1. Chapter 7: Variation of Parameters
MATH 374 Lecture 19
Chapter 7: Variation of Parameters

2. Note for this Chapter
We now look at two methods of solving non-homogeneous linear equations with variable coefficients!

3. 7.1: Reduction of Order
This method, due to D’Alembert, works for equations of the form:
b0(x)y(n) + b1(x)y(n-1) + … + bn-1(x)y’ + bn(x)y = R(x). (1)
Here is the main idea of this method …

4. Main idea of Reduction of Order
b0(x)y(n) + b1(x)y(n-1) + … + bn-1(x)y’ + bn(x)y = R(x). (1)
Main idea of Reduction of Order
If we know y1 is a solution of the homogeneous equation corresponding to (1):
b0(x)y(n) + b1(x)y(n-1) + … + bn-1(x)y’ + bn(x)y = 0, (2)
then the introduction of a new dependent variable v by the substitution
y = y1v (3)
into (1) will lead to a differential equation of order (n-1) to solve!

5. Reduction of Order - Second Order Equation case
Let’s try Reduction of Order on the equation:
y’’ + p(x) y’ + q(x) y = R(x). (4)
Suppose y1 is a solution of
y’’ + p(x) y’ + q(x) y = 0. (5)
Let
y = y1v (6)
and find the first and second derivatives of y:

6. Reduction of Order - Second Order Equation case
y’’ + p(x) y’ + q(x) y = R(x). (4)
y’’ + p(x) y’ + q(x) y = 0. (5)
Reduction of Order - Second Order Equation case
y = y1v (6)
y’ = y1’v + y1v’ (7)
y’’ = y1’’v + 2 y1’v’ + y1v’’ (8)
Substituting (6), (7), and (8) into (4) yields:
(y1’’v + 2 y1’v’ + y1v’’) + p(x)(y1’v + y1v’) + q(x)(y1v) = R(x)
or, collecting like v-terms:

7. Reduction of Order - Second Order Equation case
y’’ + p(x) y’ + q(x) y = R(x). (4)
y’’ + p(x) y’ + q(x) y = 0. (5)
Reduction of Order - Second Order Equation case
y = y1v (6)
y’ = y1’v + y1v’ (7)
y’’ = y1’’v + 2 y1’v’ + y1v’’ (8)
Substituting (6), (7), and (8) into (4) yields:
(y1’’v + 2 y1’v’ + y1v’’) + p(x)(y1’v + y1v’) + q(x)(y1v) = R(x)
or, collecting like v-terms:
y1v’’ + (2y1’ + p(x)y1’)v’ + (y1’’ + p(x)y1’ + q(x) y1)v = R(x) (9)

8. Reduction of Order - Second Order Equation case
y’’ + p(x) y’ + q(x) y = R(x). (4)
y’’ + p(x) y’ + q(x) y = 0. (5)
Reduction of Order - Second Order Equation case
y = y1v (6)
y’ = y1’v + y1v’ (7)
y’’ = y1’’v + 2 y1’v’ + y1v’’ (8)
Substituting (6), (7), and (8) into (4) yields:
(y1’’v + 2 y1’v’ + y1v’’) + p(x)(y1’v + y1v’) + q(x)(y1v) = R(x)
or, collecting like v-terms:
y1v’’ + (2y1’ + p(x)y1’)v’ + (y1’’ + p(x)y1’ + q(x) y1)v = R(x) (9)
Hence, we can rewrite (9) as
y1v’’ + (2y1’ + p(x)y1’)v’ = R(x). (10)
= 0 by (5)

9. Reduction of Order - Second Order Equation case
y1v’’ + (2y1’ + p(x)y1’)v’ = R(x). (10)
Reduction of Order - Second Order Equation case
Letting w = v’, we see that (10) is a first order linear equation in w!!
y1w’ +(2y1’ + p(x)y1)w = R(x). (11)
To solve (11) for w, we use an integrating factor.
Then since v’ = w, it follows that
v = s w dx.
Finally,
y = y1v.

10. Example 1 Use Reduction of Order to solve y’’ + y = sec3x (12)
Solution: First, note that the Method of Undetermined Coefficients will not work here!
The homogeneous equation
y’’ + y = 0 (13)
has solution yc = c1 sin x + c2 cos x.
We need a solution of (13) to use Reduction of Order, so take
y1 = sin x. (14)
(Any solution of (13) can be chosen for y1 …)

11. Example 1 (continued) y = y1v = v sin x (15)
y’’ + y = sec3x (12)
Example 1 (continued)
y = y1v = v sin x (15)
y’ = v’ sin x + v cos x (16)
y’’ = v’’ sin x + 2 v’ cos x + v (-sin x) (17)
Substituting (15), (16), and (17) into (12) gives:
(v’’ sin x + 2 v’ cos x – v sin x) + v sin x = sec3 x.
) v’’ sin x + 2 v’ cos x = sec3 x
) v’’ + 2 (cos x)/(sin x) v’ = (sec3x)/(sin x) (18)

12. v’’ + 2 (cos x)/(sin x) v’ = (sec3x)/(sin x) (18)
Example 1 (continued)
Letting w = v’, (18) becomes:
w’+2(cos x)/(sin x)w = (sec3x)/(sin x). (19)
An integrating factor for (19) is:
es (cos x)/(sinx ) dx = e2 ln(sin x) = sin2x.
Multiplying (19) by the integrating factor,
sin2x w’ + 2 sin x cos x w = sin x sec3x
) [sin2x w]’ = sin x sec3x
) sin2x w = s sin x sec3 x dx
) sin2x w = s tan x sec2x dx
Letting u = tan x for the integral,
) sin2x w = 1/2 tan2x + C1
) w = 1/2 sec2x + C1csc2x
Recall that w = v’, so we have
v’ = 1/2 sec2x + C1csc2x
) v = s (1/2 sec2x + C1csc2x) dx
) v = 1/2 tan x – C1cot x + C2
Finally, y = y1v, so it follows that
y = sin x(1/2 tan x – C1cot x + C2)
i.e.
y = 1/2 sinx tanx – C1cos x + C2sin x yp yc