Question

Solve the differential equation y"+y=(secx)^2.

Answer 1

r^2 + 1 = 0
(r+i) (r-i)

r = complex roots

r= (0+i1, 0-i1)

yh= e^0 (C1 cos(x)+C2 sin(x))

yh= C1 cos(x) + C2 sin(x)

Answer 2

yp = A cos(x) + Bsin(x)

yp' = -Asin(x) + Bcos(x) +A'cos(x) + B'sin(x)
---------------
= 0

yp' = -Asin(x) + Bcos(x)

yp'' = -Acos(x) - Bsin(x) -A'sin(x) + B'cos(x)
+ yp = A cos(x) + Bsin(x)
---------------------------------------
sec^2(x) = -A'sin(x) + B'cos(x)

Answer 3

A'cos(x) + B'sin(x) = 0
-A'sin(x) + B'cos(x) = sec^2(x)


system of equation, solve for A' and B'

Answer 4

If we cramer it

B' = sec^2 cos / cos^2 + sin^2
= sec

A' = sec^2 sin / -sin^2 - cos^2
= -tan sin

integrate A' and B' to find A and B and sub them into yp


the solution then is:

y = yh + yp

Answer 5

in the book they say that something about W.....

Answer 6

Whats George Bush got to do with this?