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.

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
 
 c = -2
 r = 2