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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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Then we try to determine u1 (t) and u2(t) so that the expression in Eq. (19) is a solution of the nonhomogeneous equation (16) rather than the homogeneous equation (18). Thus we differentiate Eq. (19), obtaining
= u\(t) 1 (t) + u^t) y1 (t) + u2(t) 2 (t) + u 2^t) y2 (t). (20)
As in Example 1, we now set the terms involving W1(t) and u'2(t) in Eq. (20) equal to zero; that is, we require that
u[(t) y1(t) + u2(t) y2(t) = 0. (21)
Then, from Eq. (20), we have
= u1(t) 1 (t) + u 2(t) 2 (t). (22)
Further, by differentiating again, we obtain
= u,1(t)y{ (t) + u1(t)y1 (t) + u2(t)y2 (t) + u2(t)y2 (t). (23)
Now we substitute for y, /, and y" in Eq. (16) from Eqs. (19), (22), and (23), respectively. After rearranging the terms in the resulting equation we find that
u1(t)[y'l (t) + p(t) (t) + q (t) y1 (t)]
+ u2(t)[y/2'(t) + p(t) y2 (t) + q (t) y2(t)]
+ u,1(t)y,1 (t) + u,2(t)y2 (t) = g(t). (24)
Each of the expressions in square brackets in Eq. (24) is zero because both y1 and y2 are solutions of the homogeneous equation (18). Therefore Eq. (24) reduces to
u[(t) (t) + u2(t) (t) = g(t). (25)
Equations (21) and (25) form a system of two linear algebraic equations for the derivatives u'1(t) and W2(t) of the unknown functions. They correspond exactly to Eqs. (6) and (9) in Example 1.
3.7 Variation of Parameters
183
By solving the system (21), (25) we obtain
/,* J2(t)g(t) /,* J1(t)g(t) ^
u i(t) = ----------------, u 2 (t) = -------------, (26)
1 W(jvj2)(t) W(jvj2)(t)
where W(J1,J2) is the Wronskian of j1 and y2. Note that division by Wis permissible since j1 and j2 are a fundamental set of solutions, and therefore their Wronskian is nonzero. By integrating Eqs. (26) we find the desired functions u1(t) and u2(t), namely,
f J2(t)g(t) ^ f J1(t)g(t) ^ ,
u. (t) = ------------------ dt+ c,, u0(t) = -f--------- dt + c,. (27)
1 J W(jvj2)(t) 1 J W(J1,J2)(t) 2 1 ;
Finally, substituting from Eq. (27) in Eq. (19) gives the general solution of Eq. (16). We state the result as a theorem.
Theorem 3.7.1 If the functions p, q, and g are continuous on an open interval I, and if the func
tions j1 and y2 are linearly independent solutions of the homogeneous equation (18)
corresponding to the nonhomogeneous equation (16),
/ + p(t ) + q (t) j = g(t),
then a particular solution of Eq. (16) is
Y() = J1(t) 1 dt + J2(t) 1 J (t)g(l) dt, (28)
1 W(jvj2)(t) }1J W(J1,y2)(t)
and the general solution is
J = C1 J1(t) + c2y2(t) + Y (t), (29)
as prescribed by Theorem 3.6.2.
By examining the expression (28) and reviewing the process by which we derived it, we can see that there may be two major difficulties in using the method of variation of parameters. As we have mentioned earlier, one is the determination of J1(t) and J2(t), a fundamental set of solutions of the homogeneous equation (18), when the coefficients in that equation are not constants. The other possible difficulty is in the evaluation of the integrals appearing in Eq. (28). This depends entirely on the nature of the functions j1, y2, and g. In using Eq. (28), be sure that the differential equation is exactly in the form (16); otherwise, the nonhomogeneous term g(t) will not be correctly identified.
A major advantage of the method of variation of parameters is that Eq. (28) provides an expression for the particular solution Y(t) in terms of an arbitrary forcing function g(t). This expression is a good starting point if you wish to investigate the effect of variations in the forcing function, or if you wish to analyze the response of a system to a number of different forcing functions.
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