Posted by Diana on November 05, 2002 at 20:55:23:
In Reply to: Re: Proving a formula is a solution posted by Joel on November 05, 2002 at 18:59:01:
: : : : 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
AHH!!! I see the light! Thank you.
: : : : 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.
: : So then this is not a solution to the original derivative y'=y^3 since the two derivatives do not equal one another?
: One more step:
: f(t) = 1 / sqrt(1/25-2t) (given)
: .... = (1/25-2t)^(-1/2) (per above)
: raise it to 3rd power:
: (f(t))^3 = (1/25-2t)^(-3/2) (same as f'(t))
AHH!! I see the light! Thank you.
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