Question 4.2.4: Solving a First-Order IVP Use the Laplace transform to solve......

We first take the transform of each member of the differential equation:

\begin{gathered} \mathscr{L}\left\{\begin{array}{l} \left.\frac{d y}{d t}\right\}+3 \mathscr{L}\{y\} = 13 \mathscr{L}\{\sin 2 t\} \end{array}\right. \end{gathered}. (12)

But from (6), \mathscr{L}\{d y / d t\} = s Y(s)-y(0) = s Y(s)-6, and from part (d) of Theorem 4.1.1, \mathscr{L}\{\sin 2 t\} = 2 /\left(s^{2}+4\right), and so (12) is the same as

s Y(s)-6+3 Y(s) = \frac{26}{s^{2}+4} \quad \text { or } \quad(s+3) Y(s) = 6+\frac{26}{s^{2}+4} \text {. }

Solving the last equation for Y(s), we get

Y(s) = \frac{6}{s+3}+\frac{26}{(s+3)\left(s^{2}+4\right)} =\frac{6 s^{2}+50}{(s+3)\left(s^{2}+4\right)} .   (13)

Since the quadratic polynomial s^{2}+4 does not factor using real numbers, its assumed numerator in the partial fraction decomposition is a linear polynomial in s :

\frac{6 s^{2}+50}{(s+3)\left(s^{2}+4\right)} = \frac{A}{s+3}+\frac{B s+C}{s^{2}+4}.

Putting the right side of the equality over a common denominator and equating numerators gives 6 s^{2}+50 = A\left(s^{2}+4\right)+(B s+C)(s+3). Setting s=-3 then yields immediately A 8. Since the denominator has no more real zeros, we equate the coefficients of s^{2} and s : 6 = A+B and 0 = 3 B+C. Using the value of A in the first equation gives B=-2, and then using this last value in the second equation gives C = 6. Thus

Y(s) = \frac{6 s^{2}+50}{(s+3)\left(s^{2}+4\right)} = \frac{8}{s+3}+\frac{-2 s+6}{s^{2}+4}.

We are not quite finished because the last rational expression still has to be written as two fractions. But this was done by termwise division in Example 2. From (2) of that example,

y(t) = 8 \mathscr{L}^{-1}\left\{\begin{array}{c} 1 \\ \hline s+3 \end{array}\right\}-2 \mathscr{L}^{-1}\left\{\begin{array}{c} s \\ \hline s^{2}+4 \end{array}\right\}+3 \mathscr{L}^{-1}\left\{\begin{array}{c} 2 \\ \hline s^{2}+4 \end{array}\right\} .

It follows from parts (c), (d), and (e) of Theorem 4.2.1 that the solution of the initial-value problem is y(t)=8 e^{-3 t}-2 \cos 2 t+3 \sin 2 t.