Question

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.

Answer 1

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