Page 260 - Schaum's Outline of Differential Equations

CHAP. 24] SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS 243

SOLUTIONS OF DIFFERENTIAL EQUATIONS
Laplace transforms are used to solve initial-value problems given by the wth-order linear differential equation
with constant coefficients

together with the initial conditions specified in Eq. (24.2). First, take the Laplace transform of both sides of
Eq. (24.6), thereby obtaining an algebraic equation for Y(s). Then solve for Y(s) algebraically, and finally take
1
inverse Laplace transforms to obtain y(x) = ^T {Y(s)}.
Unlike previous methods, where first the differential equation is solved and then the initial conditions are
applied to evaluate the arbitrary constants, the Laplace transform method solves the entire initial-value problem
in one step. There are two exceptions: when no initial conditions are specified and when the initial conditions
are not at x = 0. In these situations, c 0 through c n in Eqs. (24.2) and (24.3) remain arbitrary and the solution to
differential Eq. (24.6) is found in terms of these constants. They are then evaluated separately when appropriate
subsidiary conditions are provided. (See Problems 24.11 through 24.13.)

Solved problems

24.1. Solve / - 5y = 0; y(0) = 2.
Taking the Laplace transform of both sides of this differential equation and using Property 24.4, we obtain
££{/} - 5£{y} = £6{0}. Then, using Eq. (24.4) with c a= 2, we find

from which

Finally, taking the inverse Laplace transform of Y(s), we obtain

5x
24.2. Solve / - 5y = e ; y(0) = 0.
Taking the Laplace transform of both sides of this differential equation and using Property 24.4, we find that
5x
££{/} - 5£{y} = £{e }. Then, using Appendix A and Eq. (24.4) with c 0= 0, we obtain

from which

Finally, taking the inverse transform of Y(s), we obtain

(see Appendix A, entry 14).

24.3. Solve / + y = sin x; y(0) = 1.
Taking the Laplace transform of both sides of the differential equation, we obtain