Final answer:
The interval of existence for the solution to the initial value problem 2ty'=6y, y(-2)=-8, is all real numbers except zero, that is, t ∈ (-∞, 0) ∪ (0, +∞). This is due to the original equation being undefined at t=0. the interval of existence is t ∈ (-∞, 0) ∪ (0, +∞).
Step-by-step explanation:
The question asks about the interval of existence for the solution to the initial value problem 2ty'=6y, with y(-2)=-8. To find the interval of existence, we first solve the differential equation. This is a first-order linear differential equation that can be solved using separation of variables.
The differential equation can be rewritten as dy/y = 3 dt/t. Integrating both sides, we get ln|y| = 3 ln|t| + C, and by exponentiating, we get |y| = C|t|^3, where C is a constant. We can find C using the initial condition, which gives us |y(-2)| = C|(-2)|^3 = |-8|, hence, C = 1. The solution to the initial value problem is, therefore, y(t) = ± t^3.
Considering the specific initial condition y(-2) = -8, we can say that y(t) = -t^3 on the interval where the function is defined. Since t cannot be zero (from the original equation 2ty'=6y), the interval of existence for the solution does not include t=0. Otherwise, the solution exists for all t except zero, i.e., the interval of existence is t ∈ (-∞, 0) ∪ (0, +∞).