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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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(a) Find a differential operator H(D) with constant coefficients that annihilates g(t), that is, an operator such that H(D)g(t) = 0.
(b) Apply H( D) to Eq. (i), obtaining
H( D) L (D) y = 0, (ii)
which is a homogeneous equation of higher order.
(c) Solve Eq. (ii).
(d) Eliminate from the solution found in step (c) the terms that also appear in the solution of L (D)y = 0. The remaining terms constitute the correct form of the particular solution of Eq. (i).
22. Use the method of annihilators to find the form of the particular solution Y(t) for each of the equations in Problems 13 through 18. Do not evaluate the coefficients.
4.4 The Method of Variation of Parameters
The method of variation of parameters for determining a particular solution of the nonhomogeneous nth order linear differential equation
L [y] = y(n) + Pj(t) y(n1) + + pnj(t) y + pn (t) y = g(t) (1)
is a direct extension of the method for the second order differential equation (see Section 3.7). As before, to use the method of variation of parameters, it is first necessary to solve the corresponding homogeneous differential equation. In general, this may be difficult unless the coefficients are constants. However, the method of variation of parameters is still more general than the method of undetermined coefficients in that it leads to an expression for the particular solution for any continuous function g, whereas the method of undetermined coefficients is restricted in practice to a limited class of functions g.
Suppose then that we know a fundamental set of solutions y1, y2,..., yn of the homogeneous equation. Then the general solution of the homogeneous equation is
yc(t) = Cj yl(t) + c2y2(t) +
+ (t).
(2)
4.4 The Method of Variation of Parameters
227
The method of variation of parameters for determining a particular solution of Eq. (1) rests on the possibility of determining n functions , 2,..., such that Y(t) is of the form
Y (t) = 1(t)y1(t) + 2()2() + + (t) yn (t). (3)
Since we have n functions to determine, we will have to specify n conditions. One of these is clearly that Y satisfy Eq. (1). The other n 1 conditions are chosen so as to make the calculations as simple as possible. Since we can hardly expect a simplification in determining Y if we must solve high order differential equations for l,..., , it is natural to impose conditions to suppress the terms that lead to higher derivatives of ,..., . From Eq. (3) we obtain
Y' = (1 1 + 22 + -^ + ) + (1 1 + 22 + + 'ӄ), (4)
where we have omitted the independent variable t on which each function in Eq. (4) depends. Thus the first condition that we impose is that
1 1 + 2 2 + + 'ӄ = 0. (5)
Continuing this process in a similar manner through n 1 derivatives of Y gives
Y(m) = 1 y(m) + 2) +------------+ (), m = 0, 1, 2,..., n 1, (6)
and the following n 1 conditions on the functions 1,..., :
1 y(m1) + 2(21) + + >1) = 0, = 1, 2,..., 1. (7)
The nth derivative of Y is
Y(n) = y1n) + + )) + (1 y{n1) + + .τ)). (8)
Finally, we impose the condition that Y must be a solution of Eq. (1). On substituting for the derivatives of Y from Eqs. (6) and (8), collecting terms, and making use of the fact that L[y ] = 0, i = 1, 2,..., n, we obtain
1 y(1) + + + dJr1 = g. (9)
Equation (9), coupled with the n 1 equations (7), gives n simultaneous linear nonhomogeneous algebraic equations for t/2,..., ':
1 1 + 2 2 + + =
y1 1 + 2 2 +---------+ =0,
//1 + ӳ2 +----------+ ' = 0> (10)
y(rV)w, + + 󄄗) = g.
The system (10) is a linear algebraic system for the unknown quantities [,..., . By solving this system and then integrating the resulting expressions, you can obtain the coefficients ,..., .A sufficient condition for the existence of a solution of the system of equations (10) is that the determinant of coefficients is nonzero for each value of t. However, the determinant of coefficients is precisely W(yv y2,..., yn), and it is nowhere zero since yl,..., yn are linearly independent solutions of the homogeneous
Previous << 1 .. 121 122 123 124 125 126 < 127 > 128 129 130 131 132 133 .. 609 >> Next