Posted by Joel on November 04, 2002 at 19:03:54:
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)
: I know that 1/sqrt(1/25-2t) at y(0) = 5
: When I try to differentiate this, though, I run into problems.
: -(sqrt 1/25-2t)^-2* [(1/25-2t)^-1/2]/2 * 2
: Since the twos cancel, I end up with
: -(sqrt 1/25-2t)^-2* [(1/25-2t)^-1/2]
: That can't be right since I am sure this formula is a solution - just looking at the problems that follow this. What have I done wrong?
That's very confusing & I can't really figure out exactly what you did. Obviously you are using the quotient rule, which will work, but since the numerator is just a constant why not make your life easier & rewrite the function as:
f(t) = (1/25 - 2t)^(-1/2)
Now:
f'(t) = (-2)*(-1/2) * (1/25-2t)^(-3/2)
..... = (1/25-2t)^(-3/2)
and that's all there is to it.
If you insist on the quotient rule:
f'(t) = [-(-2)(1/2)(1/25-2t)^(-1/2)] / [(sqrt(1/25-2t))^2]
which eventually simplifies to the same thing.