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Further, by differentiating again, we obtain
y = uoy (t) + u1(t)yI (t) + u 2(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)[y1 (t) + p(t) y1 (t) + q (t) y1 (t)]
+ u2 (t) [ y2 (t) + p(t) y2 (t) + q (t)/j(0]
+ 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) y (t) + u2(t) y2 (t) = g(t). (25)
Equations (21) and (25) form a system of two linear algebraic equations for the derivatives u[(t) and u'2(t) of the unknown functions. They correspond exactly to Eqs. (6) and (9) in Example 1.
3.7 Variation of Parameters
By solving the system (21), (25) we obtain
u i(t) = ---------------, u 2 (t) = —--------------, (26)
1 W(jl,y2)(ty 2 ’ W(yvy2)(t) 1 ;
where W(y1,y2) is the Wronskian of y1 and y2. Note that division by Wis permissible since y1 and y2 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 Y2(t)g(t) ^ f Y1(t)g(t) ^ ^
u. (t) = — —y----------------- dt + c,, u(t) = —y dt + c,. (27)
1 J W(yvy2)(t) 1 2 J W(y1,Y2)(t) 2 ^ '
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 functions y1 and y2 are linearly independent solutions of the homogeneous equation (18) corresponding to the nonhomogeneous equation (16),
/ + p(t )/ + q (t ) y = g(t ),
then a particular solution of Eq. (16) is
Yt ) = y1(t ) 1 Z2(t )g(t) dt + y2W 1 y (t)g(l ) <1,. 1 i W(yi.y2)(t) }1’J W(yi,y2)(t) (28)
and the general solution is
y = ci yi(t) + c2y2(i) + 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 y1(t) and y2(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 Y1, 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.
PROBLEMS In each of Problems 1 through 4 use the method of variation of parameters to find a particular
I solution of the given differential equation. Then check your answer by using the method of undetermined coefficients.
1. /-5y + 6y = 2et 2. /-/-2y = 2e-t
3. y" + 2y + y = 3e-t 4. 4y — 4/ + y = 16et/2
Chapter 3. Second Order Linear Equations
In each of Problems 5 through 12 find the general solution of the given differential equation. In Problems 11 and 12 g is an arbitrary continuous function.
5. y" + y = tan t, 0 < t <n/2 6. y" + 9y = 9sec2 31, 0 < t <n/6
7. /' + 4/ + 4y = t—2e~2t, t > 0 8. f + 4y = 3csc2t, 0 < t < n/2
9. 4f + y = 2sec(t/2), —n< t <n 10. y" — 2? + y = et/(1 + t2)
11. / — 5/ + 6y = g(t) 12. / + 4y = g(t)
In each of Problems 13 through 20 verify that the given functions y and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and 20 g is an arbitrary continuous function.
13. t2y" — 2y = 3t2 — 1, t > 0; y1(t) = t2, y2(t) = t— 1
14. t2y" — t(t + 2)y + (t + 2)y = 2t3, t > 0; yx(t) = t, y2(t) = tet
15. tf — (1 + t)y + y = t2e2t, t > 0; y1(t) = 1 + t, y2(t) = e
16. (1 — t)y + ty — y = 2(t — 1)2e—t, 0 < t < 1; y1(t) = et, y2(t) = t
17. x2y' — 3xy + 4y = x2 lnx, x > 0; y1 (x) = x2, y2(x) = x2 ln x
18. x2 y + xy + (x2 — 0.25)y = 3x3/2 sin x, x > 0; y1(x) = x—1/2 sin x, y2(x) =
x—1/2 cos x
19. (1 — x)f + xy — y = g(x), 0 < x < 1; y1(x) = ex, y2(x) = x
20. x2y' + xy + (x2 — 0.25)y = g(x), x > 0; y1(x) = x—1/2 sinx, y2(x) =
x—1/2 cos x
21. Show that the solution of the initial value problem
L[y] = y + p(t)y + q(t)y = g(t), y(t0) = Óo, y(t0) = y0 (i)
can be written as y = u(t) + v(t), where u and v are solutions of the two initial value problems
L [u] = 0, u(y = y0, Uy) = y0, (ii)
L [v] = g(t), v(t0) = 0, v'(t0) = 0, (iii)
respectively. In other words, the nonhomogeneities in the differential equation and in the initial conditions can be dealt with separately. Observe that u is easy to find if a fundamental set of solutions of L [u] = 0 is known.