The problem involves solving an initial value problem for a second-order linear differential equation with a Dirac delta function and given initial conditions. Here's how to approach the solution:

Given:

$y'' + 8y' + 25y = \delta(t - \pi),$ with initial conditions: $y(0) = 1, \quad y'(0) = 0.$

Step 1: Solve the Homogeneous Equation

The characteristic equation for the homogeneous part is: $r^2 + 8r + 25 = 0.$ Solve for $r$: $r = \frac{-8 \pm \sqrt{8^2 - 4(1)(25)}}{2(1)} = \frac{-8 \pm \sqrt{-36}}{2} = -4 \pm 3i.$ Thus, the general solution for the homogeneous equation is: $y_h(t) = e^{-4t}(C_1 \cos(3t) + C_2 \sin(3t)).$

Step 2: Account for the Dirac Delta Function

The Dirac delta function $\delta(t - \pi)$ introduces a discontinuity in $y'$ at $t = \pi$. This requires the impulse response to be calculated.

The general solution for $y(t)$ will include the homogeneous solution and a response due to the impulse: $y(t) = y_h(t) + y_p(t),$ where $y_p(t)$ is the particular solution introduced by the delta function. Using the properties of the delta function, the jump in $y'(t)$ at $t = \pi$ is: $\Delta y'(t) = \int_{t = \pi^-}^{t = \pi^+} \delta(t - \pi) \, dt = 1.$ Thus: $y'(\pi^+) - y'(\pi^-) = 1.$

Step 3: Apply Initial Conditions and Impulse Response

1. For $t < \pi$, the solution is the homogeneous solution: $y(t) = e^{-4t}(C_1 \cos(3t) + C_2 \sin(3t)).$ Using the initial conditions: $y(0) = 1 \implies C_1 = 1,$ $y'(0) = 0 \implies y'(t) = e^{-4t}(-4\cos(3t) - 3\sin(3t)) + e^{-4t}(3C_2\cos(3t) - 4C_2\sin(3t)),$ $y'(0) = -4 + 3C_2 = 0 \implies C_2 = \frac{4}{3}.$ Thus, for $t < \pi$, $y(t) = e^{-4t}\left(\cos(3t) + \frac{4}{3}\sin(3t)\right).$

2. For $t \geq \pi$, the solution adds the impulse response. Using the jump condition: $y'(t) \text{ changes by 1 at } t = \pi.$

Thus, the final solution can be written as: