Solution:

Given ordinary differential equation (ODE):
d²y/dt² + 5dy/dt + 6y = 0

Initial conditions:
y(0) = 2
y(1) = -(1 - 3e)/e^3

To solve this ODE, we can use the characteristic equation method.

1. Characteristic Equation:
Let's assume the solution to the ODE is of the form:
y(t) = e^(rt)

Taking the first and second derivatives of y(t) with respect to t:
dy/dt = re^(rt)
d²y/dt² = r²e^(rt)

Substituting these derivatives into the ODE, we get:
r²e^(rt) + 5re^(rt) + 6e^(rt) = 0

Factoring out e^(rt), we have:
e^(rt)(r² + 5r + 6) = 0

Since e^(rt) is always positive and non-zero, the equation reduces to:
r² + 5r + 6 = 0

2. Solving the Characteristic Equation:
To solve the quadratic equation, we can factor it:
(r + 2)(r + 3) = 0

Setting each factor equal to zero, we get two solutions for r:
r₁ = -2
r₂ = -3

3. General Solution:
The general solution to the ODE is given by:
y(t) = C₁e^(-2t) + C₂e^(-3t)

where C₁ and C₂ are constants to be determined using the initial conditions.

4. Applying Initial Conditions:
Using the initial condition y(0) = 2:
2 = C₁e^(-2*0) + C₂e^(-3*0)
2 = C₁ + C₂

Using the initial condition y(1) = -(1 - 3e)/e^3:
-(1 - 3e)/e^3 = C₁e^(-2*1) + C₂e^(-3*1)
-(1 - 3e)/e^3 = C₁e^(-2) + C₂e^(-3)

5. Solving for Constants:
We now have two equations with two unknowns (C₁ and C₂). Solving these equations simultaneously will give us the values of the constants.

From equation 2, we have:
C₂ = 2 - C₁

Substituting this into equation 1, we get:
-(1 - 3e)/e^3 = C₁e^(-2) + (2 - C₁)e^(-3)

Simplifying this equation further:
-(1 - 3e) = C₁e^(-2 + 3) + 2e^(-3) - C₁e^(-3)

Expanding the exponential terms:
-(1 - 3e) = C₁e + 2e^(-3) - C₁e^(-3)

Simplifying and rearranging the terms:
C₁e - C₁e^(-3) = -(1 - 3e) - 2e^(-3)

C₁(e - e^(-3)) = -1 + 3e - 2e^(-3)

Dividing both sides by (