Apply the Laplace Transform to the differential equation.

The given differential equation is $y'' + 5y' + 6y(t) = 7$.
The initial conditions are $y'(0) = 0$ and $y(0) = 1$.

Applying the Laplace Transform to each term:
$\mathcal{L}\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0)$
$\mathcal{L}\{y'(t)\} = s Y(s) - y(0)$
$\mathcal{L}\{y(t)\} = Y(s)$
$\mathcal{L}\{7\} = \frac{7}{s}$

Substitute these into the differential equation:

$$(s^2 Y(s) - s y(0) - y'(0)) + 5(s Y(s) - y(0)) + 6 Y(s) = \frac{7}{s}$$

Substitute $y(0) = 1$ and $y'(0) = 0$:

$$(s^2 Y(s) - s(1) - 0) + 5(s Y(s) - 1) + 6 Y(s) = \frac{7}{s}$$ $$s^2 Y(s) - s + 5s Y(s) - 5 + 6 Y(s) = \frac{7}{s}$$

Group the terms containing $Y(s)$:

$$(s^2 + 5s + 6) Y(s) - s - 5 = \frac{7}{s}$$

Move the terms without $Y(s)$ to the right side:

$$(s^2 + 5s + 6) Y(s) = \frac{7}{s} + s + 5$$

Combine the terms on the right side into a single fraction:

$$(s^2 + 5s + 6) Y(s) = \frac{7 + s^2 + 5s}{s}$$

Isolate $Y(s)$:

$$Y(s) = \frac{s^2 + 5s + 7}{s(s^2 + 5s + 6)}$$

Factor the quadratic term in the denominator: $s^2 + 5s + 6 = (s+2)(s+3)$.

$$Y(s) = \frac{s^2 + 5s + 7}{s(s+2)(s+3)}$$

Set up the partial fraction expansion:

$$\frac{s^2 + 5s + 7}{s(s+2)(s+3)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+3}$$

Multiply both sides by $s(s+2)(s+3)$ to clear the denominators:

$$s^2 + 5s + 7 = A(s+2)(s+3) + B s(s+3) + C s(s+2)$$

To find $A$, set $s=0$:

$$0^2 + 5(0) + 7 = A(0+2)(0+3) + B(0) + C(0)$$ $$7 = A(2)(3)$$ $$7 = 6A \implies A = \frac{7}{6}$$

To find $B$, set $s=-2$:

$$(-2)^2 + 5(-2) + 7 = A(0) + B(-2)(-2+3) + C(0)$$ $$4 - 10 + 7 = B(-2)(1)$$ $$1 = -2B \implies B = -\frac{1}{2}$$

To find $C$, set $s=-3$:

$$(-3)^2 + 5(-3) + 7 = A(0) + B(0) + C(-3)(-3+2)$$ $$9 - 15 + 7 = C(-3)(-1)$$ $$1 = 3C \implies C = \frac{1}{3}$$

Substitute the values of $A$, $B$, and $C$ back into the partial fraction expansion:

$$Y(s) = \frac{7}{6s} - \frac{1}{2(s+2)} + \frac{1}{3(s+3)}$$

$$y(t) = \mathcal{L}^{-1}\left\{\frac{7}{6s} - \frac{1}{2(s+2)} + \frac{1}{3(s+3)}\right\}$$

Using the inverse Laplace Transform properties $\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1$ and $\mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\} = e^{at}$:

$$y(t) = \frac{7}{6}\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} - \frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\} + \frac{1}{3}\mathcal{L}^{-1}\left\{\frac{1}{s+3}\right\}$$ $$y(t) = \frac{7}{6}(1) - \frac{1}{2}e^{-2t} + \frac{1}{3}e^{-3t}$$ $$y(t) = \frac{7}{6} - \frac{1}{2}e^{-2t} + \frac{1}{3}e^{-3t}$$