y6y9y0 y02 y01Solutionfirst you solve the homogeneous equati

y\'\'+6y\'+9y=0, y(0)=-2, y\'(0)=1

Solution

first you solve the homogeneous equation:
y\'\' - 6y\' + 9y = 0
by assuming y = e^(rx)
substitution gives you:
r^2 - 6r + 9 = 0
(r-3)*(r-3) = 0
so you have a double root of r =3
This means the homogeneous solution is:
y = A*e^(3x) + B*x*e^(3x)

Then you use the initial value and the boundary condition
y(0) = A + B = -2
y\' = 3*A*e^(3x) + B[3x*e^(3x) + e^(3x)]
y\'(0) = 3*A + B = 1

Solving the system
A+B = -2
3A + B = 1
Gives A = -3/2, B = -7/2