y''-6y'-27y=0 pvi y(0)=3 y'(0)=13

1. **Find the characteristic equation**: Substitute \( y = e^{rt} \) into the differential equation to get:
r^2 - 6r - 27 = 0

2. **Solve the characteristic equation**: The characteristic equation is a quadratic, which can be solved using the quadratic formula:
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Where \( a = 1 \), \( b = -6 \), \( c = -27 \):
r = \frac{6 \pm \sqrt{36 + 108}}{2}
r = \frac{6 \pm \sqrt{144}}{2}
r = \frac{6 \pm 12}{2}
r_1 = 9, \, r_2 = -3

3. **General solution of the differential equation**:
For distinct roots \( r_1 = 9 \) and \( r_2 = -3 \), the general solution is:
y(t) = C_1 e^{9t} + C_2 e^{-3t}

4. **Apply initial conditions**:
- Using \( y(0) = 3 \):
3 = C_1 e^{0} + C_2 e^{0} = C_1 + C_2

- Using \( y'(t) = 9C_1 e^{9t} - 3C_2 e^{-3t} \) and \( y'(0) = 13 \):
13 = 9C_1 e^{0} - 3C_2 e^{0} = 9C_1 - 3C_2

5. **Solve the system of equations**:
Solve:
\[
\begin{align*}
C_1 + C_2 &= 3 \\
9C_1 - 3C_2 &= 13
\end{align*}
\]

From the first equation:
C_2 = 3 - C_1

Substitute \( C_2 = 3 - C_1 \) into the second equation:
9C_1 - 3(3 - C_1) = 13
9C_1 - 9 + 3C_1 = 13
12C_1 = 22
C_1 = \frac{22}{12} = \frac{11}{6}

Then, from \( C_2 = 3 - C_1 \):
C_2 = 3 - \frac{11}{6} = \frac{18}{6} - \frac{11}{6} = \frac{7}{6}

6. **The particular solution**:
y(t) = \frac{11}{6} e^{9t} + \frac{7}{6} e^{-3t}

Simplifying further, by multiplying throughout by 6, we get:
y(t) = 2e^{9t} + e^{-3t}

Therefore, the final solution is:

y(t) = 2e^{9t} + e^{-3t}