) + \left( {2 - 2s} \right)Y\left( s \right) + 2 & = \frac{2}{s}\\ s\left( {1 - s} \right)Y'\left( s \right) + 2\left( {1 - s} \right)Y\left( s \right) & = \frac{{2\left( {1 - s} \right)}}{s}\\ Y'\left( s \right) + \frac{2}{s}Y\left( s \right) & = \frac{2}{{{s^2}}}\end{align*}\]Once again we have a linear first order differential equation that we must solve in order to get a transform for the solution. Notice as well that we never used the second initial condition in this work. That is okay, we will use it eventually.

Since this linear differential equation is much easier to solve compared to the first one, we’ll leave the details to you. Upon solving the differential equation we get,

\[Y\left( s \right) = \frac{2}{s} + \frac{c}{{{s^2}}}\]Now, this transform goes to zero for all values of \(c\) and we can take the inverse transform of the second term. Therefore, we won’t need to use \(\eqref{eq:eq1}\) to get rid of the second term as did in the previous example.

Taking the inverse transform gives,

\[y\left( t \right) = 2 + ct\]Now, this is where we will use the second initial condition. Upon differentiating and plugging in the second initial condition we can see that \(c = -4\).

So, the solution to this IVP is,

\[y\left( t \right) = 2 - 4t\]So, we’ve seen how to use Laplace transforms to solve some nonconstant coefficient differential equations. Notice however that all we did was add in an occasional \(t\) to the coefficients. We couldn’t get too complicated with the coefficients. If we had we would not have been able to easily use Laplace transforms to solve them.

Sometimes Laplace transforms can be used to solve nonconstant differential equations, however, in general, nonconstant differential equations are still very difficult to solve.