The document solves a second order linear differential equation (DE) with initial conditions. It finds the characteristic equation, which has complex roots. The general solution is obtained as a linear combination of exponential and trigonometric terms. Applying the initial conditions yields the specific solution y(x) = (3cos1.5x+6sin1.5x)e^3x.

1. Complex roots
solve the follwong IVP
y''-6y'+11y = 0
y(0) = 3 , y'(0) = 0
Solution
characterstic equation given by
x^2-6x+11 = 0
x =6+sqrt(36-4*11) /2
x = 6+sqrt(-9) / 2
x = 6+3i/2
x =3+1.5i
x =6-sqrt(36-4*11) /2
x = 6-sqrt(-9) / 2
x = 6-3i/2x = 3-1.5i
y(x) = (Acos1.5x+Bsin1.5x)e^3x
y(0) = A = 3
y(x) = (Acos1.5x+Bsin1.5x)e^3x
y'(x) = e^3x( -1.5A sin1.5x + 1.5Bcos1.5x) +3(Acos1.5x+Bsin1.5x)e^3x

2. y'(0) = 1.5B + 3A =0
3A =1.5B
A =0.5B
B = 2A = 2*3 =6
y(x) = (3cos1.5x+6sin1.5x)e^3x