Consider the initial value problem 2ty' = 6y, y(1) =-2. Find the value of the constant C and the exponent r so that y = Ctr is the solution of this initial value problem. y = help (formulas) 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. help (inequalities) What is the actual interval of existence for the solution (from part a)? help (inequalities)

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.

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.

a. 32 degrees

step-by-step explanation:

here we have to use the rule.

m∠e = 1/2 (major arc - minor arc)

m∠e = 1/2 (212 - 148)

m∠e = 1/2 (64)

m∠e = 32 degrees

answer: a. 32 degrees

you