Posted by Soroban on November 05, 2002 at 14:55:16:
In Reply to: Proving a formula is a solution posted by Diana on November 04, 2002 at 18:36:00:
: I am trying to verify that the given formula is a solution to the initial value problem.
: Y'=y^3
: y(0) = 5
: Formula: y(t) = 1 / sqrt(1/25-2t)
Then y(t) = (1/25 - 2t)^(-1/2)
: When I try to differentiate this, though, I run into problems.
: -(sqrt 1/25-2t)^-2* [(1/25-2t)^-1/2]/2 * 2
??? There is no -2 power in the derivative.
I got it! You wrote it as [(1/25-2t)^(1/2)]^(-1) and used
the Chain Rule, didn't you?
We have: y = (1/25 - 2t)^(-1/2)
Then y' = (-1/2)(1/25 - 2t)^(-3/2) * (-2)
or, y' = 1/(1/25 - 2t)]^(3/2)
which is: [1/sqrt(1/25 - 2t)]^3
So, y' = y^3 ... There!