Respuesta :
The correct question is:
Consider the initial value problem
2ty' = 6y, y(1) = -2
(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.
b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.
c) What is the actual interval of existence for the solution obtained in part (a) ?
Step-by-step explanation:
Given the differential equation
2ty' = 6y
a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(1) = -2
Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is
2tdy/dt - 6y = 0
Implies
2td(Ct^r)/dt - 6(Ct^r) = 0
2tCrt^(r - 1) - 6Ct^r = 0
2Crt^r - 6Ct^r = 0
(2r - 6)Ct^r = 0
But Ct^r ≠ 0
=> 2r - 6 = 0 or r = 6/2 = 3
Now, we have r = 3, which implies that
y = Ct^3
Applying the initial condition y(1) = -2, we put y = -2 when t = 1
-2 = C(1)^3
C = -2
So, y = -2t^3
b) Let y = F(x,y)................(1)
Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.
Now, suppose that both F(x, y)
and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2
(possibly smaller than δ1) so that the solution y = f(x) to (1) is
the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.
c) Firstly, we write the differential equation 2ty' = 6y in standard form as
y' - (3/t)y = 0
0 is always continuous, but -3/t has discontinuity at t = 0
So, the solution to differential equation exists everywhere, apart from t = 0.
The interval is (-infinity, 0) n (0, infinity)
n - means intersection.
