1. Nonhomogeneous Linear Differential Equations (Part 2)
AP Calculus BC

2. Method of Variation of Parameters
Use the Method of Undetermined Coefficients to solve the equation ayʹʹ + byʹ + cy = G(x) when G(x) = a polynomial, Aekx, A sin kx, or A cos kx
For all other forms of G(x), use Variation of Parameters.
Method:
Solve the complementary equation first, which is ayʹʹ + byʹ + cy = 0, and write the solution as
yc = c1y1(x) + c2y2(x)
Now replace c1 and c2 with functions (vary the parameters!), so
yp = u1(x)y1(x) + u2(x)y2(x)

3. Method of Variation of Parameters
yp = u1(x)y1(x) + u2(x)y2(x)
Differentiate 
u1 and u2 are arbitrary functions, so we can impose conditions on them:
yp is a solution to the differential equation.
u1ʹy1 + u2ʹy2 = 0
So 
Therefore 
Substitute y, yʹ, and yʹʹ into the differential equation to get 

4. Method (continued) Rearranging 
y1 and y2 are both solutions to the complementary equation, so the first two terms = 0, and therefore, we are left with:

5. So What? The two key equations are: u1ʹy1 + u2ʹy2 = 0
This leaves a system of two equations for u1ʹ and u2ʹ.
After solving the system, integrate to find u1 and u2, which will give us our particular solution yp = u1(x)y1(x) + u2(x)y2(x).
Add the complementary solution and the particular solution to get the general solution, y = yc + yp.

6. Example 1 Solve the equation yʹʹ + y = tan x (0 < x < π/2)
The auxiliary equation is r2 + 1 = 0
r = ±i
Solution of complementary equation is yc = c1sin x + c2cos x
The particular solution will have the form
yp(x) = u1(x)sin x + u2(x)cos x
So ypʹ(x) = (u1ʹsin x + u2ʹcos x) + (u1cos x – u2sin x)
Set u1ʹsin x + u2ʹcos x = 0

7. Example 1 (continued)
Therefore, ypʹʹ(x) = u1ʹcos x – u2ʹsin x – u1sin x – u2cos x
Substitute back into original equation:
u1ʹcos x – u2ʹsin x – u1sin x – u2cos x + u1sin x + u2cos x = tan x
Simplify  u1ʹcos x – u2ʹsin x = tan x
Here’s our system! u1ʹsin x + u2ʹcos x = 0
u1ʹcos x – u2ʹsin x = tan x
Multiply top by sin x and bottom by cos x: u1ʹsin2 x + u2ʹsin x cos x = 0
u1ʹcos2 x – u2ʹsin x cos x = cos x tan x

8. Example 1 (continued) Adding, we get u1ʹ(sin2x + cos2x) = cos x tan x
So u1ʹ = cos x tan x  u1ʹ = sin x
Integrating  u1 = –cos x
We want a particular solution, so we don’t need “+C” here.
Since u1ʹsin x + u2ʹcos x = 0, then
Recall that
Therefore, u2 = sin x – ln(sec x + tan x)

9. Example 1 (FINAL)
So yp = –cos x sin x + [sin x – ln(sec x + tan x)] cos x.
Simplifying  yp = –cos x ln(sec x + tan x)
Finally, our general solution is
y = c1 sin x + c2 cos x – cos x ln(sec x + tan x)
WHEW!