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## Q. 2.7.6

Series Solution Method

Obtain an approximate, closed-form solution of the following problem for 0 ≤ t ≤ 0.5:

$\overset{.}{x}+x= \tan \ t \quad x(0)=0 \quad (1)$

## Verified Solution

If we attempt to use separation of variables to solve this problem, we obtain

$\frac{dx}{\tan \ t – x}=dt$

so the variables do not separate. In general, when the input is a function of time, the equation ẋ + g(x)=f(t) does not separate. The Laplace transform method cannot be used when the Laplace transform or inverse transform either does not exist or cannot be found easily. In this example, the equation cannot be solved by the Laplace transform method, because the transform of tan t does not exist.

An approximate solution of the equation ẋ + x = tan t can be obtained by replacing tan t with a series approximation. The number of terms used in the series determines the accuracy of the resulting solution for x(t). The Taylor series expansion for tan t is

$\tan \ t=t+\frac{t^3}{3}+\frac{2t^5}{15}+\frac{17t^7}{315}+… \quad |t|<\frac{\pi}{2}$

The more terms we retain, the more accurate is the series. Also, the series becomes less accurate as the absolute value of t increases. To demonstrate the series solution method, let us use a series with two terms: tan t = t + t³/3. At the largest value of t, t = 0.5, the two-term series gives 0.5417 versus 0.5463 for the true value of tan 0.5. So the two-term series is accurate to at least two decimal places over the range of t we are interested in (0 ≤ t ≤ 0.5).

Using the two-term series we need to solve the following problem:

$\overset{.}{x}+x=t+\frac{t^3}{3} \quad x(0)=0$

Using the Laplace transform, we obtain

$sX(s)+X(s)=\frac{1}{s^2}+\frac{1}{3} \frac{3!}{s^4}$

or

$X(s)=\frac{s^2+2}{s^4(s+1)}$

The inverse transform was obtained in Example 2.7.5 and is

$x(t)=\frac{1}{3}t^3-t^2+3t-3+3e^{-t}$

We can expect this approximate solution of equation (1) to be accurate to at least two decimal places for 0 ≤ t ≤ 0.5. Of course, greater accuracy can be achieved by retaining more terms in the Taylor series for tan t.