Question

Consider the initial value problem 2ty′=4y, y(1)=−2. 2ty′=4y, y(1)=−2. 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

We solve the problem by means of the method of separation of variables.
We have then:
2ty '= 4y
2t (dy / dt) = 4y
t (dy / dt) = 2y
(dy / y) = 2 (dt / t)
integrating both sides:
int (dy / y) = 2int (dt / t)
Ln (y) = 2Ln (t) + c
exp (Ln (y)) = exp (2Ln (t) + c)
exp (Ln (y)) = exp (Ln (t ^ 2) + c)
exp (Ln (y)) = exp (Ln (t ^ 2)) * exp (c)
y = c * t ^ 2 ---> y (1) = - 2.
-2 = c * (1) ^ 2
c = -2
then:
y = -2 * t ^ 2
answer:
c = -2
r = 2