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# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
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y = 0(t) = 3 e + 3 e2t. (8)
Now let us solve the same problem by using the Laplace transform. To do this we must assume that the problem has a solution y = 0(t), which with its first two derivatives satisfies the conditions of Corollary 6.2.2. Then, taking the Laplace transform of the differential equation (4), we obtain
?{/}  L{ y}  2L{ y} = 0, (9)
where we have used the linearity of the transform to write the transform of a sum as the sum of the separate transforms. Upon using the corollary to express L{y"} and L{/} in terms of L{y}, we find that Eq. (9) becomes
s2L{ y}  sy(0)  y (0)  [sL{y}  y (0)]  2L{ y} = 0,
or
(s2  s  2) 7 (s) + (1  s) y(0)  y (0) = 0, (10)
where Y(s) = L{y}. Substituting for y(0) and /(0) in Eq. (10) from the initial conditions (5), and then solving for Y(s), we obtain
Y(s) = 2s 1 2 = ( s7(\1)- (11)
s2  s  2 (s  2)(s + 1)
We have thus obtained an expression for the Laplace transform Y(s) of the solution y = \$(t) of the given initial value problem. To determine the function 0 we must find the function whose Laplace transform is Y(s), as given by Eq. (11).
302
Chapter 6. The Laplace Transform
This can be done most easily by expanding the right side of Eq. (11) in partial fractions. Thus we write
y (s) = s -1_______= -a. + -A. =a(s +1) + - 2), a2)
(s - 2)(s + 1) s - 2 s + 1 (s - 2)(s + 1)
where the coefficients a and b are to be determined. By equating numerators of the second and fourth members of Eq. (12), we obtain
s  1 = a(s + 1) + b(s - 2),
an equation that must hold for all s. In particular, if we set s = 2, then it follows that a = 3. Similarly, if we set s = 1, then we find that b = 3. By substituting these values for a and b, respectively, we have
1/3 2/3
Y(s) = -/-2 + -/-r  (13)
s - 2 s + 1
Finally, if we use the result of Example 5 of Section 6.1, it follows that 3 e2t has the transform 1 (s  2)-1; similarly, 3e~l has the transform 3(s + 1)-1. Hence, by the linearity of the Laplace transform,
y = \$(t) = 1 e2t + 3 e-t
has the transform (13) and is therefore the solution of the initial value problem (4), (5). Of course, this is the same solution that we obtained earlier.
The same procedure can be applied to the general second order linear equation with constant coefficients,
ay' + by + cy = f (t). (14)
Assuming that the solution y = \$(t) satisfies the conditions of Corollary 6.2.2 for n = 2, we can take the transform of Eq. (14) and thereby obtain
a[s2Y(s) - sy(0) - /(0)] + b[sY(s) - y(0)] + cY(s) = F(s), (15)
where F(s) is the transform of f (t). By solving Eq. (15) for Y(s) we find that
(as + b) y(0) + a/(0) F(s)
Y (s) = --------------2-------------7 + -7---- (16)
as + bs + c as + bs + c
The problem is then solved, provided that we can find the function y = \$(t) whose
transform is Y(s).
Even at this early stage of our discussion we can point out some of the essential features of the transform method. In the first place, the transform Y(s) of the unknown function y = 0(t) is found by solving an algebraic equation rather than a differential equation, Eq. (10) rather than Eq. (4), or in general Eq. (15) rather than Eq. (14). This is the key to the usefulness of Laplace transforms for solving linear, constant coefficient, ordinary differential equationsthe problem is reduced from a differential equation to an algebraic one. Next, the solution satisfying given initial conditions is automatically found, so that the task of determining appropriate values for the arbitrary constants in the general solution does not arise. Further, as indicated in Eq. (15), nonhomogeneous equations are handled in exactly the same way as homogeneous ones; it is not necessary to solve the corresponding homogeneous equation first. Finally, the method can be applied in the same way to higher order equations, as long as we assume that the solution satisfies the conditions of the corollary for the appropriate value of n.
6.2 Solution of Initial Value Problems
Observe that the polynomial as2 + bs + c in the denominator on the right side of Eq. (16) is precisely the characteristic polynomial associated with Eq. (14). Since the use of a partial fraction expansion of Y(s) to determine fi(t) requires us to factor this polynomial, the use of Laplace transforms does not avoid the necessity of finding roots of the characteristic equation. For equations of higher than second order this may be a difficult algebraic problem, particularly if the roots are irrational or complex.
The main difficulty that occurs in solving initial value problems by the transform technique lies in the problem of determining the function y = \$(t) corresponding to the transform Y(s). This problem is known as the inversion problem for the Laplace transform; 0(t) is called the inverse transform corresponding to Y(s), and the process of finding 0(t) from Y (s) is known as inverting the transform. We also use the notation L-1{ Y(s)} to denote the inverse transform of Y(s). There is a general formula for the inverse Laplace transform, but its use requires a knowledge of the theory of functions of a complex variable, and we do not consider it in this book. However, it is still possible to develop many important properties of the Laplace transform, and to solve many interesting problems, without the use of complex variables.
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