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

answer:

c = - 2

r = 2