Final answer:
The solution to the differential equation y" + 6y' + 10y = 0 is y(t) = e^{-3t}(C_1 · cos(t) + C_2 · sin(t)), where C_1 and C_2 are constants determined by the initial conditions.
Step-by-step explanation:
To find the general solution for the differential equation y" + 6y' + 10y = 0, we need to solve this second-order linear homogeneous differential equation. The characteristic equation associated with this differential equation is r^2 + 6r + 10 = 0. Solving this quadratic equation using the quadratic formula, r = (-b ± √(b^2-4ac))/(2a), we find that the discriminant (b^2-4ac) is negative, which means we will have complex roots. Let's calculate the roots:
r = (-6 ± √(36-40))/(2), which simplifies to r = -3 ± √(-4)/2. Thus, the roots are r = -3 ± i, where i is the imaginary unit. Using Euler's formula, we can then express the general solution as y(t) = e^{-3t}(C_1 · cos(t) + C_2 · sin(t)), where C_1 and C_2 are constants determined by initial conditions.