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The general solution of a differential equation combines all possible solutions to its homogeneous version (i.e., when the non-homogeneous term is set to zero). For the equation \[ y'' + 6y' + 25y = 0 \], we first determine its auxiliary equation by replacing the derivatives with powers of a variable, often 'r'. In this case, the auxiliary equation is: \[ r^2 + 6r + 25 = 0 \]. By solving this quadratic equation using the quadratic formula, we get the roots \[ -3 \pm 4i \]. We use these roots to write the general solution of the homogeneous differential equation:\[ y_h = e^{-3t}(C_1 \cos(4t) + C_2 \sin(4t)) \]. This expression represents the general solution, encompassing the complete set of solutions for the homogeneous part of the differential equation.

To find a particular solution to a non-homogeneous differential equation, we look at the non-homogeneous part of the equation. For \[ y'' + 6y' + 25y = -1 \], we guess a specific form for the solution, often related to the nature of the non-homogeneous term. Since \[ -1 \] is a constant, we try a particular solution \[y_p = A\], a constant. Substituting \[ y_p \] into the original differential equation and solving for \[ A \], we find: \[ 0 + 0 + 25A = -1 \implies A = -\frac{1}{25} \]. This gives us our particular solution \[ y_p = -\frac{1}{25} \]. When we combine this with the general solution of the homogeneous equation, we get the complete general solution: \[ y = y_h + y_p \].

Initial conditions are specific values given for a function and its derivatives at certain points, typically used to find constants in the general solution of a differential equation. For our equation, the initial conditions are \[ y(0) = -\frac{1}{25} \] and \[ y'(0) = 7 \]. These conditions help determine the constants \[C_1 \] and \[ C_2 \] in the general solution. Using the initial condition \[ y(0) = -\frac{1}{25} \], we find: \[ C_1 - \frac{1}{25} = -\frac{1}{25} \implies C_1 = 0 \]. Next, we differentiate the general solution and use the second initial condition \[ y'(0) = 7 \] to solve for \[ C_2 \]: \[ 3C_2 = 7 \implies C_2 = \frac{7}{3} \]. Substituting these values back, we get the particular values of the constants for our solution.

The quadratic formula is a vital tool for solving quadratic equations, and it is expressed as: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. For the differential equation's auxiliary equation \[ r^2 + 6r + 25 = 0 \], we identify \[ a = 1 \], \[ b = 6 \], and \[ c = 25 \]. Plugging these into the quadratic formula, we solve for 'r': \[ r = \frac{-6 \pm \sqrt{36 - 100}}{2} = \frac{-6 \pm \sqrt{-64}}{2} = \frac{-6 \pm 8i}{2} \]. This simplifies to the roots \[ -3 \pm 4i \]. These roots are then used in forming the general solution to the homogeneous part of the differential equation, essential for solving the overall problem.