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

Answer 2

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: The value of the constant c is -2 (c=-2) and the exponent r is 2 (r=2)