1. The document solves a second order non-homogeneous differential equation using the method of undetermined coefficients. It finds the general solution and the solution satisfying the given initial conditions. 2. The general solution is the sum of the complementary solution and particular solution. The complementary solution is the sum of two exponential terms and the particular solution is the sum of sine and cosine terms. 3. The initial conditions are used to find the values of constants in the general solution, resulting in the final solution being a combination of exponential, sine and cosine terms.

1. 1. (a) Use the method of undetermined coefficients to find the general solution to
y'' - 4y' + 3y = 13 cos(2x).
(b) Find the solution satisfying the initial conditions y(0) = 4/5, y'(0) = 11/5.
Solution
the characteristic equation of the given differential equation is
r 2 - 4r + 3 = 0
(r-1)(r-3)=0
r = 1, 3 (root are real and different)
its complementary function is
yc = Aex + Be3x
choose particular solution
yp=c1 cos(2x)+c2 sin(2x)
differentiate it
y'p=-2c1 sin(2x)+2c2 cos(2x)
again differentiate it
y"p=- 4c1 cos(2x) - 4c2 sin(2x)
substitute the values of yp , y'p and y"p in the given differential equation
{- 4c1 cos(2x) - 4c2 sin(2x)} - 4{-2c1 sin(2x)+2c2 cos(2x)} + 3{c1 cos(2x)+c2 sin(2x)} = 13
cos(2x)
(-c1-8c2)cos(2x) + (-c2 + 8c1)sin(2x) = 13 cos(2x)
compare the coefficients
-c1-8c2 = 13
-c2 + 8c1 = 0
this gives
c1 = - 1/5 and c2 = -8/5
hence yp=(-1/5) cos(2x)+(-8/5) sin(2x)
complete solution is
y = yc + yp
y = Aex + Be3x - (1/5) cos(2x) - (8/5) sin(2x) (1)
differentiate it
y' = Aex + 3Be3x + (2/5) sin(2x) - (16/5) cos(2x) (2)
using y(0) = 4/5, y'(0) = 11/5 in (1) and (2)
4/5 = A+B - 1/5

2. A + B = 1 (3)
11/5 = A + 3B - 16/5
A+3B = 27/5 (4)
solving (3) and(4) we get
A= -6/5 and B = 11/5
hence the solution becomes
y = (-6/5)ex + (11/5)e3x - (1/5) cos(2x) - (8/5) sin(2x)