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
