Consider the initial value problem 2ty′=4y, y (2) = - 8. 2ty′=4y, y (2) = - 8. find the value of the constant cc and the exponent rr so that y=ctry=ctr is the solution of this initial value problem.

Let's rewrite the function:

2ty ' = 4y

2t (dy / dt) = 4y

t (dy / dt) = 2y

We use the separable variables method:

(dy / y) = 2 (dt / t)

We integrate both sides:

ln (y) = 2ln (t) + C

we apply exponential to both sides:

exp (ln (y)) = exp (2ln (t) + C)

Exponential properties

exp (ln (y)) = exp (2ln (t)) * exp (C)

Log properties:

exp (ln (y)) = Cexp (ln (t ^ 2))

Exponential properties:

y = C * t ^ 2

Initial conditions y (2) = - 8:

-8 = C * (2) ^ 2

We clear C:

C = - 8 / (2 ^ 2) = - 8/4 = - 2

The function is:

y = - 2 * t ^ 2

Therefore we have to compare:

y = - 2 * t ^ 2

y = ct ^ r

The values of c and r are:

c = - 2

r = 2

Answer:

the value of the constant c and the exponent r are:

c = - 2

r = 2

so that y = - 2 * t ^ 2 is the solution of this initial value problem

2ty'=4y

Replacing y'=dy/dt in the equation:

2t (dy/dt) = 4y

Grouping terms:

dy/y=4dt / (2t)

dy/y=2dt/t

Integrating both sides:

ln (y) = 2ln (t) + ln (c), where c is a constant

Using property logarithm: b ln (a) = ln (a^b), with b=2 and a=t

ln (y) = ln (t^2) + ln (c)

Using property of logarithm: ln (a) + ln (b) = ln (ab), with a=t^2 and b=c

ln (y) = ln (ct^2)

Then:

y=ct^2

Using the initial condition: y (2) = - 8

t=2→y=-8→c (2) ^2=-8→c (4) = - 8

Solving for c:

c=-8/4

c=-2

Then the solution is y=-2t^2

Comparing with the solution: y=ct^r

c=-2, r=2

Answer: T he value of the constant c is - 2 (c=-2) and the exponent r is 2 (r=2)