Answer and Step-by-step explanation:
Solution:
The given equation is:
t (t -4)y” + 3ty’ + 4y = 2
Where,
Y (3) = 0, y’ (3) = -1
To find out the largest interval, which have unique solution,
Divide the given equation with t(t – 4):
Y” + 3y’ / t(t-4) + 4y / t(t-4) = 2 / t (t -4)
Solution exists on (0, 4)
Because:
P(t) = 3 / t -4, continuous on ( -∞, 4) and (4, ∞)
Q(t) = 4/t(t-4) , continuous on (-∞, 0), (0,4) and (4, ∞)
G(t) = 2 / t(t-4) continuous on (-∞, 0), (0,4) and (4,∞)
T0 = 3,
So solution exist on (0, 4)
Hence, (0, 4) is the largest interval for which the initial value problem:
t (t -4)y” + 3ty’ + 4y = 2
have a unique solution.
